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1 lt1228 100mhz current feedback amplifier with dc gain control u s a o pp l ic at i n very fast transconductance amplifier bandwidth: 75mhz g m = 10 i set low thd: 0.2% at 30mv rms input wide i set range: 1 m a to 1ma n very fast current feedback amplifier bandwidth: 100mhz slew rate: 1000v/ m s output drive current: 30ma differential gain: 0.04% differential phase: 0.1 high input impedance: 25m w , 6pf n wide supply range: 2v to 15v n inputs common mode to within 1.5v of supplies n outputs swing within 0.8v of supplies n supply current: 7ma the lt1228 makes it easy to electronically control the gain of signals from dc to video frequencies. the lt1228 implements gain control with a transconductance amplifier (voltage to current) whose gain is proportional to an exter- nally controlled current. a resistor is typically used to convert the output current to a voltage, which is then amplified with a current feedback amplifier. the lt1228 combines both amplifiers into an 8-pin package, and oper- ates on any supply voltage from 4v ( 2v) to 30v ( 15v). a complete differential input, gain controlled amplifier can be implemented with the lt1228 and just a few resistors. the lt1228 transconductance amplifier has a high imped- ance differential input and a current source output with wide output voltage compliance. the transconductance, g m , is set by the current that flows into pin 5, i set . the small signal g m is equal to ten times the value of i set and this relationship holds over several decades of set current. the voltage at pin 5 is two diode drops above the negative supply, pin 4. the lt1228 current feedback amplifier has very high input impedance and therefore it is an excellent buffer for the output of the transconductance amplifier. the current feed- back amplifier maintains its wide bandwidth over a wide range of voltage gains making it easy to interface the transconductance amplifier output to other circuitry. the current feedback amplifier is designed to drive low imped- ance loads, such as cables, with excellent linearity at high frequencies. s f ea t u re d u escriptio n video dc restore (clamp) circuits n video differential input amplifiers n video keyer/fader amplifiers n agc amplifiers n tunable filters n oscillators frequency response u a o pp l ic at i ty p i ca l frequency (hz) 100k ?4 gain (db) ?5 ? 3 6 1m 10m 100m lt1228 ?ta02 0 ? ? ?2 ?8 ?1 i set = 100 m a v s = ?5v r l = 100 w i set = 1ma i set = 300 m a lt1228 ?ta01 + + + + r3a 10k r2a 10k r3 100 w r2 100 w 4.7 m f r4 1.24k r6 6.19 w r5 10k i set r1 270 w r g 10 w r f 470 w 4.7 m f 15v g m cfa v out 1 8 6 3 2 7 5 4 ?5v + v in high input resistance even when power is off ?8db < gain < 2db v in 3v rms differential input variable gain amp
lt1228 2 supply voltage ...................................................... 18v input current, pins 1, 2, 3, 5, 8 (note 7) ............ 15ma output short circuit duration (note 1) ......... continuous operating temperature range lt1228c ................................................ 0 c to 70 c lt1228m ........................................ C55 c to 125 c storage temperature range ................. C65 c to 150 c junction temperature plastic package .............................................. 150 c ceramic package ............................................ 175 c lead temperature (soldering, 10 sec).................. 300 c a u g w a w u w a r b s o lu t exi t i s wu u package / o rder i for atio order part number LT1228MJ8 lt1228cj8 lt1228cn8 lt1228cs8 s8 part marking 1228 symbol parameter conditions min typ max units v os input offset voltage t a = 25 c 3 10 mv l 15 mv input offset voltage drift l 10 m v/ c i in + noninverting input current t a = 25 c 0.3 3 m a l 10 m a i in C inverting input current t a = 25 c 10 65 m a l 100 m a e n input noise voltage density f = 1khz, r f = 1k, r g = 10 w , r s = 0 w 6 nv/ ? hz i n input noise current density f = 1khz, r f = 1k, r g = 10 w , r s = 10k 1.4 pv/ ? hz r in input resistance v in = 13v, v s = 15v l 225 m w v in = 3v, v s = 5v l 225 m w c in input capacitance (note 2) v s = 5v 6 pf input voltage range v s = 15v, t a = 25 c 13 13.5 v l 12 v v s = 5v, t a = 25 c 3 3.5 v l 2v cmrr common-mode rejection ratio v s = 15v, v cm = 13v, t a = 25 c5569db v s = 15v, v cm = 12v l 55 db v s = 5v, v cm = 3v, t a = 25 c5569db v s = 5v, v cm = 2v l 55 db inverting input current v s = 15v, v cm = 13v, t a = 25 c 2.5 10 m a/v common-mode rejection v s = 15v, v cm = 12v l 10 m a/v v s = 5v, v cm = 3v, t a = 25 c 2.5 10 m a/v v s = 5v, v cm = 2v l 10 m a/v psrr power supply rejection ratio v s = 2v to 15v, t a = 25 c6080db v s = 3v to 15v l 60 db noninverting input current v s = 2v to 15v, t a = 25 c 10 50 na/v power supply rejection v s = 3v to 15v l 50 na/v inverting input current v s = 2v to 15v, t a = 25 c 0.1 5 m a/v power supply rejection v s = 3v to 15v l 5 m a/v e lectr ic al c c hara terist ics current feedback amplifier, pins 1, 6, 8. 5v v s 15v, i set = 0 m a, v cm = 0v unless otherwise noted. t j max = 175 c, q ja = 100 c/w (j) t j max = 150 c, q ja = 100 c/w (n) t j max = 150 c, q ja = 150 c/w (s) 8 7 6 5 4 3 2 1 i out ?n +in v i set v out v + gain top view n8 package 8-lead plastic dip s8 package 8-lead plastic soic lt1228 poi01 j8 package 8-lead ceramic dip + g m consult factory for industrial grade parts.
3 lt1228 e lectr ic al c c hara terist ics current feedback amplifier, pins 1, 6, 8. 5v v s 15v, i set = 0 m a, v cm = 0v unless otherwise noted. transconductance amplifier, pins 1, 2, 3, 5. 5v v s 15v, i set = 100 m a, v cm = 0v unless otherwise noted. e lectr ic al c c hara terist ics symbol parameter conditions min typ max units v os input offset voltage i set = 1ma, t a = 25 c 0.5 5mv l 10 mv input offset voltage drift l 10 m v/ c i os input offset current t a = 25 c 40 200 na l 500 na i b input bias current t a = 25 c 0.4 1 m a l 5 m a e n input noise voltage density f = 1khz 20 nv/ ? hz r in input resistance-differential mode v in ? 30mv l 30 200 k w input resistance-common mode v s = 15v, v cm = 12v l 50 1000 m w v s = 5v, v cm = 2v l 50 1000 m w c in input capacitance 3pf input voltage range v s = 15v, t a = 25 c 13 14 v v s = 15v l 12 v v s = 5v, t a = 25 c 3 4v v s = 5v l 2v symbol parameter conditions min typ max units a v large-signal voltage gain v s = 15v, v out = 10v, r load = 1k l 55 65 db v s = 5v, v out = 2v, r load = 150 w l 55 65 db r ol transresistance, d v out / d i in C v s = 15v, v out = 10v, r load = 1k l 100 200 k w v s = 5v, v out = 2v, r load = 150 w l 100 200 k w v out maximum output voltage swing v s = 15v, r load = 400 w , t a = 25 c 12 13.5 v l 10 v v s = 5v, r load = 150 w , t a = 25 c 3 3.7 v l 2.5 v i out maximum output current r load = 0 w , t a = 25 c 30 65 125 ma l 25 125 ma i s supply current v out = 0v, i set = 0v l 611 ma sr slew rate (notes 3 and 5) t a = 25 c 300 500 v/ m s sr slew rate v s = 15v, r f = 750 w , r g = 750 w , r l = 400 w 3500 v/ m s t r rise time (notes 4 and 5) t a = 25 c1020ns bw small-signal bandwidth v s = 15v, r f = 750 w , r g = 750 w , r l = 100 w 100 mhz t r small-signal rise time v s = 15v, r f = 750 w , r g = 750 w , r l = 100 w 3.5 ns propagation delay v s = 15v, r f = 750 w , r g = 750 w , r l = 100 w 3.5 ns small-signal overshoot v s = 15v, r f = 750 w , r g = 750 w , r l = 100 w 15 % t s settling time 0.1%, v out = 10v, r f =1k, r g = 1k, r l =1k 45 ns differential gain (note 6) v s = 15v, r f = 750 w , r g = 750 w , r l = 1k 0.01 % differential phase (note 6) v s = 15v, r f = 750 w , r g = 750 w , r l = 1k 0.01 deg differential gain (note 6) v s = 15v, r f = 750 w , r g = 750 w , r l = 150 w 0.04 % differential phase (note 6) v s = 15v, r f = 750 w , r g = 750 w , r l = 150 w 0.1 deg
lt1228 4 e lectr ic al c c hara terist ics transconductance amplifier, pins 1, 2, 3, 5. 5v v s 15v, i set = 100 m a, v cm = 0v unless otherwise noted. cmrr common-mode rejection ratio v s = 15v, v cm = 13v, t a = 25 c 60 100 db v s = 15v, v cm = 12v l 60 db v s = 5v, v cm = 3v, t a = 25 c 60 100 db v s = 5v, v cm = 2v l 60 db psrr power supply rejection ratio v s = 2v to 15v, t a = 25 c 60 100 db v s = 3v to 15v l 60 db g m transconductance i set = 100 m a, i out = 30 m a, t a = 25 c 0.75 1.00 1.25 m a/mv transconductance drift l C0.33 %/ c i out maximum output current i set = 100 m a l 70 100 130 m a i ol output leakage current i set = 0 m a (+i in of cfa), t a = 25 c 0.3 3 m a l 10 m a v out maximum output voltage swing v s = 15v , r1 = l 13 14 v v s = 5v , r1 = l 3 4v r o output resistance v s = 15v, v out = 13v l 28 m w v s = 5v, v out = 3v l 28 m w output capacitance (note 2) v s = 5v 6 pf i s supply current, both amps i set = 1ma l 915 ma thd total harmonic distortion v in = 30mv rms at 1khz, r1 = 100k 0.2 % bw small-signal bandwidth r1 = 50 w, i set = 500 m a 80 mhz t r small-signal rise time r1 = 50 w, i set = 500 m a, 10% to 90% 5 ns propagation delay r1 = 50 w, i set = 500 m a, 50% to 50% 5 ns symbol parameter conditions min typ max units the l denotes specifications which apply over the operating temperature range. note 1: a heat sink may be required depending on the power supply voltage. note 2: this is the total capacitance at pin 1. it includes the input capacitance of the current feedback amplifier and the output capacitance of the transconductance amplifier. note 3: slew rate is measured at 5v on a 10v output signal while operating on 15v supplies with r f = 1k, r g = 110 w and r l = 400 w . the slew rate is much higher when the input is overdriven, see the applications section. note 4: rise time is measured from 10% to 90% on a 500mv output signal while operating on 15v supplies with r f = 1k, r g = 110 w and r l = 100 w . this condition is not the fastest possible, however, it does guarantee the internal capacitances are correct and it makes automatic testing practical. note 5: ac parameters are 100% tested on the ceramic and plastic dip packaged parts (j and n suffix) and are sample tested on every lot of the so packaged parts (s suffix). note 6: ntsc composite video with an output level of 2v. note 7: back to back 6v zener diodes are connected between pins 2 and 3 for esd protection.
5 lt1228 cc hara terist ics uw a t y p i ca lper f o r c e transconductance amplifier, pins 1, 2, 3 & 5 frequency (hz) 10 10 spot noise (pa/ ? hz) 100 1000 1k 100k lt1228 ?tpc05 v s = ?v to ?5v t a = 25? 100 10k i set = 1ma i set = 100 m a small-signal bandwidth vs small-signal transconductance small-signal transconductance set current and set current vs bias voltage vs dc input voltage temperature (?) ?0 v common-mode range (v) 0.5 1.0 ?.5 v + ?5 0 25 125 lt1228 ?tpc06 50 75 100 0.5 ?.0 ?.0 1.5 2.0 v = ?v to ?5v v + = 2v to 15v input voltage (mvdc) ?00 0 transconductance ( m a/mv) 0.2 0.4 1.4 2.0 ?50 ?00 50 200 lt1228 ?tpc03 0 100 150 1.8 1.6 1.2 0.6 0.8 55? v s = ?v to ?5v i set = 100 m a 50 1.0 25? 125? set current ( m a) 10 0.1 ?db bandwidth (mhz) 1 10 100 100 1000 lt1228 ?tpc01 r1 = 100k r1 = 10k r1 = 1k r1 = 100 w v s = ?5v bias voltage, pin 5 to 4, (v) 0.01 transconductance ( m a/mv) 0.1 1 10 100 0.9 1.2 1.3 1.5 lt1228 ?tpc02 0.001 1.0 1.1 1.4 v s = ?v to ?5v t a = 25? 1.0 10 100 1000 10000 0.1 set current ( m a) total harmonic distortion vs spot output noise current vs input common-mode limit vs input voltage frequency temperature input voltage (mv p? ) 1 0.01 output distortion (%) 0.1 1 10 10 1000 lt1228 ?tpc04 i set = 100 m a v s = ?5v i set = 1ma 100 small-signal control path small-signal control path output saturation voltage vs bandwidth vs set current gain vs input voltage temperature input voltage, pin 2 to 3, (mvdc) 0 0 control path gain ( m a/ m a) 1.0 120 200 lt1228 ?tpc08 d i out d i set 40 80 160 0.2 0.4 0.6 0.8 0.9 0.7 0.5 0.3 0.1 temperature (?) ?0 v output saturation voltage (v) +0.5 +1.0 ?.0 v + ?5 0 25 125 lt1228 ?tpc09 50 75 100 0.5 2v v s ?5v r1 = set current ( m a) 10 1 ?db bandwidth (mhz) 10 100 100 1000 lt1228 ?tpc07 v s = ?v to ?5v v in = 200mv (pin 2 to 3) d i out d i set
lt1228 6 cc hara terist ics uw a t y p i ca lper f o r c e voltage gain and phase vs C3db bandwidth vs supply C3db bandwidth vs supply frequency, gain = 6db voltage, gain = 2, r l = 100 w voltage, gain = 2, r l = 1k current feedback amplifier, pins 1, 6, 8 supply voltage (?) 2 ?db bandwidth (mhz) 40 100 120 12 16 lt1228 ?tpc11 4 068101418 0 20 60 140 160 180 r f = 500 w 80 peaking 0.5db peaking 5db r f = 750 w r f = 1k r f = 2k supply voltage (?) 2 ?db bandwidth (mhz) 40 100 120 12 16 lt1228 ?tpc12 4 068101418 0 20 60 140 160 180 80 peaking 0.5db peaking 5db r f = 750 w r f = 1k r f = 2k r f = 500 w frequency (mhz) 0 voltage gain (db) 2 4 6 8 0.1 10 100 lt1228 ?tpc10 ? 1 7 5 3 1 ? phase shift (degrees) 180 90 0 45 135 225 phase gain v s = ?5v r l = 100 w r f = 750 w voltage gain and phase vs C3db bandwidth vs supply C3db bandwidth vs supply frequency, gain = 20db voltage, gain = 10, r l = 100 w voltage, gain = 10, r l = 1k w frequency (mhz) 14 voltage gain (db) 16 18 20 22 0.1 10 100 lt1228 ?tpc13 12 1 21 19 17 15 13 phase shift (degrees) 180 90 0 45 135 225 phase gain v s = ?5v r l = 100 w r f = 750 w supply voltage (?) 2 ?db bandwidth (mhz) 40 100 120 12 16 lt1228 ?tpc14 4 068101418 0 20 60 140 160 180 r f = 500 w 80 peaking 0.5db peaking 5db r f = 750 w r f = 1k r f = 2k r f = 250 w supply voltage (?) 2 ?db bandwidth (mhz) 40 100 120 12 16 lt1228 ?tpc15 4 068101418 0 20 60 140 160 180 r f = 500 w 80 peaking 0.5db peaking 5db r f = 750 w r f = 1k r f = 2k r f = 250 w voltage gain and phase vs C3db bandwidth vs supply C3db bandwidth vs supply frequency, gain = 40db voltage, gain = 100, r l = 100 w voltage, gain = 100, r l = 1k w supply voltage (?) 2 ?db bandwidth (mhz) 4 10 12 12 16 lt1228 ?tpc18 4 068101418 0 2 6 14 16 18 r f = 500 w 8 r f = 1k r f = 2k supply voltage (?) 2 ?db bandwidth (mhz) 4 10 12 12 16 lt1228 ?tpc17 4 068101418 0 2 6 14 16 18 r f = 500 w 8 r f = 1k r f = 2k frequency (mhz) 34 voltage gain (db) 36 38 40 42 0.1 10 100 lt1228 ?tpc16 32 1 41 39 37 35 33 phase shift (degrees) 180 90 0 45 135 225 phase gain v s = ?5v r l = 100 w r f = 750 w
7 lt1228 cc hara terist ics uw a t y p i ca lper f o r c e current feedback amplifier, pins 1, 6, 8 maximum capacitive load vs total harmonic distortion vs 2nd and 3rd harmonic feedback resistor frequency distortion vs frequency feedback resistor (k w ) 10 capacitive load (pf) 100 1k 10k 023 lt1228 ?tpc19 1 1 v s = 5v v s = ?5v r l = 1k peaking 5db gain = 2 frequency (hz) total harmonic distortion (%) 0.01 0.10 10 1k 10k 100k lt1228 ?tpc20 0.001 100 v s = ?5v r l = 400 w r f = r g = 750 w v o = 7v rms v o = 1v rms frequency (mhz) 1 ?0 distortion (dbc) ?0 ?0 ?0 ?0 ?0 10 100 lt1228 ?tpc21 v s = ?5v v o = 2v pp r l = 100 w r f = 750 w a v = 10db 2nd 3rd input common-mode limit vs output saturation voltage vs output short-circuit current vs temperature temperature temperature temperature (?) ?5 output short-circuit current (ma) 40 60 100 150 lt1228 ?tpc24 0 50 25 50 75 125 175 30 70 50 temperature (?) common-mode range (v) 2.0 v + 50 25 75 125 lt1228 ?tpc22 v 0 1.0 ?.0 ?.0 0.5 ?.5 1.5 0.5 ?5 50 100 v + = 2v to 15v v = ?v to ?5v temperature (?) output saturation voltage (v) v + 50 25 75 125 lt1228 ?tpc23 v 0 1.0 ?.0 0.5 0.5 ?5 50 100 r l = 2v v s ?5v spot noise voltage and current vs power supply rejection vs output impedance vs frequency frequency frequency frequency (hz) spot noise (nv/ ? hz or pa/ ? hz) 10 100 10 1k 10k 100k lt1228 ?tpc25 1 100 e n +i n ? n frequency (hz) power supply rejection (db) 40 80 10k 1m 10m 100m lt1228 ?tpc26 0 100k v s = ?5v r l = 100 w r f = r g = 750 w negative 20 60 positive frequency (hz) output impedance ( w ) 0.1 100 10k 1m 10m 100m lt1228 ?tpc27 0.001 100k 0.01 10 v s = ?5v 1.0 r f = r g = 2k r f = r g = 750 w
lt1228 8 w i spl ii f ed s w a ch e ti c cc hara terist ics uw a t y p i ca lper f o r c e current feedback amplifier, pins 1, 6 & 8 settling time to 10mv vs settling time to 1mv vs output step output step supply current vs supply voltage settling time (ns) output step (v) 60 lt1228 ?tpc28 20 0 40 80 100 ?0 10 0 ? ? ? ? 2 4 6 8 noninverting inverting v s = ?5v r f = r g = 1k inverting noninverting settling time ( m s) output step (v) 12 lt1228 ?tpc29 4 0 8 16 20 ?0 10 0 ? ? ? ? 2 4 6 8 noninverting inverting v s = ?5v r f = r g = 1k noninverting inverting supply voltage (?) supply current (ma) 12 lt1228 ?tpc30 4 0816 0 10 5 1 2 3 4 6 7 8 9 2 6 10 14 18 55? 25? 125? 175? lt1228 ?ta03 i out gain v out v + v ?n +in i set bias 1 2 3 5 6 4 7 8
9 lt1228 u s a o pp l ic at i wu u i for atio the lt1228 contains two amplifiers, a transconductance amplifier (voltage-to-current) and a current feedback am- plifier (voltage-to-voltage). the gain of the transconduc- tance amplifier is proportional to the current that is exter- nally programmed into pin 5. both amplifiers are designed to operate on almost any available supply voltage from 4v ( 2v) to 30v ( 15v). the output of the transconductance amplifier is connected to the noninverting input of the current feedback amplifier so that both fit into an eight pin package. transconductance amplifier the lt1228 transconductance amplifier has a high imped- ance differential input (pins 2 and 3) and a current source output (pin 1) with wide output voltage compliance. the voltage to current gain or transconductance (g m ) is set by the current that flows into pin 5, i set . the voltage at pin 5 is two forward biased diode drops above the negative supply, pin 4. therefore the voltage at pin 5 (with respect to v C ) is about 1.2v and changes with the log of the set current (120mv/decade), see the characteristic curves. the temperature coefficient of this voltage is about C4mv/ c (C3300ppm/ c) and the temperature co- efficient of the logging characteristic is 3300ppm/ c. it is important that the current into pin 5 be limited to less than 15ma. the lt1228 will be destroyed if pin 5 is shorted to ground or to the positive supply. a limiting resistor (2k or so) should be used to prevent more than 15ma from flowing into pin 5. the small-signal transconductance (g m ) is equal to ten times the value of i set (in ma/mv) and this relationship holds over many decades of set current (see the character- istic curves). the transconductance is inversely propor- tional to absolute temperature (C3300ppm/ c). the input stage of the transconductance amplifier has been de- signed to operate with much larger signals than is possible with an ordinary diff-amp. the transconductance of the input stage varies much less than 1% for differential input signals over a 30 mv range (see the characteristic curve small-signal transconductance vs dc input voltage). resistance controlled gain if the set current is to be set or varied with a resistor or potentiometer it is possible to use the negative tempera- ture coefficient at pin 5 (with respect to pin 4) to compen- sate for the negative temperature coefficient of the transcon- ductance. the easiest way is to use an lt1004-2.5, a 2.5v reference diode, as shown below: temperature compensation of g m with a 2.5v reference lt1228 ?ta04 lt1004-2.5 v g m 5 4 r i set i set r v be v be 2.5v 2e g the current flowing into pin 5 has a positive temperature coefficient that cancels the negative coefficient of the transconductance. the following derivation shows why a 2.5v reference results in zero gain change with tempera- ture: since g q kt i i and v e akt q where a in ct ic at c c n ic a m set set be g n = = == ? ? ? ? ? === () 387 10 19 4 27 0 001 3 100 . . ., , m e g is about 1.25v so the 2.5v reference is 2e g . solving the loop for the set current gives: i ee akt q r or i akt rq set gg set = ? ? ? ? = 22 2
lt1228 10 u s a o pp l ic at i wu u i for atio substituting into the equation for transconductance gives: g a rr m == 194 10 . the temperature variation in the term a can be ignored since it is much less than that of the term t in the equation for v be . using a 2.5v source this way will main- tain the gain constant within 1% over the full temperature range of C55 c to 125 c. if the 2.5v source is off by 10%, the gain will vary only about 6% over the same tempera- ture range. we can also temperature compensate the transconduc- tance without using a 2.5v reference if the negative power supply is regulated. a thevenin equivalent of 2.5v is generated from two resistors to replace the reference. the two resistors also determine the maximum set current, approximately 1.1v/r th . by rearranging the thevenin equations to solve for r4 and r6 we get the following equations in terms of r th and the negative supply, v ee . r r v v and r rv v th ee th ee 4 1 25 6 25 = ? ? ? ? = . . temperature compensation of g m with a thevenin voltage lt1228 ?ta05 r4 1.24k w ?5v g m 5 4 r' i set i set 1.03k v be v be v th = 2.5v r' r6 6.19k w voltage controlled gain to use a voltage to control the gain of the transconduc- tance amplifier requires converting the voltage into a current that flows into pin 5. because the voltage at pin 5 is two diode drops above the negative supply, a single resistor from the control voltage source to pin 5 will suffice in many applications. the control voltage is referenced to the negative supply and has an offset of about 900mv. the conversion will be monotonic, but the linearity is deter- mined by the change in the voltage at pin 5 (120mv per decade of current). the characteristic is very repeatable since the voltage at pin 5 will vary less than 5% from part to part. the voltage at pin 5 also has a negative tempera- ture coefficient as described in the previous section. when the gain of several lt1228s are to be varied together, the current can be split equally by using equal value resistors to each pin 5. for more accurate (and linear) control, a voltage-to- current converter circuit using one op amp can be used. the following circuit has several advantages. the input no longer has to be referenced to the negative supply and the input can be either polarity (or differential). this circuit works on both single and split supplies since the input voltage and the pin 5 voltage are independent of each other. the temperature coefficient of the output current is set by r5. lt1228 ?ta19 r5 1k r1 1m v1 v2 i out to pin 5 of lt1228 50pf r1 = r2 r3 = r4 i out = = 1ma/v + r2 1m r3 1m r4 1m (v1 ?v2) r5 r3 r1 lt1006 digital control of the transconductance amplifier gain is done by converting the output of a dac to a current flowing into pin 5. unfortunately most current output dacs sink rather than source current and do not have output
11 lt1228 u s a o pp l ic at i wu u i for atio compliance compatible with pin 5 of the lt1228. there- fore, the easiest way to digitally control the set current is to use a voltage output dac and a voltage-to-current circuit. the previous voltage-to-current converter will take the output of any voltage output dac and drive pin 5 with a proportional current. the r, 2r cmos multiplying dacs operating in the voltage switching mode work well on both single and split supplies with the above circuit. logarithmic control is often easier to use than linear control. a simple circuit that doubles the set current for each additional volt of input is shown in the voltage controlled state variable filter application near the end of this data sheet. transconductance amplifier frequency response the bandwidth of the transconductance amplifier is a function of the set current as shown in the characteristic curves. at set currents below 100 m a, the bandwidth is approximately: C3db bandwidth = 3 10 11 i set the peak bandwidth is about 80mhz at 500 m a. when a resistor is used to convert the output current to a voltage, the capacitance at the output forms a pole with the resistor. the best case output capacitance is about 5pf with 15v supplies and 6pf with 5v supplies. you must add any pc board or socket capacitance to these values to get the total output capacitance. when using a 1k resistor at the output of the transconductance amp, the output capacitance limits the bandwidth to about 25mhz. the output slew rate of the transconductance amplifier is the set current divided by the output capacitance, which is 6pf plus board and socket capacitance. for example with the set current at 1ma, the slew rate would be over 100v/ m s. transconductance amp small-signal response i set = 500 m a, r1 = 50 w current feedback amplifier the lt1228 current feedback amplifier has very high noninverting input impedance and is therefore an excel- lent buffer for the output of the transconductance ampli- fier. the noninverting input is at pin 1, the inverting input at pin 8 and the output at pin 6. the current feedback amplifier maintains its wide bandwidth for almost all voltage gains making it easy to interface the output levels of the transconductance amplifier to other circuitry. the current feedback amplifier is designed to drive low imped- ance loads such as cables with excellent linearity at high frequencies. feedback resistor selection the small-signal bandwidth of the lt1228 current feed- back amplifier is set by the external feedback resistors and the internal junction capacitors. as a result, the bandwidth is a function of the supply voltage, the value of the feedback resistor, the closed-loop gain and load resistor. the characteristic curves of bandwidth versus supply voltage are done with a heavy load (100 w ) and a light load (1k) to show the effect of loading. these graphs also show
lt1228 12 the family of curves that result from various values of the feedback resistor. these curves use a solid line when the response has less than 0.5db of peaking and a dashed line for the response with 0.5db to 5db of peaking. the curves stop where the response has more than 5db of peaking. current feedback amp small-signal response v s = 15v, r f = r g = 750 w , r l = 100 w at a gain of two, on 15v supplies with a 750 w feedback resistor, the bandwidth into a light load is over 160mhz without peaking, but into a heavy load the bandwidth reduces to 100mhz. the loading has so much effect because there is a mild resonance in the output stage that enhances the bandwidth at light loads but has its q reduced by the heavy load. this enhancement is only useful at low gain settings, at a gain of ten it does not boost the bandwidth. at unity gain, the enhancement is so effective the value of the feedback resistor has very little effect on the bandwidth. at very high closed-loop gains, the bandwidth is limited by the gain-bandwidth product of about 1ghz. the curves show that the bandwidth at a closed-loop gain of 100 is 10mhz, only one tenth what it is at a gain of two. capacitance on the inverting input current feedback amplifiers want resistive feedback from the output to the inverting input for stable operation. take care to minimize the stray capacitance between the output and the inverting input. capacitance on the inverting input to ground will cause peaking in the frequency response (and overshoot in the transient response), but it does not degrade the stability of the amplifier. the amount of capacitance that is necessary to cause peaking is a func- tion of the closed-loop gain taken. the higher the gain, the more capacitance is required to cause peaking. for ex- ample, in a gain of 100 application, the bandwidth can be increased from 10mhz to 17mhz by adding a 2200pf capacitor, as shown below. c g must have very low series resistance, such as silver mica. lt1228 ?ta08 + c g r g 5.1 w r f 510 w v out cfa v in 6 1 8 boosting bandwidth of high gain amplifier with capacitance on inverting input frequency (mhz) 1 19 gain (db) 22 25 28 31 46 49 10 100 lt1228 ?ta09 34 37 40 43 c g = 4700pf c g = 2200pf c g = 0 u s a o pp l ic at i wu u i for atio
13 lt1228 u s a o pp l ic at i wu u i for atio capacitive loads the lt1228 current feedback amplifier can drive capaci- tive loads directly when the proper value of feedback resistor is used. the graph of maximum capacitive load vs feedback resistor should be used to select the appro- priate value. the value shown is for 5db peaking when driving a 1k load, at a gain of 2. this is a worst case condition, the amplifier is more stable at higher gains, and driving heavier loads. alternatively, a small resistor (10 w to 20 w ) can be put in series with the output to isolate the capacitive load from the amplifier output. this has the advantage that the amplifier bandwidth is only reduced when the capacitive load is present and the disadvantage that the gain is a function of the load resistance. slew rate the slew rate of the current feedback amplifier is not independent of the amplifier gain configuration the way it is in a traditional op amp. this is because the input stage and the output stage both have slew rate limitations. the input stage of the lt1228 current feedback amplifier slews at about 100v/ m s before it becomes nonlinear. faster input signals will turn on the normally reverse biased emitters on the input transistors and enhance the slew rate significantly. this enhanced slew rate can be as much as 3500v/ m s! current feedback amp large-signal response v s = 15v, r f = r g = 750 w slew rate enhanced the output slew rate is set by the value of the feedback resistors and the internal capacitance. at a gain of ten with a 1k feedback resistor and 15v supplies, the output slew rate is typically 500v/ m s and C850v/ m s. there is no input stage enhancement because of the high gain. larger feedback resistors will reduce the slew rate as will lower supply voltages, similar to the way the bandwidth is reduced. current feedback amp large-signal response v s = 15v, r f = 1k, r g = 110 w , r l = 400 w settling time the characteristic curves show that the lt1228 current feedback amplifier settles to within 10mv of final value in 40ns to 55ns for any output step less than 10v. the curve of settling to 1mv of final value shows that there is a slower thermal contribution up to 20 m s. the thermal settling component comes from the output and the input stage. the output contributes just under 1mv/v of output change and the input contributes 300 m v/v of input change. fortunately the input thermal tends to cancel the output thermal. for this reason the noninverting gain of two configuration settles faster than the inverting gain of one.
lt1228 14 for example, lets calculate the worst case power dissipa- tion in a variable gain video cable driver operating on 12v supplies that delivers a maximum of 2v into 150 w . the maximum set current is 1ma. pvi i vv v r p v ma ma v v v w d s smax set s omax omax l d =+ () + () = + () [] + () =+= 235 212 7 351 12 2 2 150 0 252 0 133 0 385 . . ... w the total power dissipation times the thermal resistance of the package gives the temperature rise of the die above ambient. the above example in so-8 surface mount package (thermal resistance is 150 c/w) gives: temperature rise = p d q ja = 0.385w 150 c/w = 57.75 c therefore the maximum junction temperature is 70 c +57.75 c or 127.75 c, well under the absolute maximum junction temperature for plastic packages of 150 c. power supplies the lt1228 amplifiers will operate from single or split supplies from 2v (4v total) to 18v (36v total). it is not necessary to use equal value split supplies, however the offset voltage and inverting input bias current of the current feedback amplifier will degrade. the offset voltage changes about 350 m v/v of supply mismatch, the inverting bias current changes about 2.5 m a/v of supply mismatch. power dissipation the worst case amplifier power dissipation is the total of the quiescent current times the total power supply voltage plus the power in the ic due to the load. the quiescent supply current of the lt1228 transconductance amplifier is equal to 3.5 times the set current at all temperatures. the quiescent supply current of the lt1228 current feedback amplifier has a strong negative temperature coefficient and at 150 c is less than 7ma, typically only 4.5ma. the power in the ic due to the load is a function of the output voltage, the supply voltage and load resistance. the worst case occurs when the output voltage is at half supply, if it can go that far, or its maximum value if it cannot reach half supply. u s a o pp l ic at i wu u i for atio u s a o pp l ic at i ty p i ca l basic gain control the basic gain controlled amplifier is shown on the front page of the data sheet. the gain is directly proportional to the set current. the signal passes through three stages from the input to the output. first the input signal is attenuated to match the dynamic range of the transconductance amplifier. the attenuator should reduce the signal down to less than 100mv peak. the characteristic curves can be used to estimate how much distortion there will be at maximum input signal. for single ended inputs eliminate r2a or r3a. the signal is then amplified by the transconductance amplifier (g m ) and referred to ground. the voltage gain of the transconductance amplifier is: gr i r m set = 110 1 lastly the signal is buffered and amplified by the current feedback amplifier (cfa). the voltage gain of the current feedback amplifier is: 1 + r r f g the overall gain of the gain controlled amplifier is the product of all three stages: a r rra ir r r v set f g = + ? ? ? ? + ? ? ? ? 3 33 10 1 1 more than one output can be summed into r1 because the output of the transconductance amplifier is a current. this is the simplest way to make a video mixer.
15 lt1228 video dc restore (clamp) circuit lt1228 ?ta13 + + g m cfa v out 8 3 2 5 v 3k r f r g 4 10k 3k 5v logic input restore 0.01 m f video input 2n3906 1 6 v + 7 200 w 1000pf not necessary if the source resistance is less than 50 w the video restore (clamp) circuit restores the black level of the composite video to zero volts at the beginning of every line. this is necessary because ac coupled video changes dc level as a function of the average brightness of the picture. dc restoration also rejects low frequency noise such as hum. the circuit has two inputs: composite video and a logic signal. the logic signal is high except during the back porch time right after the horizontal sync pulse. while the logic is high, the pnp is off and i set is zero. with i set equal to zero the feedback to pin 2 has no affect. the video input drives the noninverting input of the current feedback amplifier whose gain is set by r f and r g . when the logic signal is low, the pnp turns on and i set goes to about 1ma. then the transconductance amplifier charges the capaci- tor to force the output to match the voltage at pin 3, in this case zero volts. this circuit can be modified so that the video is dc coupled by operating the amplifier in an inverting configuration. just ground the video input shown and connect r g to the video input instead of to ground. u s a o pp l ic at i ty p i ca l video fader lt1228 ?ta12 + + 1k 100 w g m lt1223 cfa v out 3 2 5 1 v in1 1k + g m 3 1k v in2 10k 5.1k 10k 5.1k 10k 100 w 2 1k 5 1 ?v v s = ?v the video fader uses the transconductance amplifiers from two lt1228s in the feedback loop of another current feedback amplifier, the lt1223. the amount of signal from each input at the output is set by the ratio of the set currents of the two lt1228s, not by their absolute value. the bandwidth of the current feedback amplifier is inversely proportional to the set current in this configuration. therefore, the set currents remain high over most of the pots range, keeping the bandwidth over 15mhz even when the signal is attenuated 20db. the pot is set up to completely turn off one lt1228 at each end of the rotation.
lt1228 16 single supply wien bridge oscillator lt1228 ?ta14 + + g m cfa v o 8 3 2 5 r f 680 w r g 20 w 4 1 6 v + 7 10k w 10k w 1.8k 160 w 1000pf 1000pf 160 w + 10 m f + 10 m f v + 470 w + 10 m f 100 w 0.1 m f 51 w 50 w 2n3906 6v to 30v f = 1mhz for 5v operation short out 100 w resistor v o = 6dbm (450mv rms ) 2nd harmonic = ?8dbc 3rd harmonic = 54 dbc in this application the lt1228 is biased for operation from a single supply. an artificial signal ground at half supply voltage is generated with two 10k resistors and bypassed with a capacitor. a capacitor is used in series with r g to set the dc gain of the current feedback amplifier to unity. the transconductance amplifier is used as a variable resis- tor to control gain. a variable resistor is formed by driving the inverting input and connecting the output back to it. the equivalent resistor value is the inverse of the g m . this works with the 1.8k resistor to make a variable attenuator. the 1mhz oscillation frequency is set by the wien bridge network made up of two 1000pf capacitors and two 160 w resistors. for clean sine wave oscillation, the circuit needs a net gain of one around the loop. the current feedback amplifier has a gain of 34 to keep the voltage at the transconductance amplifier input low. the wien bridge has an attenuation of 3 at resonance; therefore the attenuation of the 1.8k resis- tor and the transconductance amplifier must be about 11, resulting in a set current of about 600 m a at oscillation. at start-up there is no set current and therefore no attenuation for a net gain of about 11 around the loop. as the output oscillation builds up it turns on the pnp transistor which generates the set current to regulate the output voltage. 12mhz negative resistance lc oscillator lt1228 ?ta15 + + g m cfa v o 8 3 2 1k 330 w 4 1 6 v + 7 1k 30pf 51 w 50 w v o = 10db at v s = ?v all harmonics 40db down at v s = ?2v all harmonics 50db down 9.1k 750 w v 5 4.3k 4.7 m h 10k 0.1 m f v 2n3904 2n3906 this oscillator uses the transconductance amplifier as a negative resistor to cause oscillation. a negative resistor results when the positive input of the transconductance amplifier is driven and the output is returned to it. in this example a voltage divider is used to lower the signal level at the positive input for less distortion. the negative resistor will not dc bias correctly unless the output of the transcon- ductance amplifier drives a very low resistance. here it sees an inductor to ground so the gain at dc is zero. the oscillator needs negative resistance to start and that is provided by the 4.3k resistor to pin 5. as the output level rises it turns on the pnp transistor and in turn the npn which steals current from the transconductance amplifier bias input. u s a o pp l ic at i ty p i ca l
17 lt1228 u s a o pp l ic at i ty p i ca l filters single pole low/high/allpass filter lt1228 ?ta16 + + g m cfa v out 8 3 2 r f 1k r2 120 w 5 1 6 r3 120 w c 330pf f c = r3a 1k i set r g 1k r2a 1k v in lowpass input v in highpass input f c = 10 9 i set for the values shown 10 2 p i set c r f + 1 r g r2 r2 + r2a allpass filter phase response frequency (hz) 10k phase shift (degrees) ?0 ?5 0 100k 1m 10m lt1228 ?ta17 ?35 ?80 1ma set current 100 m a set current using the variable transconductance of the lt1228 to make variable filters is easy and predictable. the most straight forward way is to make an integrator by putting a capacitor at the output of the transconductance amp and buffering it with the current feedback amplifier. because the input bias current of the current feedback amplifier must be supplied by the transconductance amplifier, the set current should not be operated below 10 m a. this limits the filters to about a 100:1 tuning range. the single pole circuit realizes a single pole filter with a corner frequency (f c ) proportional to the set current. the values shown give a 100khz corner frequency for 100 m a set current. the circuit has two inputs, a lowpass filter input and a highpass filter input. to make a lowpass filter, ground the highpass input and drive the lowpass input. conversely for a highpass filter, ground the lowpass input and drive the highpass input. if both inputs are driven, the result is an allpass filter or phase shifter. the allpass has flat amplitude response and 0 phase shift at low frequen- cies, going to C180 at high frequencies. the allpass filter has C90 phase shift at the corner frequency.
lt1228 18 voltage controlled state variable filter lt1228 ?ta18 + + g m cfa lowpass output 8 3 2 1k 4 1 6 100 w 18pf ?v 3.3k 7 5 + + g m cfa bandpass output 8 3 2 1k 4 1 6 100 w 18pf ?v 3.3k 7 5 100 w v in 5v 5v 3k 3k + 100pf lt1006 1k 180 w 10k v c 51k ?v 2n3906 3.3k 3.3k f o = 100khz at v c = 0v f o = 200khz at v c = 1v f o = 400khz at v c = 2v f o = 800khz at v c = 3v f o = 1.6mhz at v c = 4v u s a o pp l ic at i ty p i ca l the state variable filter has both lowpass and bandpass outputs. each lt1228 is configured as a variable integra- tor whose frequency is set by the attenuators, the capaci- tors and the set current. because the integrators have both positive and negative inputs, the additional op amp nor- mally required is not needed. the input attenuators set the circuit up to handle 3v pCp signals. the set current is generated with a simple circuit that gives logarithmic voltage to current control. the two pnp tran- sistors should be a matched pair in the same package for best accuracy. if discrete transistors are used, the 51k resistor should be trimmed to give proper frequency response with v c equal zero. the circuit generates 100 m a for v c equal zero volts and doubles the current for every additional volt. the two 3k resistors divide the current between the two lt1228s. therefore the set current of each amplifier goes from 50 m a to 800 m a for a control voltage of 0v to 4v. the resulting filter is at 100khz for v c equal zero, and changes it one octave/v of control input.
19 lt1228 + a3 lt1006 lt1228 ?ta20 + cfa 100 w 10k rf input 0.6v rms to 1.3v rms 25mhz 300 w ?5v 15v + g m 470 w 10 w 0.01? 10k 0.01? 15v ?5v 4pf 10k 100k amplitude adjust 10k 4.7k ?5v lt1004 1.2v 10k output 2v pp 1n4148? couple thermally 3 2 7 5 1 8 4 information furnished by linear technology corporation is believed to be accurate and reliable. however, no responsibility is assumed for its use. linear technology corporation makes no represen- tation that the interconnection of its circuits as described herein will not infringe on existing patent rights. u s a o pp l ic at i ty p i ca l rf agc amplifier (leveling loop) amplitude modulator lt1228 ?ta22 + 1k + 10k ?v g m 3 5v 7 5 4 1 8 6 v out 0dbm(230mv) at modulation = 0v cfa r f 750 w r g 750 w + + 4.7 m f modulation input 8v pp 4.7 m f carrier input 30mv 2 inverting amplifier with dc output less than 5mv lt1228 ?ta21 + + 100 m f + r5 v g m 3 2 v + 7 5 4 1 8 6 v o cfa r f 1k r g 1k v s = ?v, r5 = 3.6k v s = ?5v, r5 = 13.6k v out must be less than 200mv pp for low output offset bw = 30hz to 20mhz v in includes dc
lt1228 20 linear technology corporation 1630 mccarthy blvd., milpitas, ca 95035-7487 (408) 432-1900 l fax : (408) 434-0507 l telex : 499-3977 ? linear technology corporation 1994 u package d e sc r i pti o dimensions in inches (millimeters) unless otherwise noted. j8 package 8-lead ceramic dip 0.016 ?0.050 0.406 ?1.270 0.010 ?0.020 (0.254 ?0.508) 45 0 8?typ 0.008 ?0.010 (0.203 ?0.254) so8 0294 0.053 ?0.069 (1.346 ?1.752) 0.014 ?0.019 (0.355 ?0.483) 0.004 ?0.010 (0.101 ?0.254) 0.050 (1.270) bsc 1 2 3 4 0.150 ?0.157* (3.810 ?3.988) 8 7 6 5 0.189 ?0.197* (4.801 ?5.004) 0.228 ?0.244 (5.791 ?6.197) *these dimensions do not include mold flash or protrusions. mold flash or protrusions shall not exceed 0.006 inch (0.15mm). s8 package 8-lead plastic soic n8 0594 0.045 ?0.015 (1.143 ?0.381) 0.100 ?0.010 (2.540 ?0.254) 0.065 (1.651) typ 0.045 ?0.065 (1.143 ?1.651) 0.130 ?0.005 (3.302 ?0.127) 0.020 (0.508) min 0.018 ?0.003 (0.457 ?0.076) 0.125 (3.175) min 0.009 ?0.015 (0.229 ?0.381) 0.300 ?0.320 (7.620 ?8.128) 0.325 +0.025 0.015 +0.635 0.381 8.255 () 12 3 4 87 6 5 0.250 ?0.010* (6.350 ?0.254) 0.400* (10.160) max *these dimensions do not include mold flash or protrusions. mold flash or protursions shall not exceed 0.010 inch (0.254mm). n8 package 8-lead plastic dip j8 0293 0.014 ?0.026 (0.360 ?0.660) 0.200 (5.080) max 0.015 ?0.060 (0.381 ?1.524) 0.125 3.175 min 0.100 ?0.010 (2.540 ?0.254) 0.300 bsc (0.762 bsc) 0.008 ?0.018 (0.203 ?0.457) 0??15 0.385 ?0.025 (9.779 ?0.635) 0.005 (0.127) min 0.405 (10.287) max 0.220 ?0.310 (5.588 ?7.874) 12 3 4 87 65 0.025 (0.635) rad typ 0.045 ?0.068 (1.143 ?1.727) full lead option 0.023 ?0.045 (0.584 ?1.143) half lead option corner leads option (4 plcs) 0.045 ?0.068 (1.143 ?1.727) note: lead dimensions apply to solder dip or tin plate leads. lt/gp 0694 5k rev a ? printed in usa

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