EXECSE No.1 Assuming ideal op amps, determine for each and every circuit shown below. A) 1,8k B) V 30k 1k 0,1V k 1k C) D) k K 0,5V 1V 3k 1,5V K k 4k E) F) 15V V 14k 15k 5V 3.9K 1,5k 5,1k 8V 3k 16k 15k k 15V 1
No. Assume typical op amp data for circuits A through E and worst case values for circuit F. Op amp parameters for V S ±15V minimum typical maximum O/ voltage swing ±1V ±13,5V / voltage range ±11V ±1,5V Short circuit current ±1 ±0 A) Determine V x. D) Determine. 6V 4,7k 1K k Vx 3,3k 16V 1V 1,8k 8V V 680 00 16V B) Determine for 6V and 6V. 16V 0k E) Determine. 30K 5V 1V 6,8k 16V 6,8k 1V C) Determine. 15V F) Determine the maximum value of 3 if we do no want to saturate the inputs of the op amp given and V range from 8 to10. 100k V 3V V1 3 16V 15V V 3 16V
No.3 Assume ideal op amps for each of the following circuits. Do not use any formulas, you must be able to derive them on your own. Mark up the circuit diagrams thoroughly for solutions. A) Determine B, A vf and Z iḟ B) Determine 3, A vf and Z iḟ E F 3 B C) Determine A vf and Z iḟ D) Determine and Z if seen by each /. E F V1 V E) Determine and Z if seen by each /. F) Determine B, and Z if seen by each / V1 V V1 V F B V3 1,5 1 1 3
No.4 100k () (t ) (t 1 ) 1 E C F t ( ) t1 (AC ) dt if ω 10 F C F 0,1 µf (ave) SW () ( ) for a squarewave W A) Draw the output waveform with respect to shown for frequencies of 50 Hz, 100 Hz, 1 khz and 10 khz label waveforms with AC and DC values as well as W and SW. 1V 1V 0,6 0,4 B) f is a V pp squarewave with a 50% duty cycle, calculate the frequency of that will produce 1 V pp C) epeat step B for 75% duty cycle. D) f is a 10 V pp triangular wave with a frequency of 5 khz, draw the expected O/ waveform with respect to. No.5 A) Determine the output waveform relative to an input triangular wave with a 10 V amplitude and a frequency of 50 Hz. 100 0,1 µf B) Determine the output waveform relative to an input square wave with a V amplitude and a frequency of 50 Hz. C) What is the function of the 100Ω resistor? dv F C in E dt 0,1 if ω E C E No.6 below. Determine the O/ waveform relative to an input V (rms) 1 khz sinewave for each circuit shown A) 300K B) 300K 0,1 µf 75K 75K 4
No.7 HASE SHFE sing ideal op amp analysis, prove that A VF 1 jx C 1 jx C A VF 1 C 1 / A VF arctan X C 1 No.8 HASE SHFE sing ideal op amp analysis, prove that A VF 1 jx C 1 jx C A VF 1 / A VF arctan X C 1 C / 1 No.9 Howland current source rove that the load current is given by the following equation: Vx L Hint: Work out currents in terms of and unknown V x. L L No.10 BDGE AMLFE 15V rove that is given by the following equation: F 15* 1 1 ( 1 1 ) F ² 15V 5
SOLONS No.1A ) V V 1,8k V V 1 1 k 1 1V B) 0,1V 0,1 1k 1k 0,1 30k 3V 3V 1,5V E) 15V F) 1V 14V 15k 0,133 V 14k 0,933 1,333V 1 5V 1V 1 16k 16V 1,066 15k 15V 0,133,333V 1,333 5,V 0,68V 0,1333 3.9K 5,1k 1, 1, 1,5k 3,V 8V 1,6 C) D) 0,5V 1,5V k 1,15 1,15 4,15V K 0,5V 1V,5V 11,5V 3k 1,15V K,5V 0,375 0,5V 3V 1,5V 0,15 0,5 0,15 k 4k 0,5 1,5V 0,15 1,8V 3k k 3,V 3,88V 6
No. A) 1,489 1V 0,7 4,7k 6V 7V 0,789 V x 3,3k 3.6V.6V 1,8k 1V 8V 1V C) V 3V 3V 15V 15V No feedback, the output saturates with a polarity determined by the sign of the differential / voltage: A d (V V ) (3) V sat 13,5V 13.5V B) 4,833V 6V 6,8k 6V 0k 16V 16V 0,483 14,5V ositive feedback will make the output saturate with a polarity determined by the sign of the differential / voltage. With 6V applied to the ve / of the op amp, the O/ should be of the opposite polarity, therefore assume V sat 14,5V and determine the V and verify the sign of the differential / voltage in order to validate your assumption of V sat, that is: 4,833V 0k A d (V V ) (4,836) therefore the assumption was valid. 6V 6,8k 6V 16V 0,483 14,5V Same procedure here, except now V is negative, therefore the O/ polarity is expected to be positive: V sat 14,5V Verify assumption with sign of V d (V V ) A d (V V ) (4,83(6)) 16V 7
D) 1K V 16V k 4V 6V 1,5 1K 1,5V 1,5 16V k 3,75V V 680 V 16V 3 6V 30 00 V 680 V 16V 0 3,75V 1,5 18,75 00 Output of op amp has reached current limit, E) 5V 0,5 6,8k 0,5 15V 1V 1V 30k 15V 15V 0,5 1,5 5V notice that V V and ve feedback is rendered ineffective not forcing V V 0,3875 3,875V 6,8k 1,15V 0,3875 1V 1V 30k 10,5V 10,5 0,5 1,05 mpossible voltage, therefore V sat 10,5V typical. Output of op amp has reached saturation, notice that V V and ve feedback is rendered ineffective not forcing V V F) 100k V1 3 16V V or V max 16 4 1V 3 < 1V/8,68 138 V 10 max 8,8 8,68 3 1V max 0,1 16V 3 should be less than 138 in order to keep V and V inside a safe range of ±1V. 8
9
No.3 E) VE V1 VE V VE V3 3 1,5 3 V V 0,4 V 3 F F) VE V1 VE V B F F V 0,4 V 3 3 V 3,5 1 V V 3 1 V 3,5 1 1 0,4V 3 Z if 3 V 3,5 V 1 V 3 V V F V 1 0,4V 3 V 0,4V 3 V 0,4V 3 V 1 0,4V 3 F V 1 V 0,8V 3 V F ( ) 0,4V 3 V V 0,8V 1 3 [ ] [ ] 0,4V 3 V 1 V 1,6V 3 V 3 V V 1 Z if 1 V 1 Z if V V 1 V 1 0,4V 3 V V 0,4V 3 to balance the inputs: 1 0,4V 3 V 1 1 0, 4V 3 V ( ) V F F 0 ( ) F V 1 V F 1 F V 1 F V 1 N0.3 E) Z if 1 V 1 1 Z if V B 1 to balance inputs Z if1 and Z if are not constant, but will vary as V 3 changes. Onthe other hand, Z if3 is fixed and does is not affected by either V 1 or V. 0,4 1 1,5 1 0,6 1 1,5 1 10
No.4 f F 10 π F C F 10 π 100k 0,1µ 159 Hz then () 1 E C F t ( ) t 1 ( AC) dt A) V (ave) V () V SW ( ) W (DC) 0,V 1V AEA 1V AEA 0,V DC ( ) 1000 t (AC ) dt 0,6 0,4 t 1 ( ) 1000 area ut V DC ntegral does not apply for 50 Hz and 100 Hz. ut 0,48V pp at 1 khz and 48 mvpp at 10 khz B) ( ) 1000 t t 1 (AC ) dt ( ) 1000 area 1000 1000 F 1V F 500 Hz F > 159 Hz integral OK. N O V pp 1 V pp AEA AEA DC level DC level C) ( ) 1000 t t 1 (AC ) dt ( ) 1000 area 1000 0,5 50 F 1 F 50 Hz N O / 1V AEA V pp 3 4 DC level 1 V pp DC level A E A 1V F > 159 Hz integral OK. V (ave) V () V SW ( ) (DC) 0,5V W 11
No.4 D) N 10 V pp AEA 100 µ s AEA DC level ( ) 1000 t (AC ) dt O 0,5 V pp DC level ( ) 1000 area t 1 1000 100µ 5 ( ) 0,5V O/ is a parabolic wave, not a sine wave. No.5 A) dv F C in E dt 0,1µ ± 1 ms m5v O 10 V pp N / 10 V pp 5V p 5V p ms 5V p 5V p B) On edges we have: 0,1µ ( ± ) m m V sat N 1V p ms 1V p On flat portions we have: 0,1µ ( ±0) 0 he O/ spikes settle down O V sat 50 µs V sat 50 µs in 5τ 5 E C E 50 µs C) o stabilise negative feedback in order to avoid self oscillations from the circuit. 1
No.6 A),83Vp 8,3 µap 8,3 µap 300K 8,49Vp OFF ON 8,49Vp 0,7V 75K 8,3 µap 0,7 300K 8,3 µap 8,3 µap ON,83Vp 0,7V OFF 75K 0,7V N,83V p t O O,8 3 V p t 8,4 9 V p 13
N0.6 B) DC,83Vp,83Vp 75K DC 6,5 µa DC 300K 6,5 µa DC D ON 6,5 µa DC D1 OFF 0,1 µf 106,1 µa DC 7,96V DC 79,6 µa N O O 7,96V DC (A V E A G E),8 3 V D O N,83V D1 ON 1,06 V pp 8,4 9 V t NOE: he bottom of the O/ waveform will not be exactly 8,49 practice. first iteration Assume (ave ) 8, 49V C(ave) ( ) Q C C(ave ) t C sec ond iteration (ave) (peak ) ( ) C(ave) 7,94 7,94 300K ( ) C (ave) C C(ave ) C 8,49 1,13 105,65 µa 105,65 µ 1m 0,1µ third iteration (not necessary) (ave) 8,49 1,056 7, 96V 8, 49 8,49 113, µa 300K 113, µ 1m 1,13V 0,1µ 7,94 1,056V 14
No.7 hase Shifter No.8 hase Shifter C 1 1 C 1 V jx C V V 1 V 1 V V V in 1 V 1 1 1 jx C jx C A VF V o 1 1 jx C jx 1 1 C jx C A VF A VF jx C jx C X C ( X C ) 1 1 / A VF AAN X C AAN X C / A VF AAN X C ( ) V ( in) jx C jx C V in jx C 1 1 1 jx C ( ) A VF V o 1 1 jx C jx 1 1 C jx C A VF A VF jx C jx C X C 1 ( X C ) 1 / A VF AAN X C AAN X C / A VF AAN X C 15
No.9 Howland current source No.10 Bridge Amplifier 15V VE L L in 4 F 4 3 15V From the circuit diagram, we have: From the circuit diagram, we have: L in 1 L 1 1 1 1 1 4 3 4 15 1 15 1 1 4 15 1 15 1 15 1 1 ( 1 1 ) 15 4 1 1 1 1 ( ) 0 4 F 15 1 F 1 1 1 ( ) 16
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