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EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 1 EXERCISE IDEAL OP AMP ANALYSIS No.1 Assuming ideal op amps, determine V o for each and every circuit shown below. 2k 10k 1k 30k 1k 2k 10k 2k 24k 3k 2K 2K 10K 10K 14k 16k 15k 15k 10k 3.9K 1,5k 3k 5,1k 2k -2 V Vo 1,8k Vo +0,1V Vo -0 ,5 V Vo +1V -1 ,5 V Vo +15V -15V Vo -2 V +5V +8V A) B) C) D) E) F) EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 2 No.2 Assume typical op amp data for circuits A through E and worst case values for circuit F. Op amp parameters for V SUP =±15V minimum typical maximum O/P voltage swing ±12V ±13,5V - I/P voltage range ±11V ±12,5V - Short circuit current ±12 mA ±20 mA - 30K 4,7k 3,3k 1,8k 10k V x + 6 V - 1 V - 8 V V o + 1 5 V - 1 5 V + 2 V + 3 V 10k 20k 6,8k V o + 1 6 V - 1 6 V V i n 1K 2k 680 V o + 1 6 V - 1 6 V 2 0 0 + 2 V 10K 6,8k V o 1 0 K + 1 2 V - 1 2 V + 5 V 10k 100k 10k V o + 1 6 V - 1 6 V V 1 V 2 1 0 0 K R 3 R 3 A) D e t erm i ne V x . B) Determine V o for V in =+6V and V in =-6V. C) Determine V o . D) D e t erm i ne V o . E) Determine V o . F) Determine the maximum value of R 3 if we do not want to saturate the inputs of the op amp given that V1 and V2 range from 80V to100V . EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 3 No.3 ∆ V o ( PP ) = V o ( t 2 ) − V o ( t 1 ) = − 1 R E C F () V in ( AC ) dt t 1 t 2 ∫ if ω 〉 10 R F C F V in ( ave ) = V in + PW T () + V in − SW T ( ) for a squarewave 10k 100k Vo Vin 0,1 µF A) Draw the output waveform with respect to V in shown for frequencies of 50 Hz, 100 Hz, 1 kHz and 10 kHz - label waveforms with AC and DC values as well as PW and SW. +1V -1 V 0,6T 0,4T Vin B) If V in is a 2 V pp squarewave with a 50% duty cycle, calculate the frequency of V in that will produce V o = 1 V pp C) Repeat step B for 75% duty cycle. D) If V in is a 10 V pp triangular wave with a frequency of 5 kHz, draw the expected O/P waveform with respect to V in . No.4 A) Determine the output waveform relative to an input triangular wave with a 10 V PP amplitude and a frequency of 250 Hz. B) Determine the output waveform relative to an input square wave with a 2 V PP amplitude and a frequency of 250 Hz. C) What is the function of the 100Ω resistor? 100 10k Vo Vin 0,1 µF V o =− R F C E dV in dt if ω 〈 0,1 R E C E No.5 PHASE SHIFTER A VF = P1 + jX C P1 − jX C A VF = 1 / A VF = 2arctan X C P1 Design the circuit in order to obtain a phaseshift of 20 o to 180 o with a 100K pot (P1) at a frequency of 1 kHz. R1 R1 P1 C Vin Vo EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 4 SOLUTIONS + 10V - 2k 10k 1k 30k 1k 2k 10k 2k 24k 3k 2K 2K 10K 10K 14k 16k 15k 15k 10k 3.9K 1,5k 3k 5,1k 2k -2V Vo 1,8k Vo +0,1V Vo -0,5V Vo +1V -1,5V Vo +15V -15V Vo -2V +5V +8V B) C) D) E) F) 0A 1mA 1mA- 2V + -2V -2V -12V 0A 0V 0V 0,1 mA 0,1 mA 0A 0A +3V- -3V 0A 0A -0,5V -0,5V 0,25 mA 0,25 mA +2,5V- -3V 0,125 mA 0,375 mA - 1,125V + -4,125V -1,25V -1,25V 0,125 mA 0,125 mA + 2,25V - 1,125 mA 1,125 mA + 11,25V - -12,5V 0A 0A 0A 0A 0A 0A 1mA 1mA + 16V - + 14V - +1V +1V 0,133 mA 0,933 mA 1,066 mA 0,133 mA - 1,333V + +2,333V 1,6 mA +3,2V +3,2V +1,8V- -5,2V+ - 0,68V + +3,88V 1,2 mA 1,2 mA 1,333 mA 0,1333 mA No.1 A) EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 5 4,7k 3,3k 1,8k 10k V x -3.6V +6V -1V -8V -1V -1V 0A 0A 0,7 mA 0,789 mA 1,489 mA -2.6V+ +7V- No.2 A) 10k 20k 6,8k Vo +16V -16V Vin 0A 0A+6V +6V -14,5V 0,483 mA -4,833V +- B) 10k 20k 6,8k Vo +16V -16V Vin 0A 0A 0,483 mA -6V -6V +14,5V +4,833V -+ Vo +15V -15V +2V +3V 0A +3V -13.5V C) No feedback, the output saturates with a polarity determined by the sign of the differential I/P voltage: V o = A d (V + -V - ) = (2-3) = - V o = -V sat = -13,5V Positive feedback will make the output saturate with a polarity determined by the sign of the differential I/P voltage. With +6V applied to the -ve I/P of the op amp, the O/P should be of the opposite polarity, therefore assume V o = -V sat = -14,5V and determine the V + and verify the sign of the differential I/P voltage in order to validate your assumption of V o = -V sat , that is: V o = A d (V + -V - ) = (-4,83-6) = - therefore the assumption was valid. Same procedure here, except now V - is negative, therefore the O/P polarity is expected to be positive: V o = +V sat = +14,5V Verify assumption with sign of V d = (V + -V - ) V o = A d (V + -V - ) = (+4,83-(-6)) = + EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 6 10K 30k 6,8k Vo 10K +12V -12V +5V 0A 0A E) 0V 0V 0,5 mA 0,5 mA + 15V - -15V - 15V + 0,5 mA 1,5 mA Impossible voltage, therefore V o = -V sat = -10,5V typical. 1K 2k 680 Vo +16V -16V 200 +2V 0A 0A +2V +2V 2 mA 2 mA -4V+ +6V 2mA 30 mA 32 mA + 6V - 10k 100k 10k Vo +16V -16V V1 V2 100KR3 R3 F) 100V max 0A 0A 12V max 8,8 mA 0,12 mA 8,68 mA 1K 2k 680 Vo +16V -16V 200 2V 0A 0A +2V 20 mA 18,75 mA 1,25 mA 1,25 mA 1,25 mA +1,25V + 3,75V - +3,75V + Output of op amp has reached current limit, notice that V - V + and -ve feedback is rendered ineffective not forcing V - = V + 10K 30k 6,8k Vo 10K +12V -12V +5V 0A 0A 0V 0,5 mA -10,5 - 10,5V + 1,05 mA 0,3875 mA 0,3875 mA +- + 3,875V - 1,125V Output of op amp has reached saturation, notice that V - V + and -ve feedback is rendered ineffective not forcing V - = V + V - or V + max = 16 - 4 = 12V R 3 < 12V/8,68 mA = 1382 R 3 should be less than 1382 in order to keep V - and V + in- side a safe range of ±12V. 2 D) EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 7 No.3 If F 〉 10 2 π R F C F = 10 2 π 100k × 0,1 µ = 159 Hz then ∆ V o ( PP ) =− 1 R E C F () V in ( AC ) dt t 1 t 2 ∫ A) V ( ave ) = V + PW T () + V − SW T ( ) V in (DC) = +0,2V TV areaV dtVV PP o PP o t t AC in PP o 6.08.01000 1000 1000 )( )( )()( 2 1 ××−=∆ ×−=∆ −=∆ ∫ Integral does not apply for 50 Hz and 100 Hz. 0,6T 0,4T -2V DC V in +1V -1V V out ∆V out = 0,48V pp at 1 kHz and 48 mV pp at 10 kHz +0,2V DC AREA AREA B) PPP PP o t t AC in PP o V F V T areaV dtVV 1 2 1000 1 2 1000 1000 1000 )( )()( 2 1 ==××= ×−=∆ −=∆ ∫ F = 500 Hz F > 159 Hz integral OK. DC level 2 V pp T 2 AREA AREA I N P U T DC level 1 V pp O U T P U T C) HzF V F T areaV dtVV PP PP o t t AC in PP o 375 1 375 4 3 5,01000 1000 1000 )( )()( 2 1 = ==       ×× =×−=∆ −=∆ ∫ F > 159 Hz integral OK. DC level 2 V pp 3 T 4 I N P U T 1 V pp DC level AREA A R E A O / P +1V -1V V ( ave ) = V + PW T () + V − SW T ( ) V in (DC) = +0,5V EXERCISE IDEAL OP AMP ANALYSIS Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 8 No.3 D) ∆ V o ( PP ) =− 1000 V in ( AC ) dt t 1 t 2 ∫ ∆ V o ( PP ) =− 1000 × area = 1000 × 100 µ × 5 2     ∆ V o ( PP ) = 0,25V PP O/P is a parabolic wave, not a sine wave. DC level DC level 10 V PP 0.25 V PP 100 µ µµ µ s I N P U T O U T P U T No.4 A) V o =− R F C E dV in dt V o =− 10K × 0,1 µ± 10V 2ms     V o = m 5V P 2 ms 10 V pp -5V p +5V p I N P U T 10 V pp O / P +5V p -5V p B) On edges we have: V o =− 10K × 0,1 µ ±∞ () V o = m ∞⇒ m V sat On flat portions we have: V o =− 10K × 0,1 µ ± 0 () V o = 0 The O/P spikes settle down in 5 τ = 5R E C E = 50 µs 2 ms -1V p +1V p 0V 50 µs 50 µs -V sat +V sat I N P U T O U T P U T C) To stabilise negative feedback in order to avoid self oscillations from the circuit. No. 5 Phase shifter o C X ATANP 180 0 20 1 =       ×=Φ= () () () stdnFnF TANkTANFC P FCPP X TAN P X ATAN P X ATAN oo C o C o C o 1.9026.9 1010010002 1 102 1 2 1 1010220 1 1111 ⇒= ×× = × = ==⇒         =⇒         ×==Φ ππ π The +ve input sees 0 to 100k DC wise, average of 50k, and the –ve input sees R 1 IIR 1 = R 1 /2 = 50k R 1 = 100K. This means that O/P DC offset will not be minimized for all P 1 settings.