Operational amplifiers

Bo2's Drill Exercise 3.11 (Thévenin)

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Find the Thévenin equivalent.

e,1,0,vs
r1,2,0,r1
r2,2,3,r2
o,1,2,3

Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 0 in the second — the pair of terminals you are looking into.

This one catches people out, and Symbulator says so:

with a note above the answers: the short-circuit current is unbounded, so the equivalent is a voltage source with nothing in series.

That is not a failure to solve. An ideal op amp holds its output voltage whatever current is drawn, so a short across its output carries an unbounded current, which is exactly a source with no resistance in series: R_EQ = 0 Ω. Symbulator finds it by putting a resistance across the terminals and letting it fall to zero. V_TH matches the book's answer.

AS2's Example 5.2

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Find v_o and i_o.

e,2,0,1.
r5,1,0,5'k
r4,1,o,40'k
r2,o,0,20'k
o,2,1,o

The answers you want are vo and io, in Results.

The answer is v_o = 9V and i_o = 0.65mA. This is correct.

Bo2's Figure 3.3 (Inverting)

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Find the gain of the overall circuit, v_o/v_S.

e,2,0,vs
r1,2,1,r1
r2,1,o,r2
o,1,0,o

Ask Evaluate for:

vo/vs

We get

which is correct, as can be seen in the book's answer above.

AS2's Figure 5.10 (Inverting)

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Find v_o.

e,2,0,vi
r1,2,1,r1
rf,1,o,rf
o,0,1,o

The answer you want is vo, in Results.

This problem is almost identical to the one above. The answer we get is correct:

AS2's Example 5.3 (Inverting)

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If v_i is 0.5V, calculate the output voltage v_o and the current in the 10 kΩ resistor.

e,2,0,.5
r1,2,1,10'k
rf,1,o,25'k
o,0,1,o

The answers you want are in Results:

This is correct.

TR5's Exercise 4-11 (Inverting)

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Find v_O when v_S is 2V, -4V and 6V. Notice the output of the op amp is limited to ±15V.

This may be the only non-linear problem you will see in this book, because I solved it before I realized it included the ±15V constraint. But anyway, here it goes.

e,1,0,vs
r1,1,2,10'k
r2,2,o,33'k
o,0,2,o

Put vo in Evaluate and the given value in its Conditions box:

vo

vs = 2

Then run it again with vs = -4 and once more with vs = 6.

The answers, vo = -6.6 V, 13.2 V and -19.8 V, are correct within the linear realm, but since the output is constrained to no more than 15V or less than -15V, the answer is v_O=-15V for v_S = 6V.

AS2's Practice Problem 5.3 (Inverting)

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Find the output voltage of the op amp (i.e. v_o) and calculate the current through the feedback resistor (i.e. the 15 kΩ resistor).

My answer: Notice we used the m for milli in the value of the voltage source.

e,2,0,40'm
r1,2,1,5'k
rf,1,o,15'k
o,0,1,o

The answers are:

This is correct.

AS2's Example 5.4 (Inverting)

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Determine v_o.

e6,1,0,6
r1,1,a,20'k
e2,b,0,2
rf,a,o,40'k
o,a,b,o

The answer you want is vo, in Results.

The answer, -6, is correct.

TR5's Example 4-14 (Inverting)

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Find the input-output relationship of the circuit.

The 'formal' way to symbulate this circuit would be as described below.

e,1,0,vs
r1,1,b,r1
r2,b,0,r2
r3,b,a,r3
r4,a,o,r4
o,0,a,o
rl,o,0,rl

However, since we are only interested in the ratio of the input to the output, a smarter way to symbulate it is this:

e,1,0,1
r1,1,b,r1
r2,b,0,r2
r3,b,a,r3
r4,a,o,r4
o,0,a,o
rl,o,0,1

Inverting, the quicker wayOpen in app ↗Open in split view

Ask Evaluate for vo/v1 and both approaches give the same answer:

which is correct, as can be seen by comparing it to the book's answer.

AS2's Practice Problem 5.4a (Transresistance)

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This is a current-to-voltage converter, also called a transresistance amplifier. Find v_o/i_S.

j,0,1,is1
r,1,o,r
o,0,1,o

Ask Evaluate for:

vo/is1

The answer is correct:

AS2's Practice Problem 5.4b (Transresistance)

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This is another transresistance amplifier. Again, find v_o/i_S.

j,0,1,is1
r1,1,2,r1
r2,2,0,r2
r3,2,o,r3
o,0,1,o

Ask Evaluate for vo/is1. It gives

which is equivalent to the book's answer — the same expression, gathered differently.

Bo2's Example 3.1 (Non-Inverting Amplifier)

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Find v_o/v₁.

e,p,0,v2
r1,1,0,r1
r2,1,o,r2
o,p,1,o

Ask Evaluate for:

vo/v1

We get

which is correct.

AS2's Figure 5.16 (Non-Inverting Amplifier)

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Find v_o.

e,2,0,vi
r1,0,1,r1
rf,1,o,rf
o,2,1,o

The answer you want is vo, in Results.

We get the right answer (below), an expression equivalent to the book's answer.

TR5's Example 4-13 (Non-Inverting Amplifier)

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Find V_o/V_s.

We can solve this problem in one simulation, as shown below.

e,1,0,vs
r1,1,2,r1
r2,2,0,r2
o,2,3,o
r3,o,3,r3
r4,3,0,r4

Ask Evaluate for:

vo/vs

We get

which is correct, as can be seen by comparing it to the book's answer.

We can also solve it in stages, as shown below. First, simulate the left half.

e,1,0,vs
r1,1,2,r1
r2,2,0,r2

the left halfOpen in app ↗Open in split view

Ask Evaluate for:

v2/vs

We get

which is correct for this part. Then, simulate the right half.

e,2,0,1
o,2,3,o
r3,o,3,r3
r4,3,0,r4

the right halfOpen in app ↗Open in split view

Ask Evaluate for:

vo/v2

We get

which is correct for this part. The product of these two partial answers produces the same expression shown above after the big simulation, and is the right answer.

AS2's Figure 5.17 (Voltage Follower)

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Find v_o.

e,1,0,vi
o,1,o,o

The answer you want is vo, in Results.

The answer is correct:

TR5's Figure 4-32 (Voltage Follower)

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Alright, so first we simulate the (b) circuit, to find its maximum power and load power:

e,1,0,1.5
rs,1,2,2'k

Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into. Tick the question about the load, choose DC and run it.

The answer you want is pmax, in Results, and under it prl, the power in the load as Lesson 4 describes. Ask Evaluate for it with the load in the Conditions box:

prl

load = 1000

The answers we get, pmax = 2.8125e-4 W and prl = 2.5e-4 W in the 1 kΩ load, are correct. Now we simulate the (a) circuit.

e,1,0,1.5
rs,1,2,2'k
o,2,o,o
rl,o,0,1'k

Voltage Follower, bufferedOpen in app ↗Open in split view

The answer you want is prl, in Results — the power consumed by the load resistor rl.

The answer, .00225, is correct. The apparent paradox — the load in (a) drawing more power than the source in (b) seems able to provide — evaporates once we remember that the ideal op amp in the schematic is only part of the truth: the real one has its own source of power, which makes up the difference.

AS2's Example 5.5 (Inverting)

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Find v_o.

e1,1,0,6
r4,1,a,4'k
r10,a,o,10'k
e4,b,0,4
o,b,a,o

The answer you want is vo, in Results.

The answer, -1, is correct.

AS2's Practice Problem 5.5 (Non-Inverting)

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Calculate v_o.

e,1,0,3
r4,1,2,4'k
r8,2,0,8'k
r2,3,0,2'k
r5,3,o,5'k
o,2,3,o

The answer for vo, 7, is correct.

Bo2's Example 3.2 (Adder or Summing)

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Find v_o.

ea,3,0,va
eb,2,0,vb
r31,3,1,r1
r21,2,1,r1
r1o,1,o,r2
o,0,1,o

The answer you want is vo, in Results:

which is correct, as can be seen by comparing it to the book's answer.

AS2's Figure 5.21 (Adder or Summing)

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Find v_o.

e1,b,0,v1
e2,c,0,v2
e3,d,0,v3
r1,b,a,r1
r2,c,a,r2
r3,d,a,r3
rf,a,o,rf
o,0,a,o

Ask Evaluate for expand(vo). It gives

which is correct, as can be seen by comparing it to the book's answer.

AS2's Example 5.6 (Adder or Summing)

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Find v_o and i_o.

e1,1,0,1
e2,2,0,2
r1,2,a,5'k
r2,1,a,2.5'k
r3,a,o,10'k
r4,o,0,2'k
o,0,a,o

The answers you want are in Results:

This is correct. Notice a current of 4.8mA is going into the op amp.

AS2's Practice Problem 5.6 (Adder or Summing)

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Find v_o and i_o.

e2,2,0,1.5
e1,1,0,2
e6,6,0,1.2
r2,2,8,20'k
r1,1,8,10'k
r6,6,8,6'k
r8,8,o,8'k
r4,o,0,4'k
o,0,8,o

The answers you want are in Results:

This is correct. Again, the current is going into the op amp.

Bo2's Drill Exercise 3.3 (Difference or Differential)

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ea,4,0,va
eb,3,0,vb
r1,4,1,r1
r2,1,o,r2
r3,3,2,r1
r4,2,0,r2
o,2,1,o

vo is the correct answer, equivalent to the book's answer above:

TR5's Exercise 4-13 (Difference or Differential)

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Find v_o.

e1,3,0,v1
e2,4,0,v2
r1,3,5,10'k
r2,4,6,10'k
r3,5,o,40'k
r4,6,0,15'k
o,6,5,o

vo is the correct answer:

AS2's Figure 5.24 (Difference or Differential)

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Find v_o. (And keep the answer on screen: the next three problems reuse it.)

e1,d,0,v1
e2,c,0,v2
r1,d,a,r1
r3,c,b,r3
r2,a,o,r2
r4,b,0,r4
o,b,a,o

The answer you want is in Results:

This expression is equivalent to the book's answer.

AS2's Figure 5.24 (Subtractor)

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For the same circuit of the previous problem, find v_o when R₁=R₂ and R₃=R₄.

The expression for vo is still in Results from the previous problem, so we only do this:

Ask Evaluate for expand(vo) and give it both equalities in the Conditions box:

r2 = r1
r3 = r4

The answer we get is correct:

AS2's Example 5.7 (Difference or Differential)

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Design an op amp circuit with inputs v₁ and v₂ such that v_o = -5v₁ + 3v₂.

My solution follows. The problem statement is a fancy way of saying: for the circuit of the previous problem, find the resistor values that give an output v_o = -5v₁ + 3v₂. Not strictly a Symbulator problem, but it shows how Symbulator fits into design problems.

First we take that part of v_o that is a factor of v₁, and make it equal to -5. Thus:

Conditions again, this time on the inputs rather than the resistors: v1 = 1 and v2 = 0. Then vo is

which is equivalent to the book's expression, and it is that which has to equal −5.

Symbulator will not always arrange an expression the way the book does. To rearrange it, Evaluate takes simplify(), collect(), expand(), factor() and apart().

We get

Now make that part of v_o that is a factor of v₂ equal to

1. Thus:

With v1 = 0 and v2 = 1 instead, vo is

which is equivalent to the book's expression, and it is that which has to equal 3.

the v2 factorOpen in app ↗Open in split view

We get

Now, since you have two equations, you can solve for two unknowns. Of the four resistors you get to choose, two can be whatever you want. The book recommends R₁ = 10 kΩ and R₃ = 20 kΩ. Now let's find R₂ and R₄.

Write the two equations out in the Solve card, which solves a system that is not a circuit:

-r2/r1 = -5
r4*(r1 + r2)/(r1*(r3 + r4)) = 3

with r2, r4 as the unknowns. The answer is a design rule rather than a pair of numbers:

Put the book's r1 = 10 kΩ and r3 = 20 kΩ into it and you get 50 kΩ and 20 kΩ.

Now, if this problem was part of a test, I'd like to verify that the answer is correct. To confirm this, simulate the circuit using the four values given above for the resistors.

e1,d,0,v1
e2,c,0,v2
r1,d,a,10'k
r3,c,b,20'k
r2,a,o,50'k
r4,b,0,20'k
o,b,a,o

checking the designOpen in app ↗Open in split view

vo is the desired output:

The resistor values are correct.

AS2's Practice Problem 5.7 (Difference or Differential)

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Design a difference amplifier with gain 4.

My solution follows. Again the statement is a fancy way of saying: for the circuit of the previous problem, find the resistor values that give an output v_o = 4 (v₂-v₁), that is, -4v₁ + 4v₂. Same as before.

Exactly as in the previous problem: solve the circuit twice with conditions on the inputs — v1 = 1, v2 = 0, then v1 = 0, v2 = 1 — and read vo each time. The two factors are the same as before, so the two design equations are the same shape with 4 in place of 5 and 3.

This time the book asks that you use R₁ = 10 kΩ and R₃ = 10 kΩ. So we do that.

In the Solve card:

-r2/r1 = -4
r4*(r1 + r2)/(r1*(r3 + r4)) = 4

with r2, r4 as the unknowns. It answers

which at 10 kΩ each is 40000 Ω and 40000 Ω.

AS2's Practice Problem 5.8 (Instrumentation)

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Find i_o.

e1,1,0,8.
e2,2,0,8.01
o1,1,3,3
o2,2,4,4
r1,3,5,20'k
r2,4,6,20'k
r3,5,o,40'k
r4,6,0,40'k
o3,6,5,o
r5,o,0,10'k

The answer you want is ir5, in Results.

In the schematic, the current i_o corresponds to ir5. The answer, 2e-6, is correct.

Bo2's Example 3.3 (Cascade)

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Find v_o in terms of the conductances and the applied voltage v_S.

e,1,0,vs
r12,1,2,1/g1
r14,1,4,1/g2
r4o,4,o,1/g3
r2o,2,o,1/g4
r23,2,3,1/g
r34,3,4,1/g
o1,0,2,3
o2,0,4,o

vo is:

which is correct, as can be seen by comparing it to the book's answer.

Bo2's Drill Exercise 3.4 (Cascade)

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e,1,0,vs
r1,2,o,1/1
r2,1,2,1/2
r3,2,3,1/3
r4,4,0,1/4
r5,4,o,1/5
o1,0,2,3
o2,3,4,o

vo is correct:

AS2's Example 5.9 (Cascade)

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Find v_o and i_o.

e,1,0,20'm
o1,1,2,a
o2,a,b,o
ro,o,b,10'k
r4,b,0,4'k
r2,a,2,12'k
r3,2,0,3'k

vo is .35 and iro is 2.5e-5. This is correct.

AS2's Practice Problem 5.9 (Cascade)

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Determine v_o and i_o.

e,1,0,4
o1,1,2,2
o2,2,3,o
ro,3,0,4'k
r6,3,o,6'k

The answers you want are vo and iro, in Results.

The answer is vo = 10 V and iro = 1/1000 A, that is 1mA. This is correct.

AS2's Practice Problem 5.10 (Cascade)

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If v₁ = 2V and v₂ = 1.5V, find v_o in the circuit.

e1,1,0,2
e2,2,0,1.5
o1,1,3,3
r1,2,5,10'k
r2,3,6,20'k
r5,5,4,50'k
o2,0,5,4
r3,6,4,30'k
r6,6,o,60'k
o3,0,6,o

The answer you want is vo, in Results.

The answer, 9, is correct.

TR5's Example 4-16 (Cascade)

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Derive an expression for v_o in terms of the two inputs.

e1,2,0,v1
e5,3,0,5
r1,2,4,5'k
r2,3,4,10'k
o1,0,4,a
r3,4,a,10'k
r4,4,o,20'k
o2,a,5,o
r5,5,o,20'k
r6,5,0,10'k

Ask Evaluate for expand(vo) with Rounding set to approx (full precision) in Settings. The answer is arranged differently from the book's, which is a matter of presentation rather than of arithmetic.

The answer is correct:

TR5's Exercise 4-14 (Cascade)

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Derive an expression for v_o in terms of the inputs v₁ and v₂.

e1,3,0,v1
r1,3,4,10'k
o1,0,4,5
r2,4,5,40'k
r3,5,7,20'k
e2,6,0,v2
r4,6,7,10'k
r5,7,o,40'k
o2,0,7,o

Ask Evaluate for expand(vo).

The answer is correct:

AS2's Example 5.10 (Cascade)

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If v₁ = 1V and v₂ = 2V, find v_o in the circuit.

e1,1,0,1
e2,2,0,2
r2,1,3,2'k
r4,2,4,4'k
r6,3,a,6'k
r8,4,b,8'k
r5,a,c,5'k
r15,b,c,15'k
r10,c,o,10'k
oa,0,3,a
ob,0,4,b
oc,0,c,o

The answer, 8.667, is correct.

TR5's Example 4-17 (Cascade)

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Derive an expression for v_o in terms of the inputs v₁ and v₂.

r1,5,0,r1
r2,5,a,r2
r3,a,6,r3
r4,6,o,r4
e1,3,0,v1
e2,4,0,v2
o1,3,5,a
o2,4,6,o

The answer you want is vo, in Results.

We get the right answer. Let's compare it with the book's answer.

Our answer, expanded via expand(vo), is shown below:

Playing with it by hand we get a form that, in my opinion, is prettier ;-) than the book's:

TR5's Example 4-18 (Multiple)

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Derive an expression for v_o in terms of the inputs v₁ and v₂.

e2,a,0,v2
e1,f,0,v1
o2,a,c,b
o1,f,d,e
r1,b,c,r1
r2,c,d,r2
r3,d,e,r3

To find vo, we ask Evaluate for vb-ve. We get an expression that can easily be rearranged to look like this:

This is exactly the answer from the book:

If you are ever in doubt whether two expressions are the same, subtract one from the other in Evaluate: if the answer is 0, they are the same.