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 is the one that catches people out, and Symbulator says so plainly:

• vth = \(\frac{vs \left(r_{1} + r_{2}\right)}{r_{1}}\)
• ino = ∞
• req = 0
• pmax = ∞

with a note beneath the answers explaining that the short-circuit current is unbounded, so the equivalent is a voltage source with no impedance in series with it.

None of that is a failure to solve, and none of it is guesswork. An ideal op amp holds its output voltage whatever current is drawn from it, so asking what flows through a short across that output has no answer as an equation — the two rounds Symbulator runs to build a Thévenin equivalent leave the second one unsolvable. What it does instead is put a resistance across the terminals rather than a short, and let that resistance fall to zero, which is what a short is. The current grows without bound, and an unbounded current through a short is precisely a source with nothing in the way of it: R_EQ = 0Ω.

So V_TH is correct, as can be seen by comparing it to the book's answer — and the Thévenin resistance is not merely asserted here, it is the answer Symbulator gives.

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, {9.,.00065}, is correct: v_o is 9V and i_o is 0.65mA

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 -r2/r1, 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:

-(rf/r1)vi

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 vo and ir1, in Results.

The answer, {-1.25, 5e-5}, 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

The calculator's | says "given"; version 9 says the same thing in the Evaluate card's own Conditions box. Put the answer you want in the first and the value you are giving it in the second:

vo

vs = 2

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

The answer, {-6.6,13.2,-19.8}, is 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 answer, {-.12, 8e-6}, 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)

in fullOpen in app ↗Open in split viewthe quicker wayOpen in app ↗Open in split view

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

When we evaluate vo/v1, both approaches get 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, -r, 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 -r1 - r1*r3/r2 - r3, 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 1+r2/r1, 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)

Non-Inverting, in oneOpen in app ↗Open in split viewthe left halfOpen in app ↗Open in split viewthe right halfOpen in app ↗Open in split view

Find Vo/Vs.

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

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

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, vi, is correct.

TR5's Figure 4-32 (Voltage Follower)

the source aloneOpen in app ↗Open in split viewbufferedOpen in app ↗Open in split view

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.

The answer you want is pmax, in Results.

The power in a 1 kΩ load is the expression Lesson 4 derives, with 1000 where R goes:

vth^2*1000/(req+1000)^2

The answers we get, {2.8125e-4,2.5e-4}, 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

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.

-(r2/r1)(va+vb)

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), exactly as on the calculator.

-rf v1/r1 - rf v2/r2 - rf v3/r3

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 vo and io, in Results.

The answer, {-8.,-.0048}, 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 vo and io, in Results.

The answer, {-3.8,-.001425}, 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.

((vb-va) r2)/r1

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 3 v2 − 4 v1, which is the correct answer.

AS2's Figure 5.24 (Difference or Differential)

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Find v_o. (And keep it in the memory, for you will use it in the next three problems.)

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 vo, 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.

Since we already have the expression for vo stored in the memory, we only do this:

The calculator's | is Evaluate's Conditions box. Ask Evaluate for expand(vo) and give it both equalities:

r2 = r1
r3 = r4

The answer we get, v2-v1, is correct.

AS2's Example 5.7 (Difference or Differential)

the v1 factorOpen in app ↗Open in split viewthe v2 factorOpen in app ↗Open in split viewchecking the designOpen in app ↗Open in split view

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.

Version 9 will not always arrange an expression the way the book does. If you want it another way, the Evaluate card takes simplify(), collect(), expand(), factor() and apart().

We get -r2/r1=-5. Now make that part of v_o that is a factor of v₂ equal to 3. 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.

We get r2*r4/(r1*(r3+r4))+r4/(r3+r4)=3 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.

There is no ans(1), so write the two equations out. This is what the Solve card is for — it 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 r2 = \(5 r_{1}\) and r4 = \(r_{3}\): 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.

We get r2=50000 and r4=20000. This is correct.

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

vo is the desired output, 3*v2-5*v1. The resistor values are correct.

AS2's Practice Problem 5.7 (Difference or Differential)

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 r2 = \(4 r_{1}\) and r4 = \(4 r_{3}\) — at 10 kΩ each, 40000 Ω and 40000 Ω.

We get r2=40000 and r4=40000, the correct values for the remaining resistors.

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:

((g1-g2) vs)/(g3-g4)

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: -.75 vs, which 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, {10,1/1000}, is correct. To say a 1/1000 A current is the same as 1mA.

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 — that is what the calculator's expand(approx(vo)) is doing. It is arranged differently from the book's, which is a matter of presentation rather than of arithmetic.

The answer, -2.4 v1 − 6, 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), as on the calculator.

The answer, 8 v1 – 4 v2, 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 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 — an equality sign would only be read as a comparison to solve, not as a question about sameness.