Operational amplifiers

Bo2's Drill Exercise 3.11 (Thévenin)

Find the Thévenin equivalent.

s\th("e,1,0,vs:r1,2,0,r1:r2,2,3,r2:o,1,2,3",3,0)

The th script tells us it found the Thévenin voltage, but could not find the Norton current. This is not a surprise, since an ideal op amp has zero output resistance and a fixed voltage, an infinite current when short-circuited. So the Thévenin equivalent is given by V_TH and no resistance (or R_EQ = 0Ω). Evaluating vth results in

((r1+r2) vs)/r1

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

The Thevenin resistance, as explained above, is 0Ω.

AS2's Example 5.2

Find v_o and i_o.

s\dc("e,2,0,1.:r5,1,0,5'k:r4,1,o,40'k:r2,o,0,20'k:o,2,1,o"):{vo,io}

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

Bo2's Figure 3.3 (Inverting)

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

s\dc("e,2,0,vs:r1,2,1,r1:r2,1,o,r2:o,1,0,o"):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)

Find v_o.

s\dc("e,2,0,vi:r1,2,1,r1:rf,1,o,rf:o,0,1,o"):vo

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

-(rf/r1)vi

AS2's Example 5.3 (Inverting)

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

s\dc("e,2,0,.5:r1,2,1,10'k:rf,1,o,25'k:o,0,1,o"):{vo,ir1}

The answer, {-1.25, 5e-5}, is correct.

TR5's Exercise 4-11 (Inverting)

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.

s\dc("e,1,0,vs:r1,1,2,10'k:r2,2,o,33'k:o,0,2,o"):
{vo|vs=2.,vo|vs=–4.,vo|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)

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.

s\dc("e,2,0,40'm:r1,2,1,5'k:rf,1,o,15'k:o,0,1,o"):approx({vo,irf})

The answer, {-.12, 8e-6}, is correct.

AS2's Example 5.4 (Inverting)

Determine v_o.

s\dc("e6,1,0,6:r1,1,a,20'k:e2,b,0,2:rf,a,o,40'k:o,a,b,o"):vo

The answer, -6, is correct.

TR5's Example 4-14 (Inverting)

Find the input-output relationship of the circuit.

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

s\dc("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 (and faster – as in 49 seconds instead of 55 seconds) way to symbulate it is this:

s\dc("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)

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

s\dc("j,0,1,is:r,1,o,r:o,0,1,o"):vo/is

The answer, -r, is correct.

AS2's Practice Problem 5.4b (Transresistance)

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

s\dc("j,0,1,is:r1,1,2,r1:r2,2,0,r2:r3,2,o,r3:o,0,1,o")

Asking:

expand(vo/is)

gets -r1*r3/r2-r1-r3, which is equivalent to the book's answer.

Bo2's Example 3.1 (Non-Inverting Amplifier)

Find v_o/v₁.

s\dc("e,p,0,v2:r1,1,0,r1:r2,1,o,r2:o,p,1,o"):vo/v1

We get 1+r2/r1, which is correct.

AS2's Figure 5.16 (Non-Inverting Amplifier)

Find v_o.

s\dc("e,2,0,vi:r1,0,1,r1:rf,1,o,rf:o,2,1,o"):vo

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

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

Find Vo/Vs.

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

s\dc("e,1,0,vs:r1,1,2,r1:r2,2,0,r2:o,2,3,o:r3,o,3,r3:r4,3,0,r4"):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.

s\dc("e,1,0,vs:r1,1,2,r1:r2,2,0,r2"):v2/vs

We get

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

s\dc("e,2,0,1:o,2,3,o:r3,o,3,r3:r4,3,0,r4"):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)

Find v_o.

s\dc("e,1,0,vi:o,1,o,o"):vo

The answer, vi, is correct.

TR5's Figure 4-32 (Voltage Follower)

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

s\th("e,1,0,1.5:rs,1,2,2'k",2,0):{pmax,prL|L=1000}

The answers we get, {2.8125e-4,2.5e-4}, are correct. Now we simulate the (a) circuit.

s\dc("e,1,0,1.5:rs,1,2,2'k:o,2,o,o:rl,o,0,1'k"):prL

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)

Find v_o.

s\dc("e,1,0,6:r4,1,a,4'k:r10,a,o,10'k:e4,b,0,4:o,b,a,o"):vo

The answer, -1, is correct.

AS2's Practice Problem 5.5 (Non-Inverting)

Calculate v_o.

s\dc("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)

Find v_o.

s\dc("ea,3,0,va:eb,2,0,vb:r31,3,1,r1:r21,2,1,r1:r1o,1,o,r2:o,0,1,o"):vo

-(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)

Find v_o.

s\dc("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"):expand(vo)

-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)

Find v_o and i_o.

s\dc("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"):{vo,io}

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)

Find v_o and i_o.

s\dc("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"):{vo,io}

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)

s\dc("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")

Evaluating vo we get the correct answer, equivalent to the book's answer above.

((vb-va) r2)/r1

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

Find v_o.

s\dc("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")

Evaluating vo we get 3 v2 − 4 v1, which is the correct answer.

AS2's Figure 5.24 (Difference or Differential)

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

s\dc("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"):vo

This expression is equivalent to the book's answer.

AS2's Figure 5.24 (Subtractor)

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:

expand(vo)|r2=r1 and r3=r4

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

AS2's Example 5.7 (Difference or Differential)

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:

Define v1=1:Define v2=0:expand(vo)=–5

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

Define v1=0:Define v2=1:expand(vo)=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.

solve(ans(1) and ans(2),{r2,r4})|r1=10000 and r3=20000

The expression above assumes that ans(1) and ans(2) are pointing to the two equations we found before. 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.

s\dc("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")

Evaluating vo gives us 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.

Define v1=1:Define v2=0:vo=–4

Define v1=0:Define v2=1:expand(vo)=4

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

solve(ans(1) and ans(2),{r2,r4})|r1=10000 and r3=10000

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

AS2's Practice Problem 5.8 (Instrumentation)

Find i_o.

s\dc("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"):ir5

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

Bo2's Example 3.3 (Cascade)

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

s\dc("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")

Evaluating vo we get:

((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)

s\dc("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")

Evaluating vo we get: -.75 vs, which is correct.

AS2's Example 5.9 (Cascade)

Find v_o and i_o.

s\dc("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")

Evaluating approx({vo,iro}) we get the answer, {.35,2.5e-5}. This is correct.

AS2's Practice Problem 5.9 (Cascade)

Determine v_o and i_o.

s\dc("e,1,0,4:o1,1,2,2:o2,2,3,o:ro,3,0,4'k:r6,3,o,6'k"):{vo,iro}

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)

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

s\dc("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"):vo

The answer, 9, is correct.

TR5's Example 4-16 (Cascade)

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

s\dc("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"):expand(approx(vo))

The answer, -2.4 v1 − 6, is correct.

TR5's Exercise 4-14 (Cascade)

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

s\dc("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"):expand(vo)

The answer, 8 v1 – 4 v2, is correct.

AS2's Example 5.10 (Cascade)

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

s\dc("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"):approx(vo)

The answer, 8.667, is correct.

TR5's Example 4-17 (Cascade)

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

s\dc("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"):vo

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)

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

s\dc("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, enter them both separately into the calculator and compare them with the equality sign. If the answer is 'true', they are the same.