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schematic

simulate this circuit – Schematic created using CircuitLab

I want to find the value of Vo in terms of E1, E2 , E3, E4 , R.

Note that E1, E2 , E3, E4 are voltage sources.

I don't know how to proceed for positive feedback.


EDIT 1: My answer

Apply nodal analysis at V2

\${E_4-V_2\over R}\$+ \${E_3-V_2\over R}\$= \${V_2-V_o\over R}\$

\$Vo=3V_2-E_3-E_4\$ ...(1)


Apply nodal analysis at V1

\${E_1-V_1\over R}\$+ \${E_2-V_1\over R}\$= \${V_1-0\over R}\$

\$3V_1=E_1+E_2\$

\$V_1={E_1+E_2\over 3}\$ ...(2)

Now I need to eliminate V2 from equation (1).How can I do it.

Is there a relation between V1 and V2?

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    \$\begingroup\$ I think that this circuit is like a comparator with hysteresis. \$\endgroup\$ – Martin Petrei Dec 2 '15 at 13:41
  • \$\begingroup\$ @MartinPetrei circuits with no feedback act as comparators. Right? \$\endgroup\$ – kashish Dec 2 '15 at 13:44
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    \$\begingroup\$ Not quite. Opamps with no feedback act as comparators without hysteresis. Opamps with positive feedback and no phase shift network act as comparators with hysteresis. \$\endgroup\$ – Brian Drummond Dec 2 '15 at 13:47
  • \$\begingroup\$ @BrianDrummond ok! so, how should i proceed to solve the problem? \$\endgroup\$ – kashish Dec 2 '15 at 13:48
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    \$\begingroup\$ We are not here to do your homework or test question for you. Show some effort and explain exactly where you are stuck. "Don't understand positive feedback" is too broad and just punting the whole question to us. \$\endgroup\$ – Olin Lathrop Dec 2 '15 at 13:53
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Usually I solve this kind of problems in this way.

The "positive feedback" forces the opamp to deliver either +Vcc or -Vee (i.e. the positive and negative supply voltages) to its output.

So there are two cases:

  1. Output voltage is +Vcc. This condition is held until the V+ terminal has a greater voltage than V-. So I calculate V+, V- and say "the threshold from high to low is when"...
  2. Output voltage is -Vee. This condition is held until the V+ terminal has a lower voltage than V+. I proceed exactly in the same way as above.

One numerical example.

schematic

simulate this circuit – Schematic created using CircuitLab

Let's assume the output is at +5V. It will stay at +5V until V+ > V-, so

V+ = (Vout + V2) / 2 = V2/2 + 2.5V
V- = V1/2

The value changes, consequently, when V- > V+, so when

V1/2 > V2/2 + 2.5V
V1 > V2 + 5V

So the threshold from high to low is V1 = V2 + 5V

As for the other, when the output is -Vee (0V) the condition is

V+ = (Vout + V2) / 2 = V2/2
V- = V1/2

It will stay in this condition until V+ < V-; consequently it will change status when

V+ > V-
V1 < V2

So the low-to-high threshold is V1 = V2

So, let's assume V2 constant. When V1 raises above V2 by more than 5V the output will switch to high (Vcc). Then it has to go lower than V2 in order for the output to become low.

Of course I didn't solve your exercise, because you need it in order to understand if you understood it

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  • \$\begingroup\$ If the answer completely answers your question, so, please mark it as accepted ;) \$\endgroup\$ – frarugi87 Dec 2 '15 at 15:09
  • \$\begingroup\$ No! my question is still unanswered. I'll try to solve it and then edit my question. Please then assist me. \$\endgroup\$ – kashish Dec 2 '15 at 15:11
  • \$\begingroup\$ Well, I think you will never receive a direct answer, since this really looks like a homework or a test question. You'll have to understand how it works and then apply the rules to your circuit... \$\endgroup\$ – frarugi87 Dec 2 '15 at 15:18
  • \$\begingroup\$ yes, you are right. I'll edit my question soon and come up with answer. Then , correct me. \$\endgroup\$ – kashish Dec 2 '15 at 15:25
  • \$\begingroup\$ Edited my question \$\endgroup\$ – kashish Dec 2 '15 at 15:45
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Your approach isn't correct. Given no negative feedback, your output will saturate near the positive rail or the negative rail, depending on what's going on at the input terminals of the op amp.

Because the input/output relationship is nonlinear, there is no linear way to solve it -- so, pick a starting case where you know what the output is, say the negative input terminal has a lower value than the positive terminal. The output will be near the positive rail. Based on that value of Vout, now solve for what's going on at the positive input terminal. That's the voltage for which when V- exceeds it, your comparator will change state and go to the negative rail. At that point, V+ changes. This is hysteresis if the change is such that V+ decreases, and nonsense if the change is such that V+ increases.

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TLO81 is an Operational Amplifier designed by Texas Instruments. Now, by the property of virtual ground the voltages V1 = V2. Substituting this relation in the solution you can easily calculate the Vo in terms of E1, E2, E3 and E4. Vo = E1 + E2 - E3 - E4.

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