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# OP Amp

Discussion in 'Homework Help' started by stri909, Aug 8, 2009.

1. ### stri909 Thread Starter New Member

Aug 8, 2009
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0
Hello, i have two problems that i can't seem to solve. For the first one, i know its a non-inverting OP where my Rs should be my 3.3K ohm resistor. But for my feed back, would it be the sum of the other three resistors? Also, my Vo for the top OP amp should equal my Vx correct? Also, how would i find my ia and io? thank you for your help.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If opamp work in a linear region this mean that voltage on non-invert input equal voltage on invert-input
So if the input voltage is 1.32V in a "+" then we mast get 1.32 in a "-" input.
So current that is flow through 2.2KΩ is equal
Ia=1.32V/2.2K=600μA and this current is also flow through 6.8KΩ.
So we have this situation:

And now you easily can calculate the Vx and the unknown current.

3. ### hgmjr Retired Moderator

Jan 28, 2005
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219
jony130,

The input voltage to the positive terminal of the upper opamp is -1.32 not 1.32. Other than that decrepancy your answers appear correct.

hgmjr

4. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Post your attempt at solving the problem and then we can assist you with any snags that you have encountered.

hgmjr

5. ### stri909 Thread Starter New Member

Aug 8, 2009
6
0
Thank you guys for your help.

I understand that for an ideal amp Vp=Vn as well as Ip=In = 0.
I also know that for a non-inverting amplifier, the equation is Vo = [(Rs+Rf)/Rs]Vg. In this case, my Vo should become the Vs for my inverting opamp on the bottom correct?

for my Non-inverting op amp, would my feedback be the sum of the three resistors? I.E 6.8K + 27K + 20K = 53.8K ohm.

Vo = [(2.2k + 53.8k)/2.2k]1.32 v = 33.6V

Therefore my Vx = Vo correct? Or am i approaching this problem incorrectly?

6. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Focusing on the voltages for the moment, since you know the voltage that should be present on the negative input of the non-inverting opamp, you can use that voltage together with the 2.2K and the 6.8K resistor you can compute what the voltage at the output of the inverting opamp must be.

Then knowing the behavior of an inverting opamp, you can compute what voltage must be present at the input terminals labeled Vx to produce the previously computed output voltage from the inverting opamp.

hgmjr

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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496
hgmjr, doesn't it look like that circuit has a loop through the - inputs of the two opamps? I wonder if it amounts to a Schmitt trigger type latch.

I don't have time to analyze it now, but I'll look at it later.

8. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
The information I provided was in error. I have removed it to avoid confusing future readers.

hgmjr

Last edited: Aug 9, 2009
9. ### stri909 Thread Starter New Member

Aug 8, 2009
6
0
Thank you guys for the help.
I used the voltage divider on the two lower resistors to find my Vout,
1.32v = (2.2k/6.8k)Vout.
Vout = 4.08 v

i then used Vo= (-Rf/Rs)Vs equation to find my Vx

4.08 = (-27k/20k)Vs
Vs = Vx = -3.02 v

Io = if - ia

if = -3.02/47k = -6.4 X 10^-5

Io = -6.64 X 10^-4

thank you guys for the help

10. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
You have the right idea but you have erred in your calculation of the value Vout present at the output of the bottom opamp.

HINT: The 2.2K and 6.8K resistors form a simple voltage divider. Take another look at how you calculated Vout.

hgmjr

11. ### stri909 Thread Starter New Member

Aug 8, 2009
6
0
Vout = 5.4 v?

i took another look and i noticed a calculated wrong with the voltage divider.

12. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
That is the correct value but take another look at the sign. Notice that the input applied to the terminal in the upper opamp is -1.32 not +1.32. Take this into account and then proceed to calculate what voltage would need to be applied to the terminals labeled Vx to get that voltage and you will have that value that you need for Vx. Once you have Vx and the Vout you can calculate the currents that are being asked for.

hgmjr

13. ### stri909 Thread Starter New Member

Aug 8, 2009
6
0
Thank you hgmjr, i looked at the voltage source and now know what u mean.

14. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
The circuit in Fig. P6.39 has positive feedback. It is a Schmitt trigger. The threshold voltages are ±6*(2.2k/(6.8k+2.2k))=±1.467V. Since -1.32V is within the hysteresis zone, the output is indeterminate, but will be either +6V or -6V.

EDIT: If the input never exceeds ±1.467V, the circuit just acts as an amplifier with Av=((6.8k+2.2k)/2.2k)*-(20k/27k)=-3.03.

Once the op amps saturate, it becomes a noninverting Schmitt trigger, and apparently stays that way until hell freezes over. As hgmjr said, a vey interesting circuit.

Last edited: Aug 8, 2009
15. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,923
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He, you're right it's a Schmitt trigger.
Next time I have to analysis the circuit with much more care.
So, first before any calculation we have to check the feedback loop.

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Oct 9, 2007
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17. ### Ratch New Member

Mar 20, 2007
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stri909,

Don't get hung up about whether an op-amp is inverting or not. Just follow the currents and voltages. Using conventional charge flow, you know that the voltage between the 2k2 and 6k8 resistors is -1.32. That automatically gives the current through the 2k2 resistor of -1.32/2k2 = 0.6 ma . That current has to exist in the 6k8 resistor also because of the theoretical infinite impedance of the upper opamp input. So Ia is 0.6 ma. The voltage across the 6k8 resistor is therefore -6k8*0.6 ma = -4.08 volts. Adding the two voltages gives -1.32 - 4.80 = -5.4 volts output by the lower opamp. The 27k current is -5.4/27k = 0.2 ma. Adding the currents 0.6 ma and 0.2 ma gives a Io of 0.8 ma that the lower opamp sinks. Continuing, the voltage across the 20k resistor is 0.2ma * 20k = 4 volts which is also the Vx value.

The problem does not ask to identify the circuit, or analyze its lockup characteristics.

Ratch

18. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Here are the results of a couple of simulations. With the exponentially increasing sine wave, note that node x acts as an amplifier with gain=-3.03 until the voltage exceeds ±1.467V (-1.467V in this case), at which point it begins to act as a Schmitt trigger. With the exponentially decaying sine wave, the voltage almost immediately reaches +1.467V, sending the circuit into Schmitt trigger mode, from which it never recovers.

I ran these sims on LTspice, first on my laptop, and then on my desktop. I originally got the results shown here on my laptop. I redrew the schematic on my desktop, and could not get the circuit to start up as an amplifier, even using .IC directives on the various node voltages. It always started up with the two op amp outputs stuck at the rails. I then copied the .ASC file from the laptop to the desktop, and it again ran as shown below.
Next I tried to put two copies of the circuit on the same schematic, but with the increasing amplitude into one circuit, and the decaying one on the other. Once again, it started with op amp outputs stuck at the rails.
Apparently the initial conditions are sensitive to netlist order.

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19. ### stri909 Thread Starter New Member

Aug 8, 2009
6
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Thank you all for your help.

For the 2nd problem, i am confused on how i am suppose to handle alpha and sigma resistors. with vg1 going down, do i treat is as a node and the current goes both ways?