Electronics > Beginners
Virtual ground in op amp circuit design
VinAng0811:
Hi
I have simulated 3 circuit in PSPICE. Kindly refer to the attach file.
Figure 1(a), Figure 2(a) and Figure 3(a) are the same circuit. ( a virtual ground circuit, where the output voltage of the op amp = Vdc/2)
so output voltage of the op amp in Figure 1(a), Figure 2(a) and Figure 3(a) is 2.50 V.
In Figure 2 and Figure 3, the output of the op amp (a) and (b) are connected. Based on the simulation result in PSPICE, can anyone explain why the output voltage in Figure 2(b) is 2.00 V and Figure 3(b) is 2.50 V ?
Thank you
mc172:
Because in circuits 2 and 3 you've effectively blown up the opamps. Think about what the following circuit is doing:
Also consider how much current you might expect going in or out of pin 6 on both opamps. If you aren't sure, it's on the datasheet for that device.
Then look at how you've connected yours in your circuit.
mc172:
Another thing to think about is why circuit 1(b) outputs 4.97 V. Why this number?
Alex Nikitin:
Hi and welcome to the forum. You have just learned that Spice simulations are not always useful ;) . For a start, check output currents of opamps in your simulations and see if these make any sense to you.
Cheers
Alex
VinAng0811:
circuit 1(b) the Vin=5V, and the supply voltage of the op amp is 5V, in ideal case the output voltage of a buffer should equal to 5V... why 4.97 V ? because positive supply rail.
Navigation
[0] Message Index
[#] Next page
Go to full version