Circuit analysis - DC and AC source

Discussion in 'Homework Help' started by xxxyyyba, Sep 26, 2014.

1. xxxyyyba Thread Starter Active Member

Aug 7, 2012
279
6
Hi!
Here is my circuit and graph of waveforms:

My question is, how can I find expression for output voltage vout(t)? If I use superposition method, for active AC source (DC source is short) I will get vout_ac(t)
$=Vrms\sqrt{2}\sin{(314t+\varphi )}$, where $\varphi$
is phase angle. For DC analysis (AC source is short, capacitor is removed), I will get Vout_DC=constant, so
vout(t) will be $=Vrms\sqrt{2}\sin{(314t+\varphi )}+constant$. But it is not expression for our output waveform.

Last edited: Sep 26, 2014
2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
In your simulation the DC value of V12 has not stabilized to its steady state value. Hence you are seeing a transient in the negative offset due (primarily) to the presence of the capacitor. Take your simulation out in time until the output DC offset has stabilized.

Does the schematic indicate the controlled source has a gain of 200,000???

3. xxxyyyba Thread Starter Active Member

Aug 7, 2012
279
6
That's model for some real Operational Amplifier and A (open loop gain) is always big number

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4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
OK got it - by the way, since you have an integrator your output DC offset will probably never stabilize - unless you took steps to ensure that outcome.
This isn't a real circuit in which the output would "clamp" due to the limited amplifier dynamic output range available in practice.

Last edited: Sep 26, 2014
5. xxxyyyba Thread Starter Active Member

Aug 7, 2012
279
6
I suppose it. So in order to fint complete response I have to solve diff. equations... ?

Last edited: Sep 26, 2014
6. xxxyyyba Thread Starter Active Member

Aug 7, 2012
279
6
Solved it I used Laplace transform to solve this problem and here is plot of output voltage vout(t) I got:

And here is result from Multisim:

Thanks a lot!

7. MrAl Distinguished Member

Jun 17, 2014
3,605
754
Hi,

Usually a high value resistor is used across the cap in these kinds of circuits if that change can be tolerated. For example, in this circuit even a 5 meg resistor across the cap would help to limit the output max negative voltage to around -7.5 volts instead of eventually reaching the negative rail.

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