# what circuit is it?

Discussion in 'Homework Help' started by bug13, Aug 28, 2012.

1. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
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as seen in picture, what sort of circuit is it? does it have a name? what should I google for to find out more about it? and how do I do the calculation etc..

this circuit is use to sense the voltage of a AC source, that all I know

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The intent is probably creating a DC offset AC signal. Looks very unsafe to me.

3. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
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How's that work?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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See the attachment - but don't try to build it as a real circuit. You could be electrocuted if something goes wrong or you make an error.

• ###### DC offset AC gen.pdf
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5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The large capacitor you can treat as a voltage source as long as R*C (capacitor time constant) >> 1/f_ac_signal.
So in your digram capacitor is charge to
Vc = Vdd * R4/(R3+R4). So simply replace capacitor with Vc voltage source.

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6. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
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which resistor/s do I use to calculate the time constant?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well I think that the time constant is equal to resistance seen from the capacitor terminal. t = R * C ---->R = R3||R4 + R1||R2

8. ### bug13 Thread Starter Well-Known Member

Feb 13, 2012
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ok, this confused me more, I simply don't understand why R3||R4 and R1||R2, if || here means parallel.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Simply you must find a equal resistance seeing from the capacitor terminal when the other remaining voltage source replace with a short-circuit.