# Backasswards Capacitor

Discussion in 'General Electronics Discussion' started by KEggemeyer, Jan 22, 2013.

1. ### KEggemeyer

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Jan 15, 2013
I received a circuit design from a friend that looks for the presence or absence of an AC signal and can’t figure out exactly how the capacitor in the circuit works…

The circuit has a capacitor (0.22 uF) with the upper end attached to the positive 12 v DC supply and the lower end attached to ground through a 1 mega ohm resistor. The 12 v AC signal (pulled from the same transformer before being rectified and filtered) passes through a forward biased diode for half wave rectification and attaches to the MIDPOINT between the capacitor and the resistor. The midpoint is also attached to the inverting input of a comparator. (Imaging a plus sign, capacitor on top, comparator on the right, resistor on the bottom, diode with incoming signal on the left.)

I know the capacitor is a filter, but I can’t wrap my head around how the capacitor is acting. This is nothing like any standard capacitive filter circuit I’ve seen (albeit my experience is limited). This circuit has the capacitor with the upper end continually at 12 v DC and the lower end bouncing from 0 to 12 volts. Obviously the difference in potential will form a capacitance, but how is this modeled to determine the filtering effect?

The circuit works, but I just can’t understand it and don’t know what to search for to find more information.

Thanks!

Kevin

2. ### BobK

7,682
1,688
Jan 5, 2010
Are you sure you are describing the circuit correctly? I simulted it and it does nothing, i..e. the voltage fed to the comparator is the same as the input AC voltage.

On the other hand, if the capacitor and resistor are in parallel, i.e. one side of each of them is connected to ground and the other is connected to the diode, then the voltage builds up to 12V over about 1 second, which seem much more likely to be the desired effect.

Edited: Never mind, I think I misread your circuit. How about a schematic so I won't make that mistake again?

Bob

Last edited: Jan 22, 2013
3. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
A schematic would be of great help.

4. ### KEggemeyer

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Jan 15, 2013
Thank you for the replies. I've attached a schematic.

File size:
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5. ### Timescope

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Aug 30, 2012
When ac is present, the diode and series resistors charge the bottom of the capacitor upwards, and the maximum voltage is limited by the zener diode. When there is no ac input, the capacitor discharges towards ground with a time constant of 0.22uF x 1.5 M ohms = 0.33 seconds and when it falls below 9.2 volts, the comparator output changes state.
Timescope

6. ### KEggemeyer

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Jan 15, 2013
Timescope,

Does it have to do with the capacitor not being instantly able to charge to +12 v on one plate and ground on the other----an RC time constant? At the top of the sign wave, the plates are at the same potential (near +12 volts), the half wave signal drops to ground and the capacitor starts to charge (bottom plate headed for ground), but it takes time to get there (RC time constant) and before it can the half sign wave is back. Therefore, the comparator always sees a voltage near +12 v DC. Is that it?

I guess I've just never seen a capacitor charged from the negative end...??? Of course, I'm new at this! ; )

Kevin

Last edited: Jan 22, 2013
7. ### Timescope

43
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Aug 30, 2012
Actually, you are right. The diode and resistors keep the capacitor discharged when ac is present and it charges to ground when ac is absent. The peak of a sine wave is 1.414 (root 2) times the rms voltage so if the zener diode was not in the circuit, the bottom of the capacitor would rise above 12v.
The capacitor is connected like this to put the circuit in the "ac present" state when power is first applied to the circuit.

Timescope

8. ### KEggemeyer

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Jan 15, 2013
Another one of those "Oh, duh!" moments for me. It seems easy now that I understand it. Thanks for the help!

Kevin

9. ### Timescope

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Aug 30, 2012
That's wonderful. You are welcome.

Timescope.