# Capacitor in whaetsen bridge

Discussion in 'Homework Help' started by activee, Jun 12, 2014.

1. ### activee Thread Starter Member

Jan 16, 2014
39
0
Hey, I had a question last test which was something along the line of
<< How would you smooth Vu ?>> Vu being V on the middle resistor.
Here is the drawing. I wonder why my answer I put there wasn't correct, to me it seems like it would be the same thing as the real answer..

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2. ### MrChips Moderator

Oct 2, 2009
12,651
3,460
The correct answer is that shown with the capacitor across the load Vu.

A capacitor is a charge storage device. The best analogy is a large holding tank of water to slow down any rise and fall of water pressure.

It takes time to charge and discharge a capacitor.

A capacitor in parallel with a resistor creates a low pass filter, i.e. it attenuates high frequency fluctuations.

Edit: The correct spelling is Wheatstone bridge

Last edited: Jun 12, 2014
3. ### activee Thread Starter Member

Jan 16, 2014
39
0
ok thanks so my answer was correct.

4. ### MrChips Moderator

Oct 2, 2009
12,651
3,460
That's not what I said.

5. ### activee Thread Starter Member

Jan 16, 2014
39
0
but if it takes time to charge and discharge then it won't fluctuate as much with the capacitor as I drew them and the signal will be closer to a DC signal.

Last edited: Jun 12, 2014
6. ### MrChips Moderator

Oct 2, 2009
12,651
3,460

A capacitor in series is a high pass filter. It blocks the zero and low frequencies and passes the high frequencies.

The VU meter is going to indicate the rapid changes which is the opposite of what you want.

7. ### activee Thread Starter Member

Jan 16, 2014
39
0
Hmmm I understand I think. At least I'll understand the last bit on my own. I hate to ask but I got a test in 2 hours so I'm stressed. So I would like to understand how this circuit works :

I get that the 2 first diode make it so we have the absolute value of the signal. If I'm correct the capacitor should be loaded so the current shouldn't flow through but it will polarise the signal a bit. After that Does the Zenner diode "cut" the upper edge of the signal ? Then the LED light will light when the voltage at the gate of the transistor is above 0.6v. And I guess the capacitor is to make sure the LED lights up with around the same intensity at all time.(almost)

It would help me a ton if you answered this and I'll stop posting .

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Last edited: Jun 12, 2014