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Below is a part of a schematic and I would like to understand each part of it.

enter image description here

What I think is happening:

  1. This is being fed a 48V AC source
  2. These leads are a source and a ground?
  3. I'm not sure what that is
  4. Two alternating currents on the input
  5. Not sure what this is.. a diode? An op-amp?
  6. Showing which terminal output is positive and which is negative.

These are my best "educated" guesses. I know I'm way off, so I was hoping the community could help me understand.

This is part of a schematic from steminc to drive a piezo transducer.

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up vote 14 down vote accepted
  1. Yes.

  2. No, they're just wires going to the source.

  3. It's a MOV (Metal Oxide Varistor) to absorb surges (through-hole types look like ceramic disk capacitors).

  4. Markings on the symbol for the AC input of a bridge rectifier.

  5. A bridge rectifier (four diodes).

  6. More markings on the symbol for the DC output of the bridge rectifier.

enter image description here

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Perfect response, Thank you! I'll accept once the time limit allows me to. Answer is accepted because of the format of the answer – ntgCleaner Jul 10 '14 at 17:05
and accuracy of course! – ntgCleaner Jul 10 '14 at 17:46

It's a bridge rectifier - it consists of 4 diodes and always does its best to route the most positive part of the AC power input to the + output and likewise on the negative side.

enter image description here

I wouldn't assume the power in is "source" and "ground" - it's more likely to be a floating (groundless) output from a transformer secondary.

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So this converts AC to DC current? Does the voltage still apply on the DC side? For example, does this convert 48V AC to 48V DC? – ntgCleaner Jul 10 '14 at 17:03
@ntgCleaner Yes it does and it converts with a slight loss in voltage - if you apply 48V dc on the input, the output will be more like 46.5V dc - this is due to diodes being imperfect. – Andy aka Jul 10 '14 at 17:05
Great! Thank you Andy. This alleviates a lot of my concerns. I was trying to make an oscillating circuit run by DC and a 555 timer to run a piezo transducer. This answers so much! – ntgCleaner Jul 10 '14 at 17:06
@ntgCleaner, This circuit rectifies(fullwave) the negative cycle of the AC signal. The output then uses capacitors to maintain a somewhat constant voltage (with ripple), but those are not shown here. See the wiki. If by DC you just mean current flow in one direction then yes it converts AC to DC. But if you mean constant-voltage DC then no, not quite. – sherrellbc Jul 10 '14 at 17:25
@sherrellbc - where are these assumptions you refer to? – Andy aka Jul 10 '14 at 17:27

1.This is being fed a 48V AC source

Right on.

2.These leads are a source and a ground?

Not quite. There is no specific source/ground attribution. Unless the source they get connected to is shown as grounded, they can be assumed to be floating.

3.I'm not sure what that is

That is some sort of input spike limiter, such as a transzorb.

4.Two alternating currents on the input


5.Not sure what this is.. a diode? An op-amp?

6.Showing which terminal output is positive and which is negative.


simulate this circuit – Schematic created using CircuitLab

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