# Rectifier Circuit Help

Discussion in 'General Electronics Chat' started by blah2222, Jan 3, 2013.

1. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
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33
Hi, I am trying to figure out how this attached circuit works to rectify an input voltage and output a current $Ig = 2{Vin_{ave}}{R1}$

What is the deal with the norator connected to another via a diode? Shouldn't there be another diode for full-wave rectification going in the opposite direction to rectify the negative swings?

Also, not really sure what the transistor with two emitters means. Is this a short-form for a gain of two?

Any help would be much appreciated.

JP

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2. ### GopherT AAC Fanatic!

Nov 23, 2012
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First, I am surprised this one has been sitting so long without a comment.

I understand the conceptualized circuit as follows:

The inverting op amp forces the positive half of the input to signal through the feedback loop (no resistance). The negative cycle is inverted by the op amp.

There is likely some significant circuitry covered up by the notator (constant current source) symbol.

The diode allows the un-inverted (positive cycle) AND inverted (negative cycle) to pass to a second constant current source (essentially a current mirror).

The current the flows to the capacitor-resistor filter that averages the rectified signal into a flat DC signal.

The transistor connected to the resistor becomes the master of a current control for the resister on the right (with two emitters).

The double emitters represent, in this case, another current mirror (essentially two transistors in parallel with a common collector. The two emitters allows 2x the current to pass as is allowed to pass to the collector as the single transistor on the left, a current gain of 2.

That is just my interpretation of this conceptualized circuit.

Cheers.

Last edited: Jan 6, 2013
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