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I asked a previous question about this circuit:

schematic

I am fairly confident that I understand why the transformer coil only sees something like 2V. What I am now unsure about is why the rectifier is needed in the first place.

From what I understand, if the rectifier is removed, there will still be a small voltage drop over the coil of the transformer, with the rest going to the load.

So what does the rectifier actually do here?

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  • \$\begingroup\$ Which rectifier - D1 or D2? \$\endgroup\$ – W5VO Apr 14 '15 at 7:50
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A big part of this project is that these are "parts on hand" or parts you can get at a local electronics store (possibly in bankruptcy) with poor part selection. A couple of parts are being used in non-conventional means, and are referred to in their original purpose.

Bridge rectifier D1 is effectively functioning as a current shunt. As long as current is flowing through the diodes, you will get a 1.2V AC drop, roughly independent of the magnitude of the current. An alternative would be to use just a resistor, but that would decrease the power available to the tools, waste a lot of power, and give a variable amplitude signal.

The transformer is listed as a 6VAC to 120VAC transformer, but again, that's just what you'd look for when you're shopping for parts. In reality, it is a voltage transformer with a 20:1 turns ratio (120V/6V). It's wired "backwards" to give a 1:20 ratio, meaning that the 1.2VAC signal generated by the diodes on the primary will result in roughly a 24V signal on the output. Note that the transformer is connected across the bridge rectifier, so it only "sees" 1.2V, not the full 120VAC provided by the wall outlet.

On the secondary of the transformer, the AC signal is now rectified by D2, and turned into roughly 10V DC by C2 and D3. When the tool is running with enough current, this should result in sufficient voltage to activate the relay.

Below is a CircuitLab model of the system that you can simulate to see what's going on. The saw is switched on at t=0.1s, and the vacuum will switch on shortly after that.

schematic

simulate this circuit – Schematic created using CircuitLab

Here is the simulation waveform. I have hidden the Line voltage, as it is a regular 120V AC signal. I ended up getting about 2V across the diodes mainly due to the diode model selected, so the voltages are a bit higher.

enter image description here

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  • \$\begingroup\$ Can you please explain how to run this simulation? \$\endgroup\$ – Cameron Ball Apr 14 '15 at 8:47
  • \$\begingroup\$ Hmm.... so clicking on the "simulate this circuit" link will ask you to create an account. I'm not sure that's intentional. In the mean time, if you don't want to create an account, you can edit the question, and the link changes to edit the above schematic. Click on that, and then you can get the full version without logging in. \$\endgroup\$ – W5VO Apr 14 '15 at 8:54
  • \$\begingroup\$ @CameronBall I've edited the answer to include the simulation results, as that will probably be easier. \$\endgroup\$ – W5VO Apr 14 '15 at 9:13
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D1 (the bridge rectifier) protects the transformer 6 volt winding from receiving a voltage that could be tens (or maybe hundreds) of volts from "line in". You have to consider that the load (the motor?) may go into a stall condition and without diodes you can't limit the 6V AC voltage. If this rises to say 20V AC, the output from the secondary (called 120V AC) would be stepped up by the turns ratio of the transformer.

The turns ratio is 120/6 = 20 so, with an unrestricted 20V AC fed into the 6V winding, the output voltage would be 400V AC. It could be more under stall conditions. It could fry the relay (K1), destroy the diodes D2 and D3, explode C2 and fry the transformer due to core saturation.

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  • \$\begingroup\$ Ok, but why the full bridge? Wouldn't just two parallel diodes with opposed polarity do the trick? \$\endgroup\$ – Dmitry Grigoryev Apr 14 '15 at 8:17
  • \$\begingroup\$ What exactly is meant by a "stall condition"? \$\endgroup\$ – Cameron Ball Apr 14 '15 at 8:18
  • \$\begingroup\$ @DmitryGrigoryev Two parallel diodes might do the trick but the AC voltage into the transformer would be about 1Vp-p and this might not be enough to produce enough voltage to activate the relay (hey a different relay could be chosen of course!). \$\endgroup\$ – Andy aka Apr 14 '15 at 10:23
  • \$\begingroup\$ @CameronBall motor stalling under heavy mechanical load. \$\endgroup\$ – Andy aka Apr 14 '15 at 10:23
  • \$\begingroup\$ High power bridge rectifiers are cheap and easy to mount. You could indeed use a pair of diodes rather than 4, but mounting 2 separate units would take more labor than a single bridge, and probably cost more. \$\endgroup\$ – WhatRoughBeast Apr 14 '15 at 12:37
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The rectifier limits the voltage over the 6V secondary of the transformer (misused here as primary) to ~ 1.2V (assuming it is a Si rectifier) so it doesn't have to take all the current.

You can think of a transformer as an 'impedance transformer'. In this case the impedance at the 120V side will probably be rather high, so the trasnformed impedance at the 6V 'primary' will be ( sqrt( 6 / 120 ) * high ) (or was it pow(), I don't recall for sure) which is still high :). But a high impedance in series with a load is not a good idea, hence the diodes are added to provide a low-impedance path in parallel, while still giving enough voltage to the transformer esction to do whatever it is supposed to do (what is that K for?)

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  • \$\begingroup\$ I assume K1 is the contactor (Used in electrical control panels) K is the coil, the (misused) Cap symbol is the open contacts. I personally hate that convention and mixing it with an electronics diagram is worthy of inventive punishment. \$\endgroup\$ – Spoon Apr 14 '15 at 11:31

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