Full-wave Bridge Rectifier Question

A question came up while doing some testing in our work lab. 120VAC power supply connected to a full-wave bridge rectifier. Approximately 105VDC on the output of the rectifier. Now, wiring rectifier DC output through a contactor (120VAC coil) and two normally open contacts, with no load, the voltage at the contacts of the contactor (when pulled in) was approximately 160VDC. Once we connected a load to the circuit the voltage dropped back down to where I would expect, 105VDC. But why the open circuit voltage of 160VDC? I'm not understanding that. Any thoughts?

I am more confused about the 105VDC. Do you have any DC bus capacitors? Fullwave rectified, smoothed will get you ~170VDC, 120*root(2), open circuit. What are you using to measure the voltage?

Adrenaline said:

I am more confused about the 105VDC. Do you have any DC bus capacitors? Fullwave rectified, smoothed will get you ~170VDC, 120*root(2), open circuit. What are you using to measure the voltage?

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That's the part that I actually thought I understood. An AC voltage rectified is usually around 0.9 x your AC voltage to get DC voltage (with a load). So in this case the line voltage hovers around 118VAC x 0.9 = 106.2VDC. No bus capacitors and just using a Fluke True RMS multimeter (measures AC and DC voltage).

Basically here is the setup:

If you don't have a capacitor on the output, you really don't have a DC signal... simply a full wave rectified signal. You said you're measuring 105Vdc, so I assume your voltmeter is set to DC rather than AC. If so, you would measure approximately 105Vdc... but because there is no DC, you're essentially measuring the average voltage (not rms, but average).

The output will be approximately 168.6Vpeak (because 1.4V is dropped across the diode network, and 170-1.4 = 168.6). The average of this signal is 2*Vpk/pi, or 2*168.6/pi, or 107.3V. That's about what you got (I assumed 120Vrms service but of course you're probably a bit high or low).

You can technically call what you're measuring DC since the direction of current isn't changing (it's all positive). But it needs a capacitor if you're actually wanting a stable DC voltage output.

Thanks for the responses guys. So basically, without a load and therefore no current, the meter (which was set to DC) was actually measuring Vpk. Once the load was connected, the current through the device causes a voltage drop and thus drops to the average voltage level. Am I understanding it correctly?

Where are you putting your probes? Are you measuring on either side of the contacts, or are you measuring from the line side to some system neutral?

trchambe said:

Where are you putting your probes? Are you measuring on either side of the contacts, or are you measuring from the line side to some system neutral?

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The two contacts being used had the "+" and "-" DC voltage going through them. So the probes were put on each contact to measure the voltage where the load was to be connected (though disconnected at this point for measuring).

I wonder if the contacts are behaving as a small capacitor. When open, the voltage across it with charge up to the peak AC voltage... you would then measure a 170Vdc voltage. When the contacts are closed, the "capacitor" is gone, so you would measure 105Vdc as explained above.

Out of curiosity (if you have a capacitance meter), measure the capacitance between the open contacts when they're removed from the circuit. I wonder what you would read.

Unfortunately I don't have the test setup available. This discussion here was kind of a post results analysis. An interesting thought though about the capcitance theory. I wonder what that would have read as well.

What is your target output for your circuit? Try using a a regulator or a pot to stabilize your output voltage. This way you cannot have a problem when you put a load on your DC power supply because your output voltage is now regulated. It cannot have a voltage dropped when you do this.