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I am trying to make a full wave diode bridge rectifier, as the title suggests. I have a 120-12vac transformer that I'm using to supply the circuit. However, whenever I measure the voltage across the output of the bridge, I am still getting an AC signal. Whats even weirder is that I now get readings across the terminal on both DC and AC settings of my multimeter, and the VDC measurement is ~20V, which seems a bit odd... Anyways, I have no idea what is happening, so any suggestions would be fantastic.

schematic

simulate this circuit – Schematic created using CircuitLab

Board Top View

Board Bottom View

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    \$\begingroup\$ How much AC voltage are you seeing, and are you using an average-responding or true-RMS meter? \$\endgroup\$ Commented Dec 6, 2016 at 5:31
  • \$\begingroup\$ Try placing a resistor between the DC positive and negative and try again. \$\endgroup\$
    – Bradman175
    Commented Dec 6, 2016 at 5:33
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    \$\begingroup\$ I'm pretty sure that those aren't 1N4001 diodes in the photo. What is their current rating? \$\endgroup\$ Commented Dec 6, 2016 at 10:16
  • \$\begingroup\$ If you were hoping to make a cheap 12v power supply, add a LM7812. \$\endgroup\$ Commented Dec 6, 2016 at 12:09

2 Answers 2

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it's not surprising you are seeing close to 20Vdc at the output. The nominal voltage of your transformer secondary is 12Vac (which could be even higher ~14Vac). Now, the ac voltages you have are RMS. So the capacitor is going to charge up to the peak value of the sine wave, which is 12Vac*sqrt(2) = ~17V for a secondary output of 12Vac. If this were more like 14Vac at the secondary, it would make sense why you are seeing 20Vdc at the output.

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I suggest you calculate the surge current on power up and see if the Cap ESR causes the diodes to blow. If Cap ESR = 0.1 Ohm then diodes will see initially a short circuit with peak voltage turn on with ESR*C =T fo depending DF or ESR of cap. .

Use the biggest bridge or didoes you have since peak current will be 10x average for 10% Vripple.

Bridge Caps rectify peak voltage which is 1.414 * rms and with no conduction load loss of 10% typ, your noload Vout =1.5 Vrated rms with 40% ripple .

  • but you have a large cap of 1mF which may have an ESR of 0.1 to 1 Ohm similar to the tiny diodes, which you may get from specs for rising V/I when saturated.

** The initial cap energy must be transfer on surge pulses and conducted with series losses. If initial condition is Vcap= 0 and transformer output is say 18Vp into diode bridge , what is your surge ? Vp=(18-2Vf(diode))/ESR (diodes+cap+xfmr)** This may in the 50A range for a 4A supply. 1/2CV^2 energy transferred =18V^2*1mF/2=162mW.

  • compute your diode losses and cap losses or choose bigger diodes, lower Cap ESR and use a soft start ICL NTC inrush current limiter.

Then ensure your 0V meter reference is always on V-.

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  • \$\begingroup\$ Could you please explain better the comment about ESR? \$\endgroup\$
    – Antonio
    Commented Dec 6, 2016 at 9:02
  • \$\begingroup\$ it gets repetitive. electronics.stackexchange.com/search?q=user%3A17574+esr+ \$\endgroup\$
    – D.A.S.
    Commented Dec 6, 2016 at 13:32
  • \$\begingroup\$ You say that if ESR is too low it could be the cause to diode to blow? I never thinked about this aspect. Thanks! \$\endgroup\$
    – Antonio
    Commented Dec 6, 2016 at 15:31
  • \$\begingroup\$ soft start circuits are common , or lossy with choke or power R series. \$\endgroup\$
    – D.A.S.
    Commented Dec 6, 2016 at 15:37

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