I am trying to get 5v DC by rectifying the output of a microwave transformer i rewound (the transformer gets approx 4.2V AC on the secondary winding). enter image description here I managed to get approx 5V open circuit voltage on the DC side of the rectifier but as soon as i connect a small load (tiny 6v test light which can be seen on the photos) the voltage drops from 5v to approx 4.3 volt; the AC voltage on the secondary winding of the transformer barely changes when connecting/disconnecting the light bulb). Also the voltage drop over each diode is approx 1.5v (which confuses me since i thought it should always be 0.7volt). The measurements can be seen on the photos. I have also tested with a different load (18650 charge controller ic) which gives approximately the same results. I have tried using 10A10M10 diodes bit that also gave the same results.

I am using IN4004 DC Diodes. I am using a 25v 2200uF capacitor for smoothing out the dc output (I also tried with a 4700uF cap). The capacitor is in parallel with the load as can be seen in the schematic. enter image description here Is this sort of voltage drop normal? Is there anything i can do to prevent it? Have i made a mistake somewhere in terms of circuitry or component choice? enter image description here enter image description here enter image description here enter image description here enter image description here Any input is greatly appreciated. Thank you very much!

  • \$\begingroup\$ Some drop is normal; when you meter the open-circuit output, the current flow through the windings or any other parasitic element is nearly nil, so barely any voltage drop occurs across it, but when more current flows into your intended load, this drop increases. That said, a ~0.7V diode drop is characteristic of the diodes' pn junctions, not some manufacturer choice. 1.5V doesn't make much sense. Are you sure your diodes are all working correctly and placed correctly? \$\endgroup\$
    – Bort
    Oct 30, 2019 at 17:19
  • \$\begingroup\$ You can't measure the voltage drop that way - much of the time the diode will be reverse biased so you will just get the average of the forward drop and reverse voltages. I expect if you checked the polarity you would see see that the drop is negative! \$\endgroup\$ Oct 30, 2019 at 17:26
  • 1
    \$\begingroup\$ Just a side-bar. Your secondary winding is very loosely wound. You probably want to tighten it up a lot more. Poor regulation is a common result of loosely wound secondaries. Find some appropriate "magnet wire." \$\endgroup\$
    – jonk
    Oct 30, 2019 at 18:49
  • \$\begingroup\$ You have the capacitor connected with incorrect polarity. Fix it before it blows up \$\endgroup\$
    – Justme
    Oct 30, 2019 at 19:14

3 Answers 3


I don't see anything unusual here.

You are trying to measure the dc voltage across a diode while giving the bridge an ac input. What you are actually measuring is some kind of average voltage across the diode, alternating between the forward and reverse bias cases.

If your transformer provides 5V rms then you would have a peak voltage of about 7V with no load. After the bridge rectifier a peak ac voltage with no load of about 5.5V would be reasonable. However, as soon as you start to draw current the average voltage, the dc voltage, will drop. That is simply the nature of the beast. According to www.electronics-tutorials.ws the equivalent dc voltage from a full-wave rectified ac voltage is just 0.9 times the rms value.


That is pretty much normal. The open-circuit voltage drop across the diodes will be very low, maybe 500mV but more like 700mV when drawing a more current, so that's 400mV drop for the two series diodes from no-load to a moderate load.

Your transformer voltage will drop a bit as you've seen (hopefully you've removed the magnetic shunts to decrease the leakage inductance built into a typical MOT).

And the ripple voltage across the capacitor will increase, resulting in a lower average voltage for the same AC input.

If you want a stable 5V output, you'd normally want to produce a higher voltage and regulate it down, either with a linear regulator or a switching buck regulator. For something like a 7805 you'd want around 10VDC to start with, so throw away half the power. The regulator needs 3V to work, and the other 2V is to allow for ripple and low mains voltage.

For a switching regulator you are free to use a higher voltage without incurring the same cost in efficiency, so you could produce (say) 12V nominal and regulate it down.

  • \$\begingroup\$ diode forward voltage will be close to zero (perhaps 200mV) with very low currents (like 1uA to nanoamps). This current charges filter capacitor to 5V. \$\endgroup\$
    – jalaffo
    Oct 30, 2019 at 20:05
  • \$\begingroup\$ @jalaffo The electrolytic filter cap will have some leakage, the meter will draw some current. At a current in the 10-20uA range, the voltage drop of a 1N400x will be in the 450-500mV range. \$\endgroup\$ Oct 30, 2019 at 20:09

You are not measuring diode forward voltage properly - When measuring ripply AC over diode with DC setting on multimeter, on average, you have more positive on diode cathode than diode anode. What is meant with diode forward voltage is that diode anode has more positive than cathode, when it conducts.

Capacitor being connected with wrong polarity might affect the readings as well.

Voltage dropping under load is nothing to be concerned of, transformer windings have losses. Usually, when you buy a transformer, the rated voltage is what you get at rated load current, and the voltage without any load is a bit higher than the rated voltage.


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