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I built a simple half wave rectifier using a center tap transformer and a 1N4007 diode.

I connected the 1N4007 diode to one of the live secondary wires of the transformer and measured the DC voltage across the diode and the neutral wire. It is just showing around 6 volts whereas my output AC voltage from the transformer is around 14 volts.

Also, the voltage drop across the diode is around 5 volts and not 0.7 volts.

I am a beginner in electronics and I can't figure out what's wrong.

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    \$\begingroup\$ Draw a diagram of what you have connected. \$\endgroup\$
    – JRE
    Nov 7, 2020 at 17:26
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    \$\begingroup\$ Add schematic, get more attention, get better answers. \$\endgroup\$
    – Mitu Raj
    Nov 7, 2020 at 17:27
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    \$\begingroup\$ Think about what the voltage waveform looks like before and after the diode, and what effect that could have on your multimeter when trying to measure voltage across the diode. Also, search for "smoothing capacitor". \$\endgroup\$
    – Unimportant
    Nov 7, 2020 at 17:30

1 Answer 1

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You haven't said that there's any reservoir capacitor in your circuit.

So your voltmeter is doing its best to display a DC voltage, when it's actually getting half of a sine wave. The other half is blocked by the diode, so the voltmeter is getting 0V on the negative half cycle.

So if we take a secondary voltage of 14V, knock off about 0.7V, and then halve it because half the time the voltmeter is receiving 0V, then the result will be something like 6 to 7V.

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  • \$\begingroup\$ Ok i got it but why is the voltage drop across my diode also 6v? Shouldn't it be something like 0.7? \$\endgroup\$
    – shahrOZe
    Nov 7, 2020 at 17:52
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    \$\begingroup\$ @shahrozeshahab It should be, on average, +0.7V when the diode is forward biased but -14V when reverse biased. So avound -5 to -6V over the whole cycle. \$\endgroup\$
    – Simon B
    Nov 7, 2020 at 17:54
  • \$\begingroup\$ Oh i got it all. U deserve my upvote. \$\endgroup\$
    – shahrOZe
    Nov 7, 2020 at 17:55

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