# Bridge rectifier & minimum diode voltage

The varying input voltage to a bridge rectifier spends a certain amount of time between +0.7V and 0V, as well as between -0.7V and 0V, during which there is not enough voltage across any diode to allow current to pass. Yet every representation of the output voltage shows that voltage hitting zero and then immediately rising again (rather than staying at zero until |V| > 0.7V). Is this true, and if so - why??

Thanks!

• This is probably just a matter of the scale of the drawng, and the very short time the diodes are not conducting. Commented Dec 25, 2021 at 17:05
• Makes sense, Peter! Funny how absolutely nobody shows the real case. Commented Dec 25, 2021 at 17:14
• What makes you think that "absolutely nobody shows the real case"? I believe that this is well understood among practicing electrical engineers. We use different models for diode behavior depending on the situation. For the diagram you provide, the diode model assumes an ideal model which is perfectly conducting if the forward voltage is positive. Commented Dec 25, 2021 at 17:59
• Keep in mind that the "0.7V" value is itself a crude model of diode behavior. The actual voltage will vary with the forward current, and could easily be over 1V. Commented Dec 25, 2021 at 18:09
• @rasraster What you are you really looking for, here? An s-space analysis? A time domain non-linear mathematical solution? A wide scope that covers the important regions of behavior, in detail? You have already learned that diodes are not "on" or "off" but instead non-linear devices whose current is a (exponential) function of the voltage across them. Where do you want an answer to head? It's not clear to me. Thanks.
– jonk
Commented Dec 25, 2021 at 19:54

For the rectifier to conduct it must see a voltage of at least two forward diode drops across its AC terminals.

Imagine that you have an AC source connected to your rectifier, that you can vary from 0V up to 100V p-p. As you sweep the AC range, you’d get the following:

• 0 ~ 1.4V in: nothing out
• 1.4 ~ 100V in: 0 ~ 98.6V rectified out

When there is an output at all, the unfiltered rectifier output will have a ‘dead band’ at the AC zero cross during the time that the AC voltage difference is less than two forward diode drops (that is, ~1.4V). This shows up at the bottom of the ‘trough’ between each half-wave.

Also, the peak half-wave voltage will be two diode drops less than the AC voltage peak.

The diagrams like this that you see in books and on the internet are generally idealized to make it easy for students to get the basic concept of a rectifier.

• Thanks, @GodJihyo Commented Dec 25, 2021 at 17:13

The representations are simplified. In reality there is a gap (0 V out) when the input is between -0.7 and +0.7 V.

The peaks of the output are also 1.4 V lower than the input peaks.

The R & C filter picture is approximately correct. There are some other subtleties that arise from the (leakage) inductance of a transformer (which is commonly used as the power source).

• Thank you, @jp314 Commented Dec 25, 2021 at 17:13