I'm possessed of a 12-0-12 500mA transformer.

To supply a stable 10V, I seek to connect a bridge rectifier, roughly filtered with a 1000uF/50V electrolytic capacitor and a 7810.

  • Is a bridge, or even a FWR really required with the 7810?
  • Is the 7810 capable of handling the pulsed output of a HWR?

3 Answers 3


If you use a half-wave rectifier and your output current is limited to about 140mA you should almost be okay (that's limited by the transformer winding rating of 500mA).

Nominal peak output voltage is \$12V \times 1.414 - 0.8 = 16V\$

Allow 10% for line voltage drop (not a lot), and you have 14.4V.

Allow 2.4V for the regulator to work (conservative at 140mA), and you have 2V for ripple.

If your mains frequency is 100Hz, the ripple will be \$\dfrac{I_{out} \times 0.01 \,second}{C}\$ or 1.4V peak-to-peak.

That's perhaps okay, but I'd like to see more margin for brownout on line voltage. Using a full-wave rectifier (two diodes using the center tapped winding) would be better (still only a single diode drop and half the ripple).

Using a full-wave rectifier (four diodes and half the winding) would add another diode drop and only use half the secondary, so it's a lose-lose proposition.


In this situation: 12 V at 500 mA:

After HWR (half-wave rectification) the peak-to-peak voltage is \$12\cdot \sqrt{2} - 0.7 = 16.27\,\mathrm{V}\$. The 10V regulator needs a voltage greater than 12.5V to work, so the maximum ripple shouldn't exceed 3.77V.

This means that the capacitor value should be: \$C = \dfrac{I}{fV_p} = 2650\,\mu\mathrm{F}\$ minimum.

But with a center tap transformer you can use this kind of full wave rectifier (not bridge!):

enter image description here

Source: Wikipedia - Center tap

So the minimum value capacitor has the same value because the maximum available current and frequency are double now. But if you only need max. 500 mA a capacitor of minimum \${2650 / 2} = 1325\mu \mathrm{F}\$ will be enough.


The 78XX regulators do not specify the input waveform shape. For their regulation to work, the input has to stay within a certain range at all times.

In a nutshell, the input to a 78XX regulator should be regarded as "possibly ripply DC", where the ripple is such that the voltage stays within range.

This requirement actually rules out the use the raw output of a without any reservoir capacitor, because such output periodically drops to close to 0V. (So no, the rectifier cannot handle the pulsed output of a HWR, nor the more frequently pulsed output of a FWR!) The actual waveform that is usually input into a 78XX is substantially smoothed by the reservoir capacitor, such that it is DC, with a small amount of ripple.

If a half-wave rectifier is used, the reservoir capacitor simply has to be chosen large enough (with respect to the current draw from the regulator) so that the voltage never drops too low, keeping in mind that the capacitor only charges at half the duty cycle compared to a full wave rectifier. For the 78XX regulators, "too low" means: less than about two and a half volts above the output voltage. For instance, for a 7810, the input voltage should not drop below around 12.5V at any time.

There are some simple calculations to determine the size of capacitor needed. It is also not difficult to determine by simulation, or to determine empirically (trying different capacitor values, while observing the behavior of the recitified and smoothed voltage with an oscilloscope).


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