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In either dc power converters, there must be a diode. I don't understand the need for it? Would a buck/boost converter function without a diode? Also, what is the maximum reverse voltage for a diode in both cases? Does it have to be higher than the output voltage?

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  • \$\begingroup\$ As regards the reverse voltage - yes it needs to be much higher than the output voltage. The spikes generated by the inductor can be several hundred volts. \$\endgroup\$ – F. Bloggs Dec 1 '15 at 17:34
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    \$\begingroup\$ @F.Bloggs, a properly designed regulator will not have inductive spikes to that extent. It is a good idea to have a little margin in the diode reverse voltage rating, but you needn't worry about giant voltage spikes of hundreds of volts. I am not sure where you got that idea. \$\endgroup\$ – mkeith Dec 1 '15 at 17:56
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If you look at literature, you will find that switching converters operate by charging and discharging an inductor. The main building block is composed by an inductor and two switches, connected in such a way that one switch charges the inductor and the other discharges it.

Now, you have to give a signal to those switches, so one of them must be a controllable switch (aka transistor; usually MOSFETs are used).

The other switch is usually substituted by a diode, because its usual behavior (allowing current to flow in just one direction) is enough to guarantee the proper functioning of the device.

Anyway of course you can remove the diode and substitute it with another MOSFET. Your control logic then becomes more complicated (it needs to control two switches), bit the performances increase (the voltage drop on the diode, usualli 0.3V, is now reduced to almost 0). These converters are called synchronous converters (see here for more infos).

The maximum reverse voltage depends on the device. For buck converters the reverse voltage on the diodes is roughly Vin, while on boost ones it is Vout.

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  • \$\begingroup\$ Actually the control is simpler for a synchronous converter because it can continue to operate in continuous mode down to zero (and even negative) load while a diode based converter will go into discontinuous mode under light load. \$\endgroup\$ – Peter Green Dec 1 '15 at 18:05
  • \$\begingroup\$ Yes but.. your control logic can't be a simple comparator, because it has to control two switches, avoiding overlaps and more... So the control algorithm can be simpler, but the control circuit is more complicated \$\endgroup\$ – frarugi87 Dec 2 '15 at 9:06

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