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I'm currently working on a buck converter topology with a low side switched inductor. The goal is to achieve a constant current in the inductor, therefor only the mean voltage is important and output voltage filtering isn't necessary. The following picture shows a simplified circuit diagram.

Q1 resembles the used switch, L1 the load inductor (which additionally has some serial resistance) and V1 the gate driving voltage source. The diode V1 is a standard silicon diode.

For the following, the starting point is as follows: It is assumed that a current is already flowing in the free wheeling circuitry through D1 and L1 because of previous switching events of the MOSFET. This current shall be named I1. The MOSFET is currently not conducting.

My understanding of this circuit is as follows:

When the MOSFET is switched on, it has to carry the current I1 and also the reverse recovery current of D1 before the voltage across drain and source of the MOSFET can lower. Then, D1 is reverse biased and the input voltage drops across the load L1 and D1.

My question is as follows:

The datasheet of silicon diodes specifies the reverse recovery time of a diode. When the MOSFET is switched fast, which results in a high di/dt in this case, the peak current (I1 + reverse recovery current of D1) could be reached fast and the voltage drop across the drain-source path can be lowered fast.

If the reverse recovery time of D1 is too long for the application, meaning that it can't block the reverse voltage fast enough, can there be some kind of shoot-through event where the diode still conducts while the voltage over the mosfet decreases resulting in current spikes?

Or did I just not fully understand the circuits behaviour?

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    \$\begingroup\$ I can't think of an example of a buck (or boost) regulator where there is a diode across the inductor like that. Please enlighten me. "Reverse recovery" means reverse current when reverse biased for a duration of time equal to the reverse recovery time. You could equate this to shoot-through. \$\endgroup\$
    – Andy aka
    Commented Jun 11, 2022 at 14:42
  • \$\begingroup\$ @Andyaka The inductor resembles an electromagnet, thats why only the inductor current is relevant for controlling the electromagnetic forces. The mean voltage is thereby proportional to the current in the inductor.I actually meant literal shoot-through with high short circuit currents, which to my understanding doesn't happen in the sence of destructive shoot-through, but only in the sense of short current spikes. \$\endgroup\$
    – konrad1795
    Commented Jun 11, 2022 at 18:20

2 Answers 2

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Your general understanding of the diode behaviour is correct.

When the FET turns on, the voltage at its drain doesn't begin to fall until the FET's current exceeds the inductor + diode reverse recovery current. Generally the reverse diode current immediately after switching could be more than 2x the original forward current (which was the inductor current).

When switching fast, it is easier to consider the diode's reverse recovery charge. This charge will cause a power loss (Qreverse*Vcc) on every switching cycle.

Depending on your current, voltage and efficiency needs, a Schottky diode might be more suitable than a standard rectifier.

Note your circuit doesn't have an output -- unless your goal is just to have a (triangular waveform) constant current in the inductor.

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  • \$\begingroup\$ I guess it could be a variable strength electromagnet.... \$\endgroup\$
    – John D
    Commented Jun 11, 2022 at 17:25
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    \$\begingroup\$ @JohnD Yes, it indeed is a variable strength electromagnet. \$\endgroup\$
    – konrad1795
    Commented Jun 11, 2022 at 18:22
  • \$\begingroup\$ Looking at the diodes charge seems conclusive. But one thing is still a little unclear to me. If the voltage across the mosfet is already at roughly zero volts, although the diode is still well within its reverse recovery time, will the current then be many multiples of the inductor current? In other words, can the current still rise after the voltage across the mosfet starts lowering due to a high du/dt? If I apply your description with the reverse recovery charge, this would fit. Therefore, faster diodes (e.g. schottky diodes) would solve this problem. \$\endgroup\$
    – konrad1795
    Commented Jun 11, 2022 at 19:02
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    \$\begingroup\$ @konrad1795 When inductor current is continuous, turning on the MOSFET shunts current away from the diode; this incurs reverse recovery loss. MOSFET voltage won't fall until the diode current starts dropping. Well, it can actually drop some, limited by loop inductance -- the return path from VCC to GND, hopefully through a nearby capacitor. You can consider the diode a short circuit during the instant of turn-on, until t_rr; then the MOS switches into loop inductance alone, thus it defines dI/dt. \$\endgroup\$ Commented Jun 11, 2022 at 19:49
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    \$\begingroup\$ If you use a Schottky, reverse recovery won't be a functional problem unless you are switching at few ns slew rates (which could require 100's of mA gate drive for the MOSFET). However the reverse charge (and consequential current) will require a good decoupling capacitor (0.1 uF) on the power supply -- as close as possible (<< 10 mm) between the cathode of the diode and the S of the MOSFET. Depending on your supply, additional 100's of uF may also be needed. \$\endgroup\$
    – jp314
    Commented Jun 11, 2022 at 20:00
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Instead of the diode and a mosfet you could use one half of an H-bridge. This is called synchronous rectification. You’d be basically driving the inductor with a PWM voltage source. There won’t be any diode-related losses, only mosfet conduction and switching losses. There will need to be possibly a snubber on the inductor to dissipate the dead time glitches as one bridge leg stops conducting and the other one is about to turn on.

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