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I am an Electrical Engineering student. I'm in my final year and doing a thesis about soft switching.

I am trying to grasp the calculation of diode switching losses. I'm having the impression that only the turn-off losses are calculated (which are related to the reverse recovery charge of the diode). In a book I have, Power Electronics from Mohan, Undeland and Robbins they show the voltage waveforms of power diodes (page 535 if you are interested, I don't succeed in uploading a picture). You can clearly see that at turn-on, an overshoot of the voltage occurs. During this phase, the current already builds up. How is this turn-on loss calculated? I can't really find it in diode datasheets where they only mention the parameters to calculate the reverse recovery losses.

So could anyone explain me how the turn-on losses can be calculated from the diode datasheet?

If it is negligible, could anyone explain me why?

enter image description here

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    \$\begingroup\$ Show a schematic. It's not clear what you mean by shoot thru during turn on. That makes no sense, at least not without more context. Diodes are usually pretty quick to turn on, but they look open, not shorted, during that time. \$\endgroup\$ – Olin Lathrop Sep 30 '15 at 14:29
  • \$\begingroup\$ What are switching losses ? Switching losses occur because the switches are not ideal. If the switches were ideal you'd have no switching losses right ? A diode is also a non-ideal switch. What is the difference between an ideal diode and a practical diode. When off: not much. When on or at a transition between on and off: current flows through the diode and a voltage drop occurs. Power = voltage x current so there's your loss. If you integrate these power losses over time, you would know your diode switching losses. \$\endgroup\$ – Bimpelrekkie Sep 30 '15 at 14:30
  • \$\begingroup\$ In general you cannot say of something is negligible or not, what might be negligible in your application might be limiting performance in my application ! \$\endgroup\$ – Bimpelrekkie Sep 30 '15 at 14:32
  • \$\begingroup\$ I think you're talking about overshoot on turn-on, not shoot-through. It's indeed a phenomenon not observed in signal-level diodes (according to the book). You could upload a page to imgur.com and post a link here. \$\endgroup\$ – Fizz Sep 30 '15 at 14:52
  • \$\begingroup\$ Yes indeed, I'm sorry, I made a mistake, I meant overshoot! I'll change it in the description. \$\endgroup\$ – Simon R Sep 30 '15 at 15:05
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You will struggle to find a datasheet that explicitly states the turnon information. TurnOn switching losses of a diode are typically quite small & unlike the TurnOff characteristics, ONLY affect the diode

The reason for this is the turn-off characteristics have bigger ramifications.

The key information with regards to turn-off are

  1. Reverse recovery time \$t_{rr}\$
  2. Reverse recovery charge \$Q_{rrm} \$
  3. Peak reverse recovery current \$I_{rr}\$
  4. Peak rate of fall of recovery current \$dI_{rr}/dt\$

The turn-off of the DIODE characteristic have an immediate knock on effect on the associated commutating device (eg IGBT). The reverse recovery current is additive to the turn-on switching current of the associated device and thus that devices turn-on (the IGBT). Switching 600A through an inductive load can easily have 50A of additional current

This however is turn-off of the DIODE & this information is provided in the datasheet & equally the question was associated with turn-on

Turn-OFF characteristics (for minority carriers) is dominated by die,package, installation: resistance AND inductance. As the current is rising at the \$dI_{rr}/dt\$ rate, an additional voltage is generated on-die. In practice however due to application specific installation stray inductance the \$dI_{rr}/dt\$ will be lower.

To ESTIMATE the turn-OFF switching losses of a power Diode requires three things

  1. Reverse Recovery Charge
  2. blocking Voltage
  3. switching frequency.

Depending on the size of the diode, you may or may not be provided with \$Q_R\$ against \$dI_f/dt\$ graphs.

If you are provided these graphs, the 1st step is to determine the \$Q_R\$ for your operating case: external inductance limiting the \$dI_f/dt\$ and your operating current.

The switching power can then be estimated by

\$P_s = Q_R V_r f_s\$

if you are interested in W.s/pulse (\$J_R\$) This can be estimated via: \$J_R = V_R Q_R\$

As mentioned: TurnON losses are not only tricky (due to the complex nature of interacting strays) but equally is a small contributor

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  • \$\begingroup\$ Thanks a lot, I'm already getting a better understanding. I know it's not in the question, but could you provide me a formula on how to calculate the turn-OFF losses? On IEEE explore I found already some papers that deal with this by adding modules to their simulations but a simple formula is mostly not provided. \$\endgroup\$ – Simon R Oct 1 '15 at 7:55
  • \$\begingroup\$ how is that. Sorry I can't provide analytical TurnON... it is dependent on so many things. Usual approach that I take is I do in-inverter switching losses on the IGBT & Diode as this takes into account the stiffness of the DClink and any stray inductance \$\endgroup\$ – JonRB Oct 1 '15 at 15:23

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