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I have created a test schema since I received a late response that I might be short circuiting my own push-pull setup in this post Offsetting Op-Amp output offset in combination with charge pump. It puzzled me whether that could be the reason for some longer transition times in that circuit, but moreover: I just could not answer the question whether there is a short circuit or not. So this question is more a 'back to basics'.

In my test I have a 0 to 6V square wave, driving a transistor to +6V and one to -6V in what I believe to be a textbook example that an answerer to my previous post learned me to understand. But in all honesty: I have just ordered a textbook to validate this.

enter image description here https://www.circuitlab.com/circuit/wmvfe5969ymu/mosfet-pushpull-test/

When I simulate this I indeed get something that I fear is a shortcut by both M6 and M2 switched on at the same time during every transition. Is this the case indeed? If so, why is it not the same with both transitions given that the circuit is quite symmetrical? Also, is this really 1A in reality if the power supplies can source this (it was 4A even in the simulation of my original circuit)? Am I just doing it plain wrong? Thanks!

Below the simulation result: enter image description here

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  • \$\begingroup\$ why is it not the same with both transitions given that the circuit is quite symmetrical? 1) I do not consider this circuit to be "symmetrical" 2) realize that what really matters is how fast/slow you switch of/off the gates of M2 and M6. I mean NMOS M2's gate is charged by M4, (that can be fast) but is discharged by R3 (probably much slower). M6 is switched on slow by R4 but switched off fast by M5. Note that I'm assuming that a MOSFET will have a lower on resistance than that 470 ohms you're using. I can see in your plot that this is the case (how do I see that?). \$\endgroup\$ – Bimpelrekkie Apr 8 at 11:48
  • \$\begingroup\$ by "shortcut" I think you mean a short circuit, and if the second graph is current, those spikes suggest you have one. The trick is to turn one device OFF before turning the other ON; modify the driver circuit to do that. \$\endgroup\$ – Brian Drummond Apr 8 at 12:13
  • \$\begingroup\$ Short circuit indeed. Updated it. Second chart is current indeed. \$\endgroup\$ – JeromeBu1982 Apr 8 at 13:33
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Yes, you got it!

During a certain period of time M6 and M2 conduct current simultaneously.

M2 and M6 are in their saturated region.

The current that flows is called "shoot-trough current".


It's the main mechanism of CMOS and power electronics power consumption.


One way to minimize the shoot-through parasitic effect is to slowdown the turn-on time of the high-side MOSFET M6.

This will reduce or eliminate the shoot-through current, but at the expense of higher switching losses.

Source:

https://www.ti.com/lit/an/snva590/snva590.pdf?ts=1617884587865&ref_url=https%253A%252F%252Fwww.google.com%252F

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  • \$\begingroup\$ Wow, read the source with joy. Thanks. So it is a commonly overlooked problem. I can have 100 ohm output impedance. What if I let the shoot-through for what it is, but place two 100 ohm resistors between M2 and M6 and take the output from the middle. That should reduce current to 24V/200Ohm max. Would it? \$\endgroup\$ – JeromeBu1982 Apr 8 at 13:49
  • \$\begingroup\$ If you add 2 resistors, you will limit the shoot-through current but the 2 resistors will dissipate power and the voltage edges will be smoother at the output. I did it once in the past. In order to have equal rising and falling edges I had to put 2 different resistors: 100 Ohm and 330 Ohm. \$\endgroup\$ – Enrico Migliore Apr 8 at 13:53

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