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I'm making a project to school about MOSFET transistor. But I have a problem, which refers to turning on those transistors. The problem is, drain current starts rising only when Uds fell to zero. Instead of looking like that: enter image description here

It looks like that:

enter image description here

Second picture is waveform, which I made from oscilloscope data. It's a low side switch with resistive load. I need to designate switching times and calculate power losses, but i don't know how to define turning on time and what current should I take to my equation. I was searching about this problem, but always saw pictures similar to the first one.

I've uploaded the screen from the oscilloscope. mosfet switching waveform
- CH1 is voltage from generator.
- CH2 - Ugs.
- CH3 - Uds.
- CH4 - Id.

EDIT: Ok, I read a little bit and now I understand how to define switching times and calculate power losses, so you can close this topic. And I wanna thank you guys for help and show me a way to understand that :)

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    \$\begingroup\$ How is this a problem? There's no switching loss. That's great. \$\endgroup\$ – Harry Svensson Aug 5 '17 at 20:52
  • \$\begingroup\$ Is that even possible? To have 100% efficiency? \$\endgroup\$ – Junior20 Aug 5 '17 at 20:59
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    \$\begingroup\$ If your load has some inductance, when you turn the FET on and the current was zero before, current will start ramping from zero... \$\endgroup\$ – bobflux Aug 5 '17 at 21:04
  • \$\begingroup\$ It might be small inductance from paths and wires, but should it looks like that? So, how should I measure turning on time? From Ugs rising to Uds equals 0? \$\endgroup\$ – Junior20 Aug 5 '17 at 21:11
  • \$\begingroup\$ Show your measurement setup. What's the time scale in your diagram? It's ineligible. \$\endgroup\$ – winny Aug 5 '17 at 21:19
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You say "it's a low side switch with resistive load".

It isn't.

Notice the HUGE kick in Vds when you turn the MOSFET off? That tells me it's an inductive load, or has a substantial inductive component.

The next step is to zoom in on both switch-on and switch-off events, and measure V and dI/dT for both events, and see if they give you reasonble (and similar, ideally identical) values of L, from V = -L*dI/dt.

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  • \$\begingroup\$ Yep, now I see that at this 5kHz, which can be seen on oscillogram, inductance is 100mH and it drops with frequency growing. But I'm beginner in electronics and don't understand connection between inductance and MOSFET switching. How should I define switching time and power loss in this case? \$\endgroup\$ – Junior20 Aug 5 '17 at 22:56
  • \$\begingroup\$ After all, why drain current starts to rise, when Uds voltage equals 0? Even with inductive load it should be at the same time as Uds voltage goes down... \$\endgroup\$ – Junior20 Aug 5 '17 at 23:11
  • \$\begingroup\$ Because V(supply) = Vds + V(load). So when Vds=0, V is across the inductor, so dI/dt = V(supply) / L, and V(supply) is relatively low. \$\endgroup\$ – user_1818839 Aug 5 '17 at 23:25

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