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LTSpice of IGBT not working

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Hi all I am trying to simulate an IGBT I have created in LTSpice I am not sure why it is not working can anyone provide assistance with this.

Kind Regards
Art
(Simulation attached)
 

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MikeMl

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You cannot leave the source, (emitter, whatever it is called in a IGBT) floating. By putting in a 10Ω resistor, now you can see what happens when Q1 acts as a source follower, it supplied current to pull-up V(out). When it turns off, you can see the feed through of the V1 down-edge, followed by the 10Ω resistor discharging the capacitance internal to Q1.

Why build a source-follower at all? I'm guessing that you thought V(out) would go to ~1000V (high-side-switch), which this connection is not.

335a.png
 

MikeMl

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Is this what you wanted to do?

335b.png
 

ronsimpson

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In post #3, 15 volts is across the Gate to Emitter. The transistor is on and the C-E voltage is small. This is how it should work.
(Some thing to remember, The transistor sees voltage with the Emitter being the reference. So all measurements should not be taken from ground but from Emitter.)
In post #2, The Gate toes to 15V. This turns on the transistor, which pulls up the Emitter to 9V, then turns off the transistor. (15V-9V=6V) There is not enough voltage on the gate to keep the transistor on.
If you really want to make #2 work: Remove "ground" from V1 and connect the bottom of V1 to Emitter. Now it will work.
 

MikeMl

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It is no different than this:

20.png
 

MikeMl

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...
If you really want to make #2 work: Remove "ground" from V1 and connect the bottom of V1 to Emitter. Now it will work.
But now V1 has to be a "floating" gate driver.
 

MikeMl

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Is this what you did?
 

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MikeMl

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So what do you not understand? Lookup "bootstrap".
sorry ignore me I understand now. But in real life does this still apply when I am designning would I have take the emitter back to a common or ground in reference to the gate? Or can I just apply 15V at the gate and it functionally work?
 

MikeMl

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sorry ignore me I understand now. But in real life does this still apply when I am designning would I have take the emitter back to a common or ground in reference to the gate? Or can I just apply 15V at the gate and it functionally work?
To accomplish this connection, V1 has to be totally isolated from ground. If fact, at times, it "lives" 1000V above ground. It needs "isolation" (optical, capacitive, transformer) if it is being commanded from a ground-referenced controller. Do you understand the concept of "differential" voltage between two nodes in a network? The plot shows the differential voltage between Gate and Source=V(out). The previous plot showed V(g) and V(out) with respect to GND.

335-1s.png
 
To accomplish this connection, V1 has to be totally isolated from ground. If fact, at times, it "lives" 1000V above ground. It needs "isolation" (optical, capacitive, transformer) if it is being commanded from a ground-referenced controller. Do you understand the concept of "differential" voltage between two nodes in a network? The plot shows the differential voltage between Gate and Source=V(out). The previous plot showed V(g) and V(out) with respect to GND.

View attachment 104626
ok brilliant thank you for that clarrification
 
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