Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

LTSpice of IGBT not working

Status
Not open for further replies.
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)
 

Attachments

  • Draft335.asc
    717 bytes · Views: 145
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
 
Is this what you wanted to do?

335b.png
 
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.
 
It is no different than this:

20.png
 
...
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.
 
Is this what you did?
 

Attachments

  • 335-1.png
    335-1.png
    27.2 KB · Views: 156
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?
 
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
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top