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I want to measure the current of my FET while it blows up, but the Rds(on) of th...
markwuflux

markwuflux

@copilot I want to measure the current of my FET while it blows up, but the Rds(on) of the FET is like 30mOhms. Would I be able to find 1mOhm shunts and measure it with a scope?
2y

copilot

Yes, it is possible to measure the current through the MOSFET using a 1mOhm shunt resistor, but you need to consider several factors:
  1. Selection: You'll need to find a power shunt resistor that can handle the amount of current you're planning to measure. Shunt resistors are usually rated for a particular maximum current. Make sure your selected resistor's rating matches or exceeds the current you're expecting.
  2. Accuracy: Shunt resistors have a tolerance, which can affect the accuracy of current measurement. A higher tolerance results in less accurate measurements.
  3. Amplification: Since the voltage drop across a 1mOhm resistor will be very small, you may need a differential amplifier to amplify this voltage before measuring it with an oscilloscope.
  4. Placing: The shunt resistor should ideally be installed very close to the MOSFET you're measuring to minimize the voltage drop in the interconnecting wires.
Please make sure you're taking the necessary safety precautions while performing these tests, as a "blowing up" MOSFET can produce a lot of heat and potentially harmful energy discharges.
Let me know if you need help with modifying your flux.ai project to include a shunt resistor and differential amplifier.
2y
4
https://storage.googleapis.com/flux-comment-files/files/commentFileId-c737236169...
markwuflux

markwuflux

There are two routes: you can ensure that the transistor disspates more power than the capacitor by keeping Rds(on) higher.
Or you can keep Rds(on) low, drastically increasing the current in the circuit and thus how fast the transistor will blow up.
Seems like low ESR capacitors might be better for the job. It's hard to calculate without knowing how much energy it takes to blow up the FET.
2y
markwuflux

markwuflux

P=I^2*R and I is the same for both the ESR of the supercapacitors and that of the FETs. This means that R determines which device dissipates more power. However, the other half of the equation is how much power each device can handle before failure. Using a small FET whose package cannot handle much heat will ensure that it fails before the large capacitors.
2y
markwuflux

markwuflux

3 Supercaps in series. ##Transistor:##
Vgs=2V, Ids = 20A, Vds = 8V, Pdiss= 160W.
##SuperCap:##
ESR=15mR I=20A Pdiss= 6W each.
2y
markwuflux

markwuflux

Image

Image

https://storage.googleapis.com/flux-comment-files/files/commentFileId-c73723616981c8d47e53df23690a637854d6a6217ffead324c040108ba538b6a.png?GoogleAccessId=graviton-mvp%40appspot.gserviceaccount.com&Expires=4860347994&Signature=yXvQmUPZRKhx4G%2FQ9EUMGX1gYSTBtGMdbpmh7WopoZeKR00BEUyEPhdlarhcmJlP0fzlHRsYpSSjC4zUXIBTFuTSGr249fw6VFVUPKqmcrq87OF4hckt9bbbgm0nUjAHRl2E7YVcUFreP9WHXFEjM2gMPW2f0WMDeqqG83t6KzJLZG0xBwK%2FaUuUUZVWZhEhx33x2Ps3xfj5s3CEPjHf83hxzLF7yWoSucoaYXiGuAc7F8%2BGiqD3ZNerU%2FOXuI7%2FBVS4Y5olJZgxOnLbij%2Fq2MEL6nqyLzPIvvi0YxnIsm4T%2BDMQjXMYHv7q0JmbaTjM6hzz6LYNNRBVYHbZ8kmfiw%3D%3D
2y
3
Q1
Q2
GND
Gate
R1
Resistance
500uΩ
C3
ESR
15mΩ
Capacitance
50 F
C1
ESR
15mΩ
Capacitance
50 F
Gate
Gate
C2
ESR
15mΩ
Capacitance
50 F
+8V
TP2
Gate
Q3
TP1

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FET Explosion Board

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I want to blow up a MOSFET gloriously.
Basically supercapacitor onto MOSFET with a gate flipping setting.

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