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Electronics => Projects, Designs, and Technical Stuff => Topic started by: harrimansat on April 19, 2024, 06:49:53 pm
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Trying to measure H parameters with a curve tracer, I have seen that the Vbe drops when transistor is active region, BC547. I dont find any explantion looking the Ebers-Moll transistor model,equations and theory, and Gummel-Poon.
Anyone can explain why it drops?
First photo is Ic vs Vce, second Ic vs Vbe, and third Ic vs Vbe x10
I have tried the same with other transistors and others curve tracers with the same result.
Thanks
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Heating perhaps? 10mV decrease can be explained by 5°C junction temperature rise.
What's the Vce in those tests?
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Hi, 10v is the first photo. Is easy to see thermal effects at curve tracer, but they are slowly and affect at the whole curve, not only the edge
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I think the Ic vs Vbe traces are the same as the Ic vs Vce except that Vbe replaces Vce. If that's the case then you have constant base input current and stepped Vce voltage for a number of different base currents. If you notice the first trace you will see that for higher currents the Vce voltage cuts short earlier, that is because BJT breakdown voltage is current dependent, and when the tester detects this problem it ceases that test. For the Vbe vs Ic you have the same condition except it is a bit more pronounced. There is special overcurrent protection built into the tester it might be a side effect of that.
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The voltage Vbe definitely dropped a few millivolts in the active zone, I tested it today with two multimeters, measuring the base current to verify that it is constant, and another measuring Vbe
The only explanation that I find is hre
Unigor multimeter is for check Ib (constant)
https://www.youtube.com/watch?v=m0UW1Jzp89g (https://www.youtube.com/watch?v=m0UW1Jzp89g)
Someone making the same answer 6 years ago
https://electronics.stackexchange.com/questions/331319/can-reverse-voltage-gain-for-a-bjt-be-negative/710384#710384 (https://electronics.stackexchange.com/questions/331319/can-reverse-voltage-gain-for-a-bjt-be-negative/710384#710384)
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The transistor is breaking down, the conduction is transistioning to avalanche breakdown so the current is not increasing because of base injection but because of avalanche current which standard SPICE models don't do real well e.g. in LTSpice I put 500V across an 2N3904 and it only conducts 500pa. Don't expect classic BJT behaviour during breakdown.
Please note the SOA picture, its scales are log(Ic ) vs log(Vce) and you have a linear decrease in Vce for Ic just like the first picture you posted.
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Hi, thanks for your answer, but transistor is at 10v Vce, there isn't possible breakdown at that voltage
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Hi, thanks for your answer, but transistor is at 10v Vce, there isn't possible breakdown at that voltage
And MJE15031, a 150V transistor, cannot breakdown at 15V, but wait the SOA curve says they can. :-//
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This whole thing makes no sense.
What causes Ic and Vbe to change if you say that Ib and Vce are held constant?
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Vce is sweeping, not constant
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Above Vbe, below, Vce, ib=6uA , constant
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I guess it's the Early effect, then.
It's well known that Vbe and Ib required for any given Ic decrease with increasing Vce.
Apparently, the Vbe for any given Ib decreases too.
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Does it increase or decrease?
In that graphic I see that increases, in my experimental measurements, and in the curve tracer it decreases
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Yes, it seems a thermal effect, tested today with a low speed ramp, with a constant 20uA base to emitter. Graph is Vbe, and Vce(60 secons slow ramp)
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I think the mystery is solved, for testing power semiconductors without heatsink the 576 has a pulsed base supply function, at 300us and 80us. It hadn't occurred to use it for low power semiconductors. Without dissipation the return effect practically disappears. I leave you a video, I also heat the transistor with my fingers to see the effect.
Oscilloscope is connected to Vbe to see pulse duration
https://www.youtube.com/watch?v=88r2NBlYLSU (https://www.youtube.com/watch?v=88r2NBlYLSU)
Thanks to all!
Best regards
Miguel