# How accurately does Vbe determine collector current?

#### dannyf

Joined Sep 13, 2015
2,197
" it gets worse"

Not as dire.

As the temperature goes up, the power being dissipated also goes up. This leads to slower rise in temperature, everything else being equal.

The net impact is that for some transistors they will reach thermal equilibrium. Otherwise, you would have any working transistors now.

The opposite is also true: with the right combination, any bjt can be driven into thermal run away by a ccs.

#### cabraham

Joined Oct 29, 2011
82
" it gets worse"

Not as dire.

As the temperature goes up, the power being dissipated also goes up. This leads to slower rise in temperature, everything else being equal.

The net impact is that for some transistors they will reach thermal equilibrium. Otherwise, you would have any working transistors now.

The opposite is also true: with the right combination, any bjt can be driven into thermal run away by a ccs.
I already covered bjt driven from CCS and ask you reread my post. If the CCS value exceeds the bjt rating, of course it will die. A 5 amp CCS driving a part whose Ie is rated at 0.50 amps will surely destroy iy. But a 0.40 amp CCS driving a part rated at 0.50 amps is stable.

But for CVS it is different. Suppose I drive a bjt with a 0.40 amp CCS with a 0.50 amp rated part. I measure Vbe at 0.75 volts. So all is well, it runs indefinitely at stable temp. But if I force a CVS across Vbe at 0.75 volts, the device does not likely survive. Even though 0.75V/0.40A/0.30W is inside the part rated limits, thermal runaway still happens.

Back in the mid 70's in my 1st electronics lab I had to plot I-V behavior for a p-n junction diode. I could easily put several hundred mA through the diode, and measure the forward drop up to 0.80 volts. But placing a CVS directly across the junction incurred thermal runaway at voltages well below 0.80 volts. That is my point. With CCS drive, 300 mA and 0.80 volts do not hurt the diode. But with CVS drive, runaway happens well below 0.80 volts.

Of course CCS drive can destroy a part. But you have to exceed the rated limit to do that. Not so with CVS. Thermal runaway is why CCS is used, not CVS. Those who say that one can current limit the CVS (resistor or whatever), monitor the forward voltage drop, then adjust the CVS value to keep steady current and avoid thermal runaway, are unwittingly driving the junction with a de facto CCS. What is a CVS, and CCS?

A CCS adjusts the voltage to maintain constant current. A CVS vice versa. Anyway, these are interesting things to ponder. I welcome comments and/or questions. Best regards.

Claude

#### dannyf

Joined Sep 13, 2015
2,197
So all is well, it runs indefinitely at stable temp. But if I force a CVS across Vbe at 0.75 volts, the device does not likely survive.
You may have too simplistic definition on "ratings". A device may have multiple specs whose maximum ratings vary based on the conditions the device is subject to. For example, soa and pulse width or case / die templerature, to name a couple.

It is quite easy to destroy a transistor, for example, by driving its Ic within its max rating, ccs or CVS.

Fundamentally, a device will fail not because how it is driven - the device has zero knowledge whether it is driven by a ccs or CVS.