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I was reading this page about thermal run away.

enter image description here

And it mentions:

Due to increase in the temperature of collector junction, Ico will increase, Vbe will decrease.

But for Ico to increase Ic and Ie, the Ico has to flow through the base-emitter junction and eventually to ground. This would amplify Ico (aka Icbo) and create Iceo = (1+beta) * Iceo.

But all these would happen if R2 would not exist. I mean the above circuit is not the same circuit here.

Since there is R2 which provides a path to ground, would Ico still flow through base-emitter junction or flow through R2 to ground? Because, to me if it flows through R2 it wouldn't cause thermal runaway. (?)

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Revised short answer

With base open Icbo flows thru emitter with current gain until Vcb max is reached then it becomes a zener then if Rc is low like a voltage source, leads to thermal runaway unless there is current limiting to prevent PD and temperature rise.

Another REF


As I recall, the normal diode reverse bias leakage on the collector is on the order of 1 uA and roughly doubles every 10'C rise.

If I understand it right, Icbo is measured with the emitter open and Iceo is measured with the base open which creates some equivalent R'ce. Then under the conditions of high Vce across this leakage resistance can cause some self heating.

Now if the collector and emitter R's are much smaller than this equivalent Rce leakage then Vce does not drop much due to leakage. Thus the leakage current creates more rise in Pd in the transistor than the external resistors.

I think the runaway condition comes when the transistor is more thermal insulated and higher Vce such that the leakage current is a significant source of heat, which does not conduct externally.

I have never personally encounter this problem as I recall because I never used very high Vce thermally insulated parts that cause the temperature to rise 10'C from twice the leakage current.

Normally H biased parts are very stable and only drift very little with temp rise unless you have very high DC gain, but not to the point of thermal runaway.

What's your experience?

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More details

Now look at a photo transistor with no input current. These are designed for high current gain but very low leakage of Iceo and there is no base bias current. (Iceo means base open)

enter image description here

Iceo= 0.1 μA @ Vce=20V where (β+1)Icbo=Iceo thus leakage R'ce = 20V/0.1 μA = 200 MΩ @ 20V but BVceo = 60V min @ 0.1mA then R'ce = 60V/100μA drops to 0.6 MΩ the threshold of avalanche condition.

Thus if Rc was say 10kΩ much lower than R'ce, Pd= 60V* 0.1mA = 6mW the opto would be quickly followed by thermal runaway if Vce exceeds 60V. ( depending on tolerance and process quality of the part.)

I hope this answers every question, otherwise feel free to ask more.

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  • \$\begingroup\$ bo means base open. eo means emitter open. \$\endgroup\$ – Marla Dec 31 '18 at 13:35
  • \$\begingroup\$ Hi @Marla I believe Icb means between collector and base and with open emitter , Icbo , at least that's what the textbooks agree with. \$\endgroup\$ – Sunnyskyguy EE75 Dec 31 '18 at 17:28
  • \$\begingroup\$ Now the more I investigate, I see it used both ways. Check this definition : ecetutorials.com/question-answers/… \$\endgroup\$ – Marla Dec 31 '18 at 17:35
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    \$\begingroup\$ @TonyEErocketscientist I see. I wrote that because your sentence starts with "If emitter is open". It seems if R2 is much smaller than hFE*Re then the thermal effect of Icbo and Iceo will be mitigated correct? \$\endgroup\$ – panic attack Jan 1 at 4:36
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    \$\begingroup\$ MY question was clear why are you answering something Im not asking? very strange \$\endgroup\$ – panic attack Jan 1 at 4:44

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