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I'm wondering what happens in a BJT transistor if we force a voltage bigger than 0.7V on Vbe.

A colleague of mine said that if we put on Vbe, let's say, 5V the transistor would be saturated.

I haven't seen the case in question but I suppose that happens if Vbe > Vce.

Still is there any harm to the transistor if we force a bigger voltage than 0.7 there?

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    \$\begingroup\$ The base emitter junction of a BJT acts like a diode. Ask your colleague what will happen if you apply 5 V to a diode. If that's hard to visualise then what will happen if you apply 5 V to an LED with a forward voltage of 2 V? \$\endgroup\$
    – Transistor
    Commented May 4, 2019 at 17:24
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    \$\begingroup\$ Every increase by \$60\:\text{mV}\$ leads to an increase of current by a factor of 10. (See the Shockley diode equation for why.) \$5\:\text{V}-700\:\text{mV}\approx 72\cdot 60\:\text{mV}\$ so you might expect \$10^{72}\times\$ the diode current, assuming other things didn't interfere: (1) inability to supply the current; (2) current crowding effects; (3) over-heating and/or explosive results; etc. \$\endgroup\$
    – jonk
    Commented May 4, 2019 at 17:41
  • \$\begingroup\$ I have put a couple of volts positive bias on BJT without harming it. Of course it was a bit nippy where the transistor was located. \$\endgroup\$ Commented May 4, 2019 at 18:30

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You can't make Vbe larger than 0.7V without destroying the transistor.

As long as the current is low enough to not damage the transistor, the voltage drop on Vbe will be 0.7V.

If you measure more than 0.7V from the base to the emitter of a bipolar junction transistor (BJT,) then you have destroyed it.

Between the base and the emitter of a BJT is the equivalent of a diode. That's why Vbe is 0.7V - it is the foward voltage of a silicon diode.

Just like any diode, the forward voltage stays around 0.7V unless it is damaged.

So, no, you can't "force" Vbe to be higher than 0.7V if you want to actually use it as a transistor afterwards.


I've been reminded that not everyone will realize that "0.7V" is a sort of shorthand for "the rated Vbe of your transistor."

Depending on how the transistor is made, Vbe can be higher or lower than 0.7V. It also varies depending on forward current and temperature just like in any other diode.

In any case, the actual Vbe is inherent to the diode and the current through it. If you try to force any higher voltage onto it then you will destroy it.

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As 'Transistor' says - a BJT base-emitter junction IS a diode. The normally seen voltage of around 0.6 - 0.7 V is what is adequate to provide a base current that causes the transistor to operate in a normally desired manner.

If you increase Vbe the Ib also increases - up to some maximum allowable Vbe limit.
At some point above that the transistor will be damaged or destroyed.

Some transistor data sheets provide Vbe/Ib curves.
This is reasonably uncommon.
In most cases Vbe of <= 1V would meet all needs for small signal transistors.
In some power applications, with Ibe in the amps range, Vbe MAY be more than 1 Volt.


Some data sheets specify Veb REVERSE bias breakdown limits of the base-emitter junction where it acts as a zener diode.
This datasheet for a BC337 transistor specifies Vbe_max of 5 Volts.

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  • \$\begingroup\$ that 5v would be for reverse-bias Zener operation \$\endgroup\$ Commented May 5, 2019 at 2:15
  • \$\begingroup\$ @analogsystemsrf " .. reverse bias zener ..." -> True. That's a different parameter to what I mention and another issue. What I said about "some maximum allowable Vbe limit" and damage and/or destruction is correct BUT not usually specified in the data sheet. | What you (correctly) point out is that the parameter given is Veb (reverse polarity) and not Vbe as I stated. Vbe max save/sensible still exists but is less often specified. It will be constrained by device dissipation limits if by nothing else - but in practice some lower value will probably result in damage. \$\endgroup\$
    – Russell McMahon
    Commented May 5, 2019 at 16:59
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Isn't that like asking what happens if you put 10 Amps thru a 100mA diode?

You can ignore the Shockley equation above the rated max current, because then it behaves like a heater resistor and it will never come close to the giga-amp calculated current at 5V.

For BJT's I call it saturated when the bulk resistance dominates the VI curve when it becomes linear. For the Vce(sat) it is similar except what happens is the hFE drops towards 10% of its linear hFE as the junction saturates. Then increasing load current follows the Rce load line and not the hFE current ratio.

I have a rule of thumb for Rbe bulk resistance R=5/Pd which is the device max power rating Pd at 85'C. THe Rce saturation resistor is typically Rce = 0.5/Pd with a 50% but there are special device designs much lower.

There are some transistors however which are rated for 1A Max. by design and IC is rated for 3A max or 6A pulse. But attention to Joules of heating or the Safe Operating Area (SOA) are necessary. enter image description here enter image description here

So the bottom line is damage to the junction by overtemperature will result if the Safe Operating Area or the Absolute max current is exceeded. enter image description here

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