Maximum Emitter Base Voltage spec

Thread Starter

Gibson486

Joined Jul 20, 2012
355
I am making a simple on-off led driver. The device that holds the led only allows you to excite with a high side driver. Simple enough. I decided to just do a simple 2 transistor driving method with an npn and pnp (pic attached). My question is the the Maximum Emitter Base Voltage spec. For the transistors I use, they spec it at 6V. Does this mean that my circuit will not work? I am guessing that I need to add another resistor in series with the 10K to ensure the voltage never goes too low and that divider would go to the base of the pnp?

1572977608937.png
 

Audioguru again

Joined Oct 21, 2019
6,671
When the NPN transistor turns on and causes all its current to destroy the base-emitter junction of the PNP then which gets destroyed first? The NPN or the PNP?

Of course, a resistor is needed to limit the base current of the PNP.

Do not worry about a reverse voltage (max allowed is 6V) of the emitter-base because it will never occur in your circuit. How can the base voltage of the PNP become higher than +30V?

Do you have a resistor or current limiter circuit to limit the LED current?
 

Thread Starter

Gibson486

Joined Jul 20, 2012
355
Understood....however, it says the emitter base can not exceed 6V. In my circuit, when the npn is on, the base of the pnp will be approx 0V while the emitter is 24V. Current into the base aside, that would kill the pnp because the emitter-base voltage is now 24V. Why would it have to exceed 30V? I think I am misunderstanding something.

edit....wait...I see what you are saying...it's emitter minus base.

and yes, I added a resistor as well.
 
Last edited:

ericgibbs

Joined Jan 29, 2010
18,766
hi,
Consider that the 2k7 is connected directly from the PNP Base to 0V [ no NPN]
This means the PNP is switched ON, ie conducting.
Assume the LED requires 100mA and the Gain of the PNP is say 10 to 20 , this means the Base requires approx 10mA of current

So 24V/2k7 = ~ 9mA.

Do you follow OK.?
E
 

SteveSh

Joined Nov 5, 2019
109
I believe the 6V max spec for the transistors is the maximum base-emitter reverse voltage, which has little to do with the forward base-emitter voltage. In the circuit shown, neither the 2N2222A nor the '2907A will have the Vbe(reverse) limit exceeded, assuming the MCU doesn't put out anything less than -6V.
 

Thread Starter

Gibson486

Joined Jul 20, 2012
355
hi,
Consider that the 2k7 is connected directly from the PNP Base to 0V [ no NPN]
This means the PNP is switched ON, ie conducting.
Assume the LED requires 100mA and the Gain of the PNP is say 10 to 20 , this means the Base requires approx 10mA of current

So 24V/2k7 = ~ 9mA.

Do you follow OK.?
E
I see. makes perfect sense!
 

Audioguru again

Joined Oct 21, 2019
6,671
The base-emitter of a transistor is a diode. When the transistor is turned on then the base to emitter forward bias voltage is the diode voltage drop of about 0.7V. When the base-emitter has a reverse bias voltage which never happens in your circuit then the base-emitter has an avalanche breakdown (like a Zener diode) at around 6V to 7V of reverse voltage that damages the base-emitter a little each time it happens. A Zener diode is designed for the heat caused when it has voltage and current at the same time but the base-emitter of a transistor is not designed to survive making itself hot like a Zener diode.
 
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