NPN* Active/Linear Mode Question

Thread Starter

Msnydz85

Joined Sep 24, 2021
9
I am looking for help analyzing the circuit shown below. It looks to me that the Zener diode is used to set the bias voltage at the base to 5.7V but I can't figure out what the output voltage would be. Any help would be appreciated
1632503641478.png
 
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BobTPH

Joined Jun 5, 2013
3,769
Do you know any rule that tells you the base to emitter voltage for a BJT in active mode?

By the way, that is an NPN, not a PNP.

Bob
 

Thread Starter

Msnydz85

Joined Sep 24, 2021
9
Do you know any rule that tells you the base to emitter voltage for a BJT in active mode?

By the way, that is an NPN, not a PNP.

Bob
hi M85,
Welcome to AAC.
Is this Homework or a College assignment.?
E
This is an NPN I should've written that correctly I apologize. I don't know how to determine the base to emitter voltage for a BJT in active mode.

Mod: Corrected.E
 
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MrChips

Joined Oct 2, 2009
24,398
The base is clamped to the zener voltage at 5.7V.
The base-emitter voltage is one P-N diode voltage in forward bias, 0.7V.
Work out the output voltage from there.
 

Thread Starter

Msnydz85

Joined Sep 24, 2021
9
hi M85,
Check the datasheet for a typical Silicon transistor.
E

Use the Typical value.
So the base emitter voltage would be typically 0.7V like a single PN junction would be. Then the 5.7V at the base because of the Zener diode. So the Vout would be 23.6?
 

ericgibbs

Joined Jan 29, 2010
13,825
hi M.
Think about it you have a 5.7V Zener voltage and a Vbe of 0.7V ...Subtract, so what do you make Vout now.?
E

Corrected typo
 
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Papabravo

Joined Feb 24, 2006
16,938
Vbe is a "voltage drop" from the base to the emitter. A voltage drop means the voltage at the emitter will be lower than the voltage at the base.
Vce will be large enough so that any appreciable current through the transistor will cause it to get "finger-burning" hot. I'm not in favor of using this circuit in practice with the specified conditions.
 

ElectricSpidey

Joined Dec 2, 2017
1,868
There is nothing wrong with that circuit, it's a simple Zener regulator with a series pass transistor, wouldn't get any hotter than any other type of linear regulator.

(the 30 volt input is a bit high ;) }
 

Audioguru again

Joined Oct 21, 2019
3,632
1) The question askes about using a 5.7V zener diode which nobody makes. It should of said a standard 5.6V one.
2) The question did not say if the transistor is germanium or silicon.
3) The question shows a input that is 30VAC instead of a lower DC voltage.
 
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