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

 

hi M85,
Welcome to AAC.
Is this Homework or a College assignment.?
E

 

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

 

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

 

hi M85,
Welcome to AAC.
Is this Homework or a College assignment.?
E

This isn't being graded for any assignment or anything.

 

hi M85,
Check the datasheet for a typical Silicon transistor.
E

Use the Typical value.

 

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.

 

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?

 

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

 

So the Vout would be 23.6?

Not even close.

Hint: The transistor is acting as an emitter follower.

 

Any help would be appreciated

Can you write the voltage equation for this loop:

 

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.

 

Can you write the voltage equation for this loop:
View attachment 248779

It would have to be 5V there. Right?

 

Not even close.

Hint: The transistor is acting as an emitter follower.

5V.

 

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 }

 

So the base emitter voltage would be typically 0.7V like a single PN junction would be.

That's the approximate voltage for silicon transistors.

It would have to be 5V there. Right?

That's what I get...

Can you see how the output voltage would vary with load current?

 

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.

 

hi agu,
Its not clear but it does say 'dc'.

E

 

I also thought it said "ac" at first. Then I thought then why have the +/- signs. Closer inspection revealed the "d".

 

I also thought it said "ac" at first. Then I thought then why have the +/- signs. Closer inspection revealed the "d".

I thought it was AC at first as well. Then it just made zero sense to me at that point.