# Zener Diode

#### celect

Joined May 31, 2005
18
I was wondering how this zener got it voltage, I attached the problem I working on, I assume this to be a voltage regulator circuit.

If this zener is regulating 15V is my output to the load going to be 15V.

How does this work, why are the transistors in the circuit if this zener will work
with out them.

#### Brandon

Joined Dec 14, 2004
306
Originally posted by celect@Jun 8 2005, 10:02 AM
I was wondering how this zener got it voltage, I attached the problem I working on, I assume this to be a voltage regulator circuit.

If this zener is regulating 15V is my output to the load going to be 15V.

How does this work, why are the transistors in the circuit if this zener will work
with out them.
[post=8321]Quoted post[/post]​
Zener's are used a bit differently than a typical diode. They are normally used in reverse, like the one in the circuit is designed, but this zener is not regulating the output at all.

This zener is indirectly regulating the top xtr which is the actual voltage source through its effect on the bottom xtr.

In typical AC/DC converters the zener is used to regulate along with a cap.

#### David Bridgen

Joined Feb 10, 2005
278
Originally posted by celect@Jun 8 2005, 04:02 PM
I was wondering how this zener got it voltage, I attached the problem I working on, I assume this to be a voltage regulator circuit.

If this zener is regulating 15V is my output to the load going to be 15V.

How does this work, why are the transistors in the circuit if this zener will work
with out them.
[post=8321]Quoted post[/post]​
The zener is in series with the base-emitter of Q2. The voltage at the junction of the 2k5 and 1k5 resistors will therefore be 15 + 0.7V, so the current through the 2k5 is 6.28mA.
This current also flows through the 1k5. Also flowing through the 1k5 will be the base-emitter current (negligible) and the current through the resistor between the base and emitter of Q2. Since its value is not stated all we can say is that the output voltage will be the sum of the voltage at the junction of the 2k5 and 1k5 resistors (15.7) and at least 9.42 (given by at least 6.28mA x 1k5.) i.e. output voltage is at least 25.12.

#### davidand

Joined Jun 2, 2005
43
Originally posted by David Bridgen@Jun 8 2005, 04:28 PM
The zener is in series with the base-emitter of Q2. The voltage at the junction of the 2k5 and 1k5 resistors will therefore be 15 + 0.7V, so the current through the 2k5 is 6.28mA.
This current also flows through the 1k5. Also flowing through the 1k5 will be the base-emitter current (negligible) and the current through the resistor between the base and emitter of Q2. Since its value is not stated all we can say is that the output voltage will be the sum of the voltage at the junction of the 2k5 and 1k5 resistors (15.7) and at least 9.42 (given by at least 6.28mA x 1k5.) i.e. output voltage is at least 25.12.
[post=8329]Quoted post[/post]​
Thanks for explaining this to me. So if R4 shorts or is removed and they since no vaule was given for R4 how would I determine the voltage output?

#### mozikluv

Joined Jan 22, 2004
1,437
hi

if you remove R4 then Q1 will not conduct since that resistor is you base bias of Q1. likewise with R3 to Q2

#### David Bridgen

Joined Feb 10, 2005
278
Originally posted by davidand@Jun 9 2005, 02:55 AM
Thanks for explaining this to me. So if R4 shorts or is removed and they since no vaule was given for R4 how would I determine the voltage output?
[post=8340]Quoted post[/post]​
The labelling of the resistors is unreadable, so I don't know which one you mean.

#### celect

Joined May 31, 2005
18
Originally posted by David Bridgen@Jun 9 2005, 09:32 AM
The labelling of the resistors is unreadable, so I don't know which one you mean.
[post=8354]Quoted post[/post]​

I re-labeled the drawing, in this voltage regulating circuit how would I determine the Voltage out if R4 is open or shorted do I need to know the vaule of R4?
or is this resistor a standard for circuits

#### David Bridgen

Joined Feb 10, 2005
278
Originally posted by celect
in this voltage regulating circuit how would I determine the Voltage out if R4 is open or shorted do I need to know the vaule of R4?
R4 provides the base current for Q1, the series pass transistor (the actual regulator.)
If R4 was replaced by a short circuit, Q1 would be permanently "on", like a switch, and the output voltage would be full and unregulated output of the basic supply.
If R4 were removed, the Q1 would effectively be open circuit and there would be no output.
The value of R4 has no direct influence on the output voltage and you don't need to know it.

The resistor to which I referred, without a value, is the one between base and emitter of Q2, apparently R3.
Q2 will have about 0.6V across its base-emitter junction. The current through R3, due to the 0.6V, will also flow through the 1k5 and therefore have an influence on the voltage across the 1k5.

The information you need to state the output voltage of the circuit with any accuracy is incomplete.
The best you can do with the information which is given is to say that it is about 25.
My guess is that it is a poor attempt at a 24V supply.

#### celect

Joined May 31, 2005
18
Originally posted by Brandon@Jun 8 2005, 03:50 PM
Zener's are used a bit differently than a typical diode. They are normally used in reverse, like the one in the circuit is designed, but this zener is not regulating the output at all.

This zener is indirectly regulating the top xtr which is the actual voltage source through its effect on the bottom xtr.

In typical AC/DC converters the zener is used to regulate along with a cap.
[post=8327]Quoted post[/post]​
if xtr on the bottom Q2 conducts more, what would happen to xtr on top Q1 would it conduct more also?

#### pebe

Joined Oct 11, 2004
626
Originally posted by celect@Jun 11 2005, 12:59 AM
if xtr on the bottom Q2 conducts more, what would happen to xtr on top Q1 would it conduct more also?
[post=8389]Quoted post[/post]​
What is xtr?

#### celect

Joined May 31, 2005
18
Originally posted by pebe@Jun 11 2005, 04:38 AM
What is xtr?
[post=8407]Quoted post[/post]​
I assumed it to mean transitor

#### pebe

Joined Oct 11, 2004
626
Originally posted by celect@Jun 13 2005, 11:53 PM
I assumed it to mean transistor
[post=8471]Quoted post[/post]​
I've never heard of it before. Why not just refer to Q1 or Q2?

#### celect

Joined May 31, 2005
18
Originally posted by pebe@Jun 13 2005, 07:16 PM
I've never heard of it before. Why not just refer to Q1 or Q2?
[post=8473]Quoted post[/post]​
I will for now on, somebody replied back to me and I thought that was how they were refering to the transistors.
thanks

#### David Bridgen

Joined Feb 10, 2005
278
Originally posted by celect@Jun 14 2005, 03:59 AM
I will for now on, somebody replied back to me and I thought that was how they were refering to the transistors.
thanks
[post=8482]Quoted post[/post]​
That's the danger of using stupid jargon and silly buzzwords.

I've said it on other forums and I'll no doubt say it again:

If I use jargon or buzzwords and you don't understand, it will be my fault.

If I use plain English and you don't understand, it will be yours.

#### David Bridgen

Joined Feb 10, 2005
278
Originally posted by celect@Jun 11 2005, 12:59 AM
if ... Q2 conducts more, what would happen to ... Q1 would it conduct more also?
Let's take it one step at a time.
When and/or why will Q2 conduct more?

#### davidand

Joined Jun 2, 2005
43
Originally posted by David Bridgen@Jun 14 2005, 11:49 PM
Let's take it one step at a time.
When and/or why will Q2 conduct more?
[post=8513]Quoted post[/post]​
I think if the zener were to short then Q2 would conduct more,

#### David Bridgen

Joined Feb 10, 2005
278
Originally posted by davidand@Jun 15 2005, 10:35 PM
I think if the zener were to short then Q2 would conduct more,
[post=8542]Quoted post[/post]​
I thought it might be more beneficial if you were thinking of normal operation.
Do you understand how the circuit works?

#### davidand

Joined Jun 2, 2005
43
Originally posted by David Bridgen@Jun 15 2005, 08:56 PM
I thought it might be more beneficial if you were thinking of normal operation.
Do you understand how the circuit works?
[post=8547]Quoted post[/post]​

No I'm sorry celect
I was just going over some of the posting.
and put my foot in my muth again.