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Series Pass Regulator question 2..

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randolfo

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In a series pass regulator, if you measure 25.7 V at the collector of common-emitter transistor at the sense circuit for a supply of 30V . Could you say the circuit is operating normally or there is a short in emitter-to-base ? Please elaborate..

Thank you..:)
 

crutschow

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Most Helpful Member
You should have posted this in your previous thread.

Post a schematic of the regulator.
 

randolfo

New Member
Schematic..

Sorry, here is the circuit schematic. Attached..thank you.
 

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MikeMl

Well-Known Member
Most Helpful Member
Why didn't you just post this in the original thread?

Vc=30, Ve=25, Vb=25.7.
 

randolfo

New Member
Two additional questions..

In the circuit shown , a technician measures the output voltage of the power supply at 30 V. Which is the correct statements ?
1) A collector-to-emitter short is present in Q1.
2) Resistor R2 is shorted.
3) Resistor R4 is open.
4) Resistor R1 is open.
I am choosing answer no. 2 in this case. Not sure though if i am correct. Because if R2 is shorted there is no voltage drop that will limit the base bias voltage at Q1.

When a load is applied to the power supply in the circuit , the output voltage tends to decrease. What tends to counteracts this action?
1) The conduction of Q2 decreases.
2) The conduction of Q1 decreases.
3) The voltage at the emitter of Q2 increases.
4) The voltage at the junction of R3 and R4 increases.

I am choosing answer no. 2 in this case, because the output voltage depends on the conduction of Q1.

Please help me to understand this circuit. Are my answers correct ?:)
 

randolfo

New Member
Why didn't you just post this in the original thread?

Vc=30, Ve=25, Vb=25.7.

Sorry, i forgot to...

Based on the values, does it mean there is a short in emitter-to-base, why the Vc is going to 25.7 , is there any explanation to that behavior ?
 
In the circuit shown , a technician measures the output voltage of the power supply at 30 V. Which is the correct statements ?
1) A collector-to-emitter short is present in Q1.
2) Resistor R2 is shorted.
3) Resistor R4 is open.
4) Resistor R1 is open.
I am choosing answer no. 2 in this case. Not sure though if i am correct. Because if R2 is shorted there is no voltage drop that will limit the base bias voltage at Q1.

When a load is applied to the power supply in the circuit , the output voltage tends to decrease. What tends to counteracts this action?
1) The conduction of Q2 decreases.
2) The conduction of Q1 decreases.
3) The voltage at the emitter of Q2 increases.
4) The voltage at the junction of R3 and R4 increases.

I am choosing answer no. 2 in this case, because the output voltage depends on the conduction of Q1.

Please help me to understand this circuit. Are my answers correct ?:)

Just a quick guesstimate:

A short across CE puts the VCC at the output.

A shorted R2 will put a voltage at the output = to (VCC - Vbe) around 29.3volts approx.

An open R4 will possibly saturate the feedback transistor dropping the VB of the series pass to a much lower value possibly putting the series pass close to cutoff, lowering output to a very low value. Until it reaches an equilibrium because the series pass also biases the feedback transistor.

An open R1 may not have any affect on the output as long as the Zener diode is in full conduction, however if the zener is not in full conduction due to R1 not biasing it on properly, then the feedback transistoer will conduct less, allowing VB of series pass to rise higher putting a higher voltage output, but there will still be some regulation depending on the amount of feedback current that is flowing through the zener, however the full VCC at the output will never be realized due to the CE saturation voltage that is inherent with the transistor.



Question 2.

Remember transistors are more like current sources, which means the voltage needs to change with respect to a change in resistance in order to keep the same amount of current flowing.

So with a specific current flowing at the output, what happens to current when another resistance is put in parrallel with the original circuit path, the current must now divide, which means less current avaiolable to the base of Q2.
So what will Q2 need to do to get the voltage at it's base high enough to supply the original current it had before the load resistance was used.?

Remember the output of the transistor Q2 is feeding back on (Q1) is feeding it (Q2) it's base voltage.
 
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randolfo

New Member
Thank you "12345678901234567890 " , it really helps. Now it's clearer to me to where would i go for the answer. :) . Question 1, it should be No. 1 and Question 2, should be no. 4 . :d . Wheww, i wish i am more adept in analyzing circuits, still lot to learn. I love electronics..
 
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