common emitter amp fault finding

Discussion in 'Homework Help' started by bug13, Oct 16, 2012.

1. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62 as seen in picture

DC Voltage
Vb=1.8v, Ve=1.1v, Vc=6v

AC voltage
Vb=0.6mV, Ve=0mV, Vc=73mV

Fault No.5
DC Voltage
Vb=3.4v, Ve=2.7v, Vc=2.8v

AC voltage
Vb=0mV, Ve=0mV, Vc=0mV

I can't figure out what cause this, I believe the closest I get is B and C short circuit, as a result Vb increases, but other than that, I don't have a clue.

2. vk6zgo Active Member

Jul 21, 2012
677
85
Why is Vb= 1.8v in the "no fault" condition?

What voltage would Vb = if you disconnected the base connection in the "no fault" condition?

What do you notice about Ve in the "fault" condition?

What do you remember about the relationship between Ic & Ie?

3. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62
Vb=1.8V is given out in the question, in a simulation it gives Vb=1.77V, my lab section shows Vb=1.8V as well.

Vb is still 1.8V if disconnected in no fault condition

Ve in the "fault" condition is high than no fault condition, but I can't figure what cause it? I first suspect B-E s/c, but then Vb should equal Ve.

Ic=Ie-Ib? is it the answer you got in mind?

4. vk6zgo Active Member

Jul 21, 2012
677
85
What would Vb read if you disconnect it from the base in the fault condition?

What is the function of R1?,R2?,both?

Ic=Ie-Ib,hence Ib= Ie-Ic.

With Ie= Ve/Re,& Ic = (10-Vc)/Rc what is the change in Ib between "no-fault" & "fault"conditions?

What effect would this have?

5. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62
I don't know what would Vb read if it was disconnected from the base in fault condition, as I am not able to create this specific fault so far

R1 and R2 is to bias the transistor to a quiescent condition where Vc should be 6V with no input.

Ie=Ve/Re=2.7/470=5.74mA
Ic=(10-Vc)/Rc=(10-2.8)/2.2k=3.27mA
Ib=Ie-Ic=5.47mA-3.27mA=2.47mA

so I continue:

I1=(10-Vb)/R1=(10-3.4)/10K=660uA
I2=Vb/R2=3.4v/2.2k=1.55mA

and I get:

Iin=Ib+I2-I1=2.47mA+1.55mA-660uA=3.36mA

So the fault is the signal generator?

6. JoeJester AAC Fanatic!

Apr 26, 2005
4,048
1,774
You can add switches to each of the legs of the transistor ... some to illustrate an open, some to indicate an leg to leg short.

Your answer will become more evident as you show each switch open or close .... to simulate a problem.

The quiesent voltages certainly indicate a problem.

I don't see the Vb rising to 3.4 ... even though the problem stated it would. (10*2200)/(10000+2200) does not equal 3.4

Last edited: Oct 17, 2012
7. vk6zgo Active Member

Jul 21, 2012
677
85
Going back to the "no fault" condition.
Ib << than the current through R1,R2,so may be safely ignored.

By the Voltage Divider formula :
Vb=10*R2/(R1+R2)=1.8V (as we were given).

Now,in the "fault" condition,the supply voltage is still 10V,but now Vb=3.4V.
What does that imply about the voltage divider,R1,R2?

Hint: Resistors usually don't fail low in value.

PS:There is nothing to indicate that the signal source has any DC component,so that tends to absolve C1.

8. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62
I have tried different possibilities in my simulation, but I am not able to create this fault, maybe there is something I haven't tried.

9. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62
I am not sure if I get what you mean, but these are what I have done:

R1 = O/C, S/C (open circuit, short circuit)
R2 = O/C, S/C

and I am not able to create the fault condition in my simulation.

but if I replace the signal generator with 3.36mA current with C1 S/C, my simulation is pretty close to the given fault condition.

But I haven't tried increasing the supply voltage in my simulation, maybe I can try that when I have access to the lab

10. vk6zgo Active Member

Jul 21, 2012
677
85

Don't use the simulation, use the maths!!

You are all over the answer.

Just for laughs,try increasing the value of R2 to 5kΩ,while leaving R1 at 10kΩ

It won't be exactly the same result as the "fault",but close!

11. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62
Oh I see, thanks for your time and patient, I am to rely on the simulation software and forget about how to do it by maths, thanks a lot

Last edited: Oct 17, 2012
12. vk6zgo Active Member

Jul 21, 2012
677
85

No,that's exactly the opposite of what I said! It doesn't matter in this case,as if you change the value of R2 as suggested,the result should be that Vb is very close to what it is in the
"fault " case,whether you work out the maths,or simulate it.

Remember what I said about resistors not decreasing in value if they are faulty.
Assume R1 has not changed.

I would urge you to learn to use the maths,as you won't have a simulator in the exam room!

13. JoeJester AAC Fanatic!

Apr 26, 2005
4,048
1,774
We might consider the totality of our experiences but unfortunately, those in the academic setting can have resistors decrease in value.

The only resistor I've clearly identified as a decreased in value was in an UHF transceiver, the AN/URC-80 ... but that was a long time ago.

More recently another tech found a 39k resistor burnt down to about 30 ohms and that took out a 470 ohm 2W resistor.

So yes, resistors can decrease in value. I would not discount the ability to decrease R1 to a value to more closely match the parameters set forth in this academic exercise ...

Here's an interesting tidbit on resistor failures....

• ResistorsFailure.pdf
File size:
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Last edited: Oct 18, 2012
14. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62
OK, I am a bit lost again, I think I get it because the maths works out the I2+Ic=Ie, so I believe the fault is R2 o/c, but what were you trying to lead me to?

Last edited: Oct 18, 2012
15. vk6zgo Active Member

Jul 21, 2012
677
85
IF R2 has increased in value,say to 5kΩ as I suggested,ignoring Ib <<<I2
Then Vb becomes 3.333V --very close to 3.4V.
Everything else about the transistor works--note that Ve has increased to be
0.7V less than Vb.

16. JoeJester AAC Fanatic!

Apr 26, 2005
4,048
1,774
Here's what I did ...

First ... we know from KCL that the base current is the emitter current minus the collector current. We know the emitter current is (2.7/470) and the collector current is ((10-2.8)/2200). That leaves the the base current exiting the base terminal.

Next we know that the current through R1 equals the current through R2 (3.4/2200) plus the base current.

Next following ohms law, we know the resistance is (10-3.4) / (IR2 + Ib).

Electron current flow does not allow R2 to increase in value for this problem.

In the probability game ... the transistors fail before capacitors and then the resistors. Only about 9% of the resistors decrease, so the odds were against that. Personally I played the odds when I first looked at this circuit, and failed. It wasn't until I tested everything against KCL did the answer reveal itself.

Ib can not be ignored as it exceeds IR2.

Last edited: Oct 18, 2012
17. JoeJester AAC Fanatic!

Apr 26, 2005
4,048
1,774
One problem I had with this circuit was the initial no fault conditions seemed out of whack. Vc of 6 volts stated the IRC drop was only 4 volts.

Below is a chart juxtaposing the Voltages and Currents across and through each resistor in the circuit. There is an inordinate error between the "paper" circuit and the "simulated" circuit.

File size:
5 KB
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18. bug13 Thread Starter Senior Member

Feb 13, 2012
1,644
62

from what I understand it, is either R1 decrease in value or R2 increase in value, but you said resistor not decrease in value when faulty (while Joe said it could happen but very rare)

Said if i2 increase in value, I can't figure out where that Ib=2.476mA from (Ie-Ic=5.47mA-3.27mA=2.47mA)

so I would like to believe R1 decrease in value (Vcc-Vb)/(I2+IB)=(10-3.4)/(1.55mA+2.47mA)= ~1.6K (thanks Joe for pointing this out)

I am totally confused now

19. JoeJester AAC Fanatic!

Apr 26, 2005
4,048
1,774
I was told many years ago when troubleshooting transistor circuits to "follow the currents". I reminded myself of that when I created the spreadsheet and appropriate formula's to follow Ohms Law and Kirchoff's Law.

In electron current flow it's from negative to positive, or from less positive to more positive. R2 did not change unless you figured out a way for the electroncs to flow from more positive to less positive.

I see that was fault number 5 ... put up the others .....

20. vk6zgo Active Member

Jul 21, 2012
677
85
I give up!! 