PLEASE HELP! Transistor Problem

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
PLEASE HELP!

The following circuit is given and I need to solve for VB, VE, and VC. I know it's not a super hard problem but my professor isn't great and I think he confused me more. I think I have VB and VE but verification of these would be nice too.

Thanks!

upload_2019-9-27_23-31-8.png
 

Attachments

Picbuster

Joined Dec 2, 2013
990
How to start?
First of all produce the formulae for:
IC in relation to IB
Same for IE in relation to IB and IC
Next step is to use Ohms laws to resistors.
Combine the two.

We are not going to do your work!
Yes, we like to help you but it should roll, at the end, out off your brains.

Show us what you have don so far.

Picbuster
 

LvW

Joined Jun 13, 2013
890
With reference to Picbusters answer...I am afraid, the most important quantitty to start the calculation is missing:
The voltage Vbe=Vb-VE that controls the collector current Ic.
You have no other chance than to assume a suitable value for Vbe within a relatively small range which can be found in each textbook (for transistors working in the active region).

Hint: It is not a "super hard" problem - however, do not underestimate the calculation. It is not too simple! However, you can simplfy it (as a first step).
Therefore, I will give you the following hints:
* As a first step (iteration) you can neglect the base current IB. IB is not a very important parameter and - more than that - you would have to assume an estimated value because the exact value for IB is unknown (very large tolerances).
* Now you can calculate all the needed currents and voltages - starting with VBE=const (estimated value) and IB=0. It is the base-emitter voltage VBE which determines the collector current IC (Shockley equation).
* If you want, you can do a second iteration based on the calculated values - now with a suitable value for IB=IC/B (B=assumed DC curent gain).

Comment: You may be surprised that the calculation should (must !) start with an assumed/estimated value for VBE. However, due to the resistor RE all the currents are not very sensible to the exact (unknown) value for VBE (provided it is in the "normal" range). This is the desired result of negative DC feedback (provided by RE).
 
Last edited:

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
The following equations are what I have done to attempt to solve. The values obtained for Vb and Ve are of very minimal error when compared to my experimental values. (This problem is part of a lab report). I can't wrap my head around any way to find Vc. Even Vcc-VR3 doesn't work because I don't know VR3, nor do I know how to get Ic with the multiple power sources the way they are.
upload_2019-9-28_14-51-35.png
 

WBahn

Joined Mar 31, 2012
25,067
PLEASE HELP!

The following circuit is given and I need to solve for VB, VE, and VC. I know it's not a super hard problem but my professor isn't great and I think he confused me more. I think I have VB and VE but verification of these would be nice too.

Thanks!

View attachment 187001
So what do you think you have for Vb and Ve? We are not mindreaders, so there's no way we can tell you if you are right or not.
 

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
How to start?
First of all produce the formulae for:
IC in relation to IB
Same for IE in relation to IB and IC
Next step is to use Ohms laws to resistors.
Combine the two.

We are not going to do your work!
Yes, we like to help you but it should roll, at the end, out off your brains.

Show us what you have don so far.

Picbuster
I'm not asking you to do my work. I want to know how to do this problem. I have the option to find these values using Multisim, but I am choosing to do the hand calculations instead because I want to know how to do them. My professor just doesn't explain it well. If you'll look at my latest reply to my own thred, you'll see the calculations I've completed thus far.
 

WBahn

Joined Mar 31, 2012
25,067
The following equations are what I have done to attempt to solve. The values obtained for Vb and Ve are of very minimal error when compared to my experimental values. (This problem is part of a lab report). I can't wrap my head around any way to find Vc. Even Vcc-VR3 doesn't work because I don't know VR3, nor do I know how to get Ic with the multiple power sources the way they are.
View attachment 187040
Let's set aside the question of whether Vb and Ve are correct or not for just a bit and focus on what you know about Ic.

If you know Ve, can you find Ie?

What do you know about the relationship between Ic, Ie, and Ib for the transistor (regardless of which operating mode it is in)?

If the transistor is acting in the active region, what do you know about the relationship between Ic and Ib?
 

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
Let's set aside the question of whether Vb and Ve are correct or not for just a bit and focus on what you know about Ic.

If you know Ve, can you find Ie?

What do you know about the relationship between Ic, Ie, and Ib for the transistor (regardless of which operating mode it is in)?

If the transistor is acting in the active region, what do you know about the relationship between Ic and Ib?
I tried this:
upload_2019-9-28_15-3-9.png
But this yields a value that is way off from my experimental value which was +4.34 V
 

WBahn

Joined Mar 31, 2012
25,067
The following equations are what I have done to attempt to solve. The values obtained for Vb and Ve are of very minimal error when compared to my experimental values. (This problem is part of a lab report). I can't wrap my head around any way to find Vc. Even Vcc-VR3 doesn't work because I don't know VR3, nor do I know how to get Ic with the multiple power sources the way they are.
View attachment 187040
Okay, now let's look at your calculations for Ve and Vb.

Your first equation has R1 and R2, but you haven't indicated which resistors are R1 and R2.

IF you mean R1 and R2 to be the left hand resistors, they are not in series and so they don't have the same current flowing in them. IF you make a certain assumption, you can claim that they are ALMOST in series, but you will need to justify that assumption at the end of the day.

IF you mean R1 and R2 to be the right hand resistors, then you are implicitly assuming that there is 0 V across the transistor collector-emitter terminals, which is not a good assumption at all.

Your second equation starts off with ΔV, which your first equation concluded was 20 V, but in your second equation it is apparently -10 V. You need to be consistent in your definitions and use of your terms, otherwise your work becomes a jangled mess that no one (not us, not you, not the grader) can follow.

Your second and third equations, looking at the actual values (and not the setup), are consistent with the assumption I indicated earlier (that places the two left-hand resistors almost in series). Can you identify what that assumption is?

There is no need (or justification) for reporting results to six decimal places. The resistors almost certainly have tolerances that justify only two or three at the most. Then you are using an assumed value of Vbe that only has a single significant digit and you are neglecting the transistor current gain, which even if reasonable still puts you into the two- to three- sig fig only range.
 

WBahn

Joined Mar 31, 2012
25,067
I tried this:
View attachment 187042
But this yields a value that is way off from my experimental value which was +4.34 V
You have worse problems then that. Last I checked 20 - (-0.00139)(10000) was 33.9.

You need to ask yourself, always, whether your answers make sense?

Does it make sense that you have a negative collector current?

If not, go back and look at why you are using a negative voltage for the voltage ACROSS the emitter resistor. Remember, Ohm's Law deals with voltages ACROSS resistors and currents THROUGH resistors. You can't just throw the nearest voltage value at the equation and expect meaningful results.
 

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
I'm only looking for help regarding how to do the problem, not a critique of my nomenclature or rounding. These are simply equations that I whipped up in word to try to make communication with this community easier. I understand that I should have been more clear with my resistor naming and with my ΔV identification, and thus I have clarified below:
upload_2019-9-28_15-15-25.png
Thus, Equation 4 has a typo (it should be R4 not R3). Additionally, the first use of "ΔV" refers to the voltage from source to source when only analyzing R1 and R2 (Assuming Ib is 0), and the second use of "ΔV" refers to the reference voltage source of -10V when only analyzing Vb. I'll try to be more clear in future replies.
 

WBahn

Joined Mar 31, 2012
25,067
A good place to always start is to annotate your diagram so that you have a clear picture of what you are doing (and so that others are working from that same picture so that you can communicate clearly with them).

upload_2019-9-27_23-30-42.png

Then you need to use these definitions in your work so that you can document and communicate exactly what you are doing.

For instance, if you want the emitter current, then you apply Ohm's Law to the emitter resistor.

\(I_E \; = \; \frac{(V_E \; - \; V_{EE}\)}{R_E}\)

Do you see the mistake you made and how taking the time to clearly define what you are doing can help avoid it?
 

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
You have worse problems then that. Last I checked 20 - (-0.00139)(10000) was 33.9.

You need to ask yourself, always, whether your answers make sense?

Does it make sense that you have a negative collector current?

If not, go back and look at why you are using a negative voltage for the voltage ACROSS the emitter resistor. Remember, Ohm's Law deals with voltages ACROSS resistors and currents THROUGH resistors. You can't just throw the nearest voltage value at the equation and expect meaningful results.
It was simply a typo. I entered the result of the multiplication without subtracting it from 20. Thank you for catching that. I understand to check my answers. I also understand that the negative collector current is a red flag, but nevertheless, 33.9 is still far off from my experimental value. It seems you're using this site to condescend rather than help, so I would appreciate a bit more understanding.
 

WBahn

Joined Mar 31, 2012
25,067
It was simply a typo. I entered the result of the multiplication without subtracting it from 20. Thank you for catching that. I understand to check my answers. I also understand that the negative collector current is a red flag, but nevertheless, 33.9 is still far off from my experimental value. It seems you're using this site to condescend rather than help, so I would appreciate a bit more understanding.
I'm trying to help you see how to (1) do you work in a way that minimizes making mistakes in the first place, and (2) review your work so that you have a better chance of catching the mistakes you will still inevitably make (since you, like the rest of us, are only human).

If you understand that the negative collector current is a red flag, then you have a red flag telling you that something is WRONG at a step BEFORE you get either -13.91 V or 33.9 V, so it shouldn't be surprising that either of those results is meaningless. You need to focus on getting rid of the red flag before proceeding.

But, since you feel that trying to help you do your work in a clear and organized way that allows you to communicate that work accurately and effectively, or that trying to help you set up your work in a way that reduces the likelihood of making mistakes, or that trying to help you learn how to evaluate your work so as to find and fix the errors that you make constitutes condescension, I won't condescend your work any further. Good luck and goodbye.
 

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
A good place to always start is to annotate your diagram so that you have a clear picture of what you are doing (and so that others are working from that same picture so that you can communicate clearly with them).

View attachment 187045

Then you need to use these definitions in your work so that you can document and communicate exactly what you are doing.

For instance, if you want the emitter current, then you apply Ohm's Law to the emitter resistor.

\(I_E \; = \; \frac{(V_E \; - \; V_{EE}\)}{R_E}\)

Do you see the mistake you made and how taking the time to clearly define what you are doing can help avoid it?
This makes more sense. Thanks. But I still get a number that is much higher than the measured voltage (+4.34V). I understand the next step would be to question if the measured voltage is wrong, but these, measured values were confirmed by the professor during lab. The following are my most recent calculations. Given that the answer is about 10V away from my measured value, is there some way in which I am supposed to incorporate the +10V source at the top?

upload_2019-9-28_15-35-15.png
 

Thread Starter

ElectroJake

Joined Sep 27, 2019
11
I'm trying to help you see how to (1) do you work in a way that minimizes making mistakes in the first place, and (2) review your work so that you have a better chance of catching the mistakes you will still inevitably make (since you, like the rest of us, are only human).

If you understand that the negative collector current is a red flag, then you have a red flag telling you that something is WRONG at a step BEFORE you get either -13.91 V or 33.9 V, so it shouldn't be surprising that either of those results is meaningless. You need to focus on getting rid of the red flag before proceeding.

But, since you feel that trying to help you do your work in a clear and organized way that allows you to communicate that work accurately and effectively, or that trying to help you set up your work in a way that reduces the likelihood of making mistakes, or that trying to help you learn how to evaluate your work so as to find and fix the errors that you make constitutes condescension, I won't condescend your work any further. Good luck and goodbye.
I apologize for reading your messages as condescension. I understand that you are trying to help. Please help me finish this problem.
 

BobTPH

Joined Jun 5, 2013
2,198
You are making 2 errors as far as I can see. My previous post, which you seem to have ignored, pointed out the first one.

BTW, I get 4.12V when I do the calculation.

Bob
 
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