Need help with a homework!

Thread Starter

Matherios

Joined Feb 25, 2018
74
Hello,i got a homework from a friend but im stuck and i would apprieciate if i could have some help.
The homework asks how much the value of RE should be to have the transistor in saturation for β=10.It gives VEB=0.7V and VEC(SAT)=0.2V
Im really amateure and sorry for my bad english.
pnp transistor.jpg
 
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ericgibbs

Joined Jan 29, 2010
8,886
hi Matherios,
Welcome to AAC.
Could you post your attempt at solving the problem so that we can suggest ways of getting you 'unstuck'.
E
 

WBahn

Joined Mar 31, 2012
24,854
Given the information provided, can you determine what the voltage is at the emitter of the transistor?

Can you determine what the voltage is at the collector?

Can you determine the various currents from there?

Make your best attempt and show your work.
 

WBahn

Joined Mar 31, 2012
24,854
Hi ericgibbs,
I haven't done similar problem,that's why i don't know how to approach it.
Throughout your career you will face lots of problems that you haven't done before; that's what engineering is all about -- solving other people's problems.

So you need to apply what you DO know to the information that is given and see if you can tease out additional information.

For instance, IF you knew the voltage at the collector of the transistor, could you find the current in the 1 kΩ resistor? If so, then can you figure out how to find the voltage at the collector of the resistor?
 

Thread Starter

Matherios

Joined Feb 25, 2018
74
Given the information provided, can you determine what the voltage is at the emitter of the transistor? 0,7V

Can you determine what the voltage is at the collector? 0.5V

Can you determine the various currents from there? Ic=-5,5mA

Make your best attempt and show your work.
If i have something wrong please tell me.
 

WBahn

Joined Mar 31, 2012
24,854
If i have something wrong please tell me.
I think you've got it so far. I can't tell for sure because you don't define a direction for your collector current, so when you say that it is -5.5 mA, I don't know if that is 5.5 mA entering the collector or leaving the collector because there's no reference for the direction of the current.

So take the original schematic and annotate it with the voltages and currents that you will be referring to in your work.

What do you think the next step should be?

Remember, your end goal is to find the proper value for the emitter resistor. So if you can find the voltage across that resistor and the current through that resistor, you have what you need to find the resistance.
 

Thread Starter

Matherios

Joined Feb 25, 2018
74
I think you've got it so far. I can't tell for sure because you don't define a direction for your collector current, so when you say that it is -5.5 mA, I don't know if that is 5.5 mA entering the collector or leaving the collector because there's no reference for the direction of the current.

So take the original schematic and annotate it with the voltages and currents that you will be referring to in your work.

What do you think the next step should be?

Remember, your end goal is to find the proper value for the emitter resistor. So if you can find the voltage across that resistor and the current through that resistor, you have what you need to find the resistance.
I found Ib from Ib=Ic/β
Then with Ib and Ic i found Ie and with ie i found Re which if im not wrong it's 710ohm
 
Last edited:

WBahn

Joined Mar 31, 2012
24,854
Please, show your work!

Don't just give us the key equations you used and the final result. Show your work!

Remember, you are asking strangers to help you for free. So make it easy for them to do so. Show your work!

It is MUCH easier to follow your well-laid out work and verify that it is correct than to have to solve it ourselves and compare to your answers. Furthermore, if you DO make a mistake (even if it's just adding 10 and 20 and coming up with 50) then all we can do is tell you that you are wrong, where if you show your work we can tell you that you might want to check your math in line #4.
 

WBahn

Joined Mar 31, 2012
24,854
The only thing that you have wrong is that your Re is negative. You have

Re = (5 V - 0.7 V) / (-6.05 mA) which is -710 Ω

That should raise a red flag. NEVER just flip the sign because you know the answer you want must be positive. Always assume that the reason it came out negative is because you messed something up.

In this case you have defined the collector current to be the current flowing into the collector pin, which is fine (it is often defined the other way for a PNP, but at the end of the day it's arbitrary).

But that means that your Ie = Ic + Ib equation requires that Ib and Ic be the current flowing INTO the base and collector, respectively, while the Ie is the current flowing OUT of the emitter.

But your voltage in your Re equation yields the current flowing INTO the emitter.

You need to try to be a bit more precise in your work -- but you are well on the way.

Also, if you are reporting results to three sign figs, the final result should be 711 Ω.

Good job.
 

Thread Starter

Matherios

Joined Feb 25, 2018
74
The only thing that you have wrong is that your Re is negative. You have

Re = (5 V - 0.7 V) / (-6.05 mA) which is -710 Ω

That should raise a red flag. NEVER just flip the sign because you know the answer you want must be positive. Always assume that the reason it came out negative is because you messed something up.

In this case you have defined the collector current to be the current flowing into the collector pin, which is fine (it is often defined the other way for a PNP, but at the end of the day it's arbitrary).
Thank you for your help sir.

But that means that your Ie = Ic + Ib equation requires that Ib and Ic be the current flowing INTO the base and collector, respectively, while the Ie is the current flowing OUT of the emitter.

But your voltage in your Re equation yields the current flowing INTO the emitter.

You need to try to be a bit more precise in your work -- but you are well on the way.

Also, if you are reporting results to three sign figs, the final result should be 711 Ω.

Good job.
 
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