Transistor hybrid bias calculations help

Bernie564

Joined Nov 28, 2014
7
I was given the problem for homework to calculate Ib for this circuit.

Vcc=9V
βdc=300

I am happy with collector feedback bias, emitter feedback bias and voltage divider bias but this seems to combine collector and voltage divider boas and i am stuck with combining the working out how to combine or derive the formula for it.
So i thought if i could work out Ie i could then divide that by βdc to get Ib.
I tried this by removing the Re term and subtracting R1, but it doesn't seem right i think the subtracting R1 bit is wrong, but i'm not sure how to make it work.

Ie=(Vcc-Vbe)/(Rc-R1+(Rb/βdc)

Ie=(9-0.7)/(2700-1300+(2400/300)
Ie=(8.3)/(1400+8)
Ie=5.89mA and then Ib would be 19.65uA

Have i missed something when trying to combine these two types of bias? Is there a better way to think about this so i can look at the circuit and work out the right method?

Best,
Bernie

WBahn

Joined Mar 31, 2012
26,398
Forget about trying to apply some memorized formula that applies to this configuration or that configuration. Just analyze the circuit that you are given.

With a β this high, you can get close to the answer by first assuming that it is infinite and finding the collector current and then dividing by the actual beta. That will give you a value that is almost certainly within 1% unless the base circuit is particularly high impedance. Then you can calculate the actual value and have something to compare it to as a sanity check.

Assuming β=∞ :

Q1) What is the voltage at the base (relative to the emitter, which we will call 0V)?
Q2) What is the current in R1?
Q3) What is the current in Rb?
Q4) What is the voltage at the collector?
Q5) What is the current in Rc?
Q6) What is the collector current?
Q7) What is the base current if β=300?

Now perform the analysis taking into account the finite β from the beginning?

Bernie564

Joined Nov 28, 2014
7
Forget about trying to apply some memorized formula that applies to this configuration or that configuration. Just analyze the circuit that you are given.

With a β this high, you can get close to the answer by first assuming that it is infinite and finding the collector current and then dividing by the actual beta. That will give you a value that is almost certainly within 1% unless the base circuit is particularly high impedance. Then you can calculate the actual value and have something to compare it to as a sanity check.

Assuming β=∞ :

Q1) What is the voltage at the base (relative to the emitter, which we will call 0V)?
Q2) What is the current in R1?
Q3) What is the current in Rb?
Q4) What is the voltage at the collector?
Q5) What is the current in Rc?
Q6) What is the collector current?
Q7) What is the base current if β=300?

Now perform the analysis taking into account the finite β from the beginning?

Thanks WBahn, I just don't feel very confident with these things and get stuck with where to start.

So
Q1 0.7V (ish)
Q2 IR1=0.7V/1300R = 0.538mA
Q3 IRb = Ib+0.538mA
Q4 I am thinking this should be a straightforward three resistor divider but i am confused by the 0.7V i think should be across R1...can you explain this further?
Q5 if i can get my head around Q4, this is V/Rc
Q6 IRc - IRb
Q7 (IRC - IRb)/300

I like that, it has started the seed of the overall solution in my head. I am a bit stuck with Q4. I am thinking that V across R1 i should just take as 0.7 V and then work out the other two resistors as if between 0v and 9-0.7 as they only see the the potential difference between those two points...is that right? Or am i barking up the worng tree?

Thanks for getting me started, much appreciated.

Best
Bernie

WBahn

Joined Mar 31, 2012
26,398
You answer to Q3 is correct, but remember that we are assuming infinite beta right now, so what would Ib be? See if that helps you along.

WBahn

Joined Mar 31, 2012
26,398
FYI: The answer assuming infinite beta to get the collector current and the actual answer assuming finite beta differ by about 0.63%.

Bernie564

Joined Nov 28, 2014
7
You answer to Q3 is correct, but remember that we are assuming infinite beta right now, so what would Ib be? See if that helps you along.
Oh yes, so with infinite gain, the transistor will be saturated?

So assuming VC = VE

Then VC = 0V
So V across VB is 0.7V so
Q3 IRB =0.7/2400 =0.291mA

I need sleep, i will pick this up again in the morning and hopefully it will become clear! Thanks for your help so far WBahn.

Bernie

WBahn

Joined Mar 31, 2012
26,398
Oh yes, so with infinite gain, the transistor will be saturated?

So assuming VC = VE

Then VC = 0V
So V across VB is 0.7V so
Q3 IRB =0.7/2400 =0.291mA

I need sleep, i will pick this up again in the morning and hopefully it will become clear! Thanks for your help so far WBahn.

Bernie
No, infinite beta does NOT mean the transistor will be saturated -- these are two completely separate concepts. The infinite beta means that the base current will be zero, that's all. If it helps, think of it as not "infinite", but rather just bitching huge. Instead of 300, what if it were a 300 million? Then if you had a collector current of 300 mA you would only have a base current of one nanoamp, which would almost certainly be completely negligible for any computations that are performed.

Bernie564

Joined Nov 28, 2014
7
No, infinite beta does NOT mean the transistor will be saturated -- these are two completely separate concepts. The infinite beta means that the base current will be zero, that's all. If it helps, think of it as not "infinite", but rather just bitching huge. Instead of 300, what if it were a 300 million? Then if you had a collector current of 300 mA you would only have a base current of one nanoamp, which would almost certainly be completely negligible for any computations that are performed.
Ahhh ok, so we are thinking along the lines of the base current being so small when compared to the c-e current that we can disregard the base current flow for initial analysis? I will come back to this after i have finished cooking and having a think about it.

WBahn

Joined Mar 31, 2012
26,398
Ahhh ok, so we are thinking along the lines of the base current being so small when compared to the c-e current that we can disregard the base current flow for initial analysis? I will come back to this after i have finished cooking and having a think about it.
Now you're getting it. Though it isn't so much that the base current is small compared to the c-e current as much as it is that the base current is small compared to the currents in R1 and Rb. This allows us to assume that these currents are the same and, therefore, that R1 and Rb form a classic voltage divider and get a result that is very, very close to the actual case. This would be a completely invalid assumption if the current in R1 were comparable to or, worse, smaller than the base current. Note that large beta doesn't, by itself, guarantee that the assumption is valid since for any value of collector current and beta, we could find values of R1 and/or Rb that would be large enough to invalidate the assumption.

But that's fine. First we make what we hope is a reasonable assumption. Second we apply that assumption to get an estimated answer. Third we evaluate whether the assumption really was reasonable. It's easy to forget that last step, but don't.

Bernie564

Joined Nov 28, 2014
7
Assuming β=∞ :

Q1) What is the voltage at the base (relative to the emitter, which we will call 0V)?
Q2) What is the current in R1?
Q3) What is the current in Rb?
Q4) What is the voltage at the collector?
Q5) What is the current in Rc?
Q6) What is the collector current?
Q7) What is the base current if β=300?

sorry still editing....will fill in in a few mins

Bernie564

Joined Nov 28, 2014
7
Assuming β=∞ :

Q1) What is the voltage at the base (relative to the emitter, which we will call 0V)?
Q2) What is the current in R1?
Q3) What is the current in Rb?
Q4) What is the voltage at the collector?
Q5) What is the current in Rc?
Q6) What is the collector current?
Q7) What is the base current if β=300?

sorry still editing....will fill in in a few mins
Assuming β=∞ :

Q1) What is the voltage at the base (relative to the emitter, which we will call 0V)?1.82V
Q2) What is the current in R1? =Collector V / Rb+R1 =1.407mA
Q3) What is the current in Rb? ditto as in series and disregarding Ib
Q4) What is the voltage at the collector? 5.205 V by divider of Rc and Rb+R1
Q5) What is the current in Rc? = 9 - 5.205/2700 = (rail - Rc drop)/Rc= 1.4055
Q6) What is the collector current? iRc - I Rb+R1 = 1.4055 - 1.407mA = -1.5uA....that can't be right!
Q7) What is the base current if β=300? ans to Q6/300

i am treating all three resistors as voltage divider ladder. Not sure this is the right way. i want base volts to be 0.7 but doing this doesnt let me.

not sure this is right

WBahn

Joined Mar 31, 2012
26,398
Assuming β=∞ :

Q1) What is the voltage at the base (relative to the emitter, which we will call 0V)?1.82V
Q2) What is the current in R1? =Collector V / Rb+R1 =1.407mA
Q3) What is the current in Rb? ditto as in series and disregarding Ib
Q4) What is the voltage at the collector? 5.205 V by divider of Rc and Rb+R1
Q5) What is the current in Rc? = 9 - 5.205/2700 = (rail - Rc drop)/Rc= 1.4055
Q6) What is the collector current? iRc - I Rb+R1 = 1.4055 - 1.407mA = -1.5uA....that can't be right!
Q7) What is the base current if β=300? ans to Q6/300

i am treating all three resistors as voltage divider ladder. Not sure this is the right way. i want base volts to be 0.7 but doing this doesnt let me.

not sure this is right
If you don't show your work, you don't get partial credit. So all I can do is start from your unsupported answer to Q1 and say that it is wrong. With no work shown, I have no basis upon which to tell why it is wrong.

But consider this: What is the (assumed) base-emitter voltage on a transistor that is in the active region of operation?

Bernie564

Joined Nov 28, 2014
7
If you don't show your work, you don't get partial credit. So all I can do is start from your unsupported answer to Q1 and say that it is wrong. With no work shown, I have no basis upon which to tell why it is wrong.

But consider this: What is the (assumed) base-emitter voltage on a transistor that is in the active region of operation?
Sorry WBahn, i thought my little explanation would be ok. I treated the three resistors as in series as a voltage divider network to get the answer for Q1 but my gut says it should be ~0.7V as that would be the b-e voltage of the tansistor in the active region. If that is right then the PD seen by Rc and Rb in series is 9.0V - VR1 = 9-0.7 = 8.3V. I think my next step should be then to rework the potential divider (only Rc and Rb) to work out Vc etc etc....am i finally on the right route? I hope so and thank you for you patience.

WBahn

Joined Mar 31, 2012
26,398
Sorry WBahn, i thought my little explanation would be ok. I treated the three resistors as in series as a voltage divider network to get the answer for Q1 but my gut says it should be ~0.7V as that would be the b-e voltage of the tansistor in the active region. If that is right then the PD seen by Rc and Rb in series is 9.0V - VR1 = 9-0.7 = 8.3V. I think my next step should be then to rework the potential divider (only Rc and Rb) to work out Vc etc etc....am i finally on the right route? I hope so and thank you for you patience.
In order to treat the three resistors as in series, what has to be true? They must all have the same (or very nearly the same) current. By assuming no base current for the transistor, we can treat Rb and R1 as being in series, but are Rb and Rc in series, even approximately? If they are, then that requires that there be little to no collector current in the transistor and we have no basis for such an assumption.

But we do know that if the transistor is in the active region, that the base-emitter voltage is going to be about 0.7V and so we can start with that. Keep in mind that we are assuming that the transistor is in the active region and so this will be something we will need to verify when all is said and done.

Reworking things using a potential divider consisting of Rc and Rb won't work because Rc and Rb are not in series.

No need to thank me for my patience, but it is appreciated. Note that I am forcing you to struggle with things instead of just explaining everything because we usually learn best by fighting through our own misconceptions about things.