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2 transistor circuit input output question.

brandonb916

Jan 26, 2016
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Let me start by saying thank you for reading this. I hate signing up just to ask a question, but I would really appreciate any help. I am studying for an employment test and it is on stuff I learned in the Marine Corps more than a decade ago, so I'm a little rusty.

I actually have the answers to the problem I have, but I don't quite understand how they came about the 8 volts. I get that the transister lowers voltage by .7 and that's why the second answer is 7.3. I know this is probably fairly simple, but I just can't make sense of it. I have written the answers from the key in the blanks. I am more than willing to put in the work myself and I don't expect to be spoon fed. So if any of you have a suggestion on where I can look, I'd very much appreciate it.homework.jpg
 
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dorke

Jun 20, 2015
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If you do a Tevenin equivalent of the 5k and 20K bias resistors, you will get a voltage Vbb =8V and a series resistor of 4k.



i10K=VE/10k=(VB-0.7)/10K
ib=(8-VB)/4k
i10k=2*iE=2*(beta+1)*ib=2*(beta+1)*(8-VB)/4k

thus
2*(beta+1)*(8-VB)/4k=(VB-0.7)/10K
solve for VB and make an assumption on beta(if not given)

for beta=30 we get VB=7.95v
for beta=100 we get VB=7.98v
 

brandonb916

Jan 26, 2016
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Thank you very much for the detailed answer. Certainly would not have come up with that on my own. Now to practice. Thanks again!
 

Tha fios agaibh

Aug 11, 2014
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If you do a Tevenin equivalent of the 5k and 20K bias resistors, you will get a voltage Vbb =8V and a series resistor of 4k.
Dorke: Pardon my ignorance, but how did you arrive at the 4k series equivalence?

John
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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Dorke: Pardon my ignorance, but how did you arrive at the 4k series equivalence?

John
That would be the parallel resistance of 5 kΩ and 20 kΩ. When calculating the Thevinin equivalent of the bias circuit at the base, you assume there is zero impedance in the 10 V voltage source which, along with the ground, means the two resistors are effectively in parallel. Base current is considered negligible for this bias calculation, so the Thevinin equivalent voltage is just the divided voltage at the junction of the two bias resistors, i.e., 8 V.
 

Tha fios agaibh

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Ok, I think I get it. But I can't completely get my head wrapped around why you would parallel the 5k and 20k resistors (using thevenin eq) when calculating ib.

I want to think of the base current traveling primarily down thru the 5k into the base, then out the emitter and thru the 10k resistor. Though a portion does flow down thru the 20k
So Why is the 20k even considered since current is not being drawn up from ground thru the 20k and into the base?

Should I not think of current direction but rather the fact that the 5k and 20k are paralleled thru the power supply rendering 4k at the base?

John
 

dorke

Jun 20, 2015
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@john,
Here is a trivial Thevenin theorem explanation (not an accurate nor a very good one,just for starters).
Notes:
1.Don't understand from the above that the theorem only holds for resistive networks-it holds for any linear network (any Z combinations)!

2.The voltage/current sources are replaced by their equivalent internal resistances(only for an ideal voltage source it is 0 ohms)-look at the blue in the below pic.

3.@Hop, I need to correct you here(the first time ever ;)):
We don't neglect iB when finding the equivalent circuit.
The Thevenin voltage-Vth, is calculated at open circuit conditions ("no load").

tev.jpg
 
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Merlin3189

Aug 4, 2011
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Just wondered why we can't just say the 5k and 20k form a potential divider across the 10V?
Then Vbase = 10 x 20/25 = 8V

Only problem with that would be any base current, which you already decided to neglect.
 

Ratch

Mar 10, 2013
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Just wondered why we can't just say the 5k and 20k form a potential divider across the 10V?
Then Vbase = 10 x 20/25 = 8V

Only problem with that would be any base current, which you already decided to neglect.

That is the correct way to do it. The reason is that the 20k bias resistor is so much smaller that the effective emitter resistance of beta*10k that the transistor base current is negligible compared to the base bias current. In other words, the bottom base bias resistor will control the bias almost completely with the base current having little influence over the 8 volts base voltage. Now if you raised the bias resistors to lower the current, then you might have to take the base current into consideration and do Thevenin calculations.

Ratch
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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...

3.@Hop, I need to correct you here(the first time ever ;)):
We don't neglect iB when finding the equivalent circuit.
The Thevenin voltage-Vth, is calculated at open circuit conditions ("no load").
But you did neglect the base current, iB, when finding the equivalent circuit. You assumed there was no base current and calculated the Thevenin equivalent for the junction of the two bias resistors. If there were significant bias current, compared to the current through the bias resistor network, then that current would have to be considered as part of the Thevenin equivalent. The effect would be to lower the Thevenin equivalent voltage at the junction of the two bias resistors.

Okay, ignoring the other transistor's contribution for a moment, the voltage across the common emitter resistor plus the forward-biased base-emitter junction voltage of the second transistor has to equal the voltage at the junction of the two bias resistors. This is true no matter what the base current or the "gain" of the transistor is. However, the base current very much depends on the transistor "gain" and serves to reduce the bias voltage below the nominal 8 V that the open-circuit Thevenin equivalent circuit predicts. Let's not even consider that the base-emitter voltage is a function of the collector current and stick in an "average" value of 0.7 V if these are silicon transistors, as you did in post #2.

@dorke this may be the first time ever for you correcting me, but I have made plenty of mistakes in postings to this forum that others have been quick to point out. I just don't consider this to be one of them, for the reasons described above. Anyway, mistake or not, ignoring the base current is not going to cause any huge errors in the circuit analysis unless the transistors have very low gain.

So Why is the 20k even considered since current is not being drawn up from ground thru the 20k and into the base?
You can bias the base with a single resistor, but you don't want to. The whole idea with this circuit is to maintain the base of the second transistor at a constant voltage, against which the input is "compared" to produce a differential gain in the collector circuit. OTOH, with a single resistor, and reasonably high gain in the transistors, the base would be "pulled" almost to the supply voltage with a single 5 kΩ resistor. A lot more information on differential amplifiers can be found here.

Hop
 

dorke

Jun 20, 2015
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Hop,
You are surprising me here!
No,No,No, to think of it as neglecting anything is conceptually wrong.
NO assumptions whatsoever are made about the "load" in Thevnin's theorem.
You should go through it again,I'm sure you will see that.

As for our case here, let me prove that by an example.
Let us change both bias resistor values (5k,20k),
from K-ohm to M-ohm(5M,20M),G-ohm if you wish, it makes no difference at all ,
by your assumption iB isn't "neglactable" anymore...;)

Do we get any different value for Vth(open circuit-no load condition)?
Nop, exactly the same 8v
What would change? Rth only ,from 4k to 4M or 4G ohm...
 
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Merlin3189

Aug 4, 2011
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Isn't it simply that you ignore any load resistance in calculating the Thevinin equivalent and , here, get 8V in series with 4k. Then when you attach a load, such as the base current, you get a voltage drop in the series resistance, which causes the voltage at the base to fall below the Thevinin 8V. (You would only ever see a Thevinin voltage when there was no load.)
So here, if beta is say 100, then the base will draw roughly 7.3 microamps and drop 29mV in the 4k Thevinin series resistance, making the base voltage about 7.97V instead of 8.
 

Ratch

Mar 10, 2013
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Hop,
You surprising me here!
No,NO,NO, to think of it as neglecting anything is conceptually wrong.
NO assumptions whatsoever are made about the "load" in Thevnin's theorem.
You should go through it again,I'm sure you will see that.

As for our case here, let me prove that by an example.
Let us change both bias resistor values (5k,20k),
from K-ohm to M-ohm(5M,20M),G-ohm if you wish, it makes no difference at all ,
by your assumption iB isn't "neglactable" anymore...;)

Do we get any different value for Vth(open circuit-no load condition)?
Nop, exactly the same 8v
What would change? Rth only ,from 4k to 4M or 4G ohm...
Yes, the open circuit voltage would still be 8 volts, but now the Thevenin resistance would be comparable to 10k*beta. That would cause the bias to jump all over the place depending on what the base current is. It is recommended that the bottom base bias resistor be no more than 10% of the emitter resistor times beta (10k*beta in this case). This stabilizes the base voltage against variations of bias current. General Rule: Make the bias current much greater than the base current.

Ratch
 

dorke

Jun 20, 2015
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Isn't it simply that you ignore any load resistance in calculating the Thevinin equivalent and , here, get 8V in series with 4k. Then when you attach a load, such as the base current, you get a voltage drop in the series resistance, which causes the voltage at the base to fall below the Thevinin 8V. (You would only ever see a Thevinin voltage when there was no load.)
So here, if beta is say 100, then the base will draw roughly 7.3 microamps and drop 29mV in the 4k Thevinin series resistance, making the base voltage about 7.97V instead of 8.

Everything is fine except using the term "ignore ".
Nothing is ignored!
The "Load" is disconnected for the purpose of calculating the Vth and Rth according to Thevinin's Theorem.

See this video for a simple explanation (from 3:20 on).
and this one too,better I think.
 
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Merlin3189

Aug 4, 2011
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Yeah, fine. You say disconnect, I say potato.

But seriously, Th equivalent cct is a concept, not a real cct. If we wanted to measure VTh on a real cct, yes we would have to disconnect anything other than the part of the cct we wanted the Th equivalent for.
(But of course we would also have to disconnect our meter as well! So in reality we can't measure VTh. Nor ITh, because whatever we put in place to measure the current, is not a short cct. Hair splitting? Of course!)

Since Th equiv cct is actually a concept, we apply it to a conceptual circuit, made up of the mesh of components for which we want the Th equiv. The rest we can disconnect in our imagination (we do not need to erase them from the diagram) or we could simply ignore.
In fact, when we are trying to find the Th equiv for a certain mesh of components, we think about those components only. We don't need to even ignore the other components, because we never thought of them in the first place. To think about them, then imagine them not connected, seems a bit of a waste of effort!

And I still think bringing Th into this is a sledgehammer to crack a nut. My potential divider works even better if I disconnect the base current rather than ignoring it! I'm surprised you guys aren't using LvW's model of a transistor as well.
 

dorke

Jun 20, 2015
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@Merlin3189,
The Thevnin Theorem is a very important one In circuit theory.
Both a concept and a powerful practical lab tool(not in all cases though).
It's aim is to simplify circuit analysis.

Don't use it ,don't understand it, or whatever if you don't need to ...
But,if you do use it, you should change the way of thinking about it.
BTW,
ignorance and potatoes wont get you too far ...:(
 
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Tha fios agaibh

Aug 11, 2014
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You can bias the base with a single resistor, but you don't want to.
Thanks for the explanation but it doesn't answer my question.
I understand the Thevenin resistance calculation, but not necessarily why its used in this case. I also get that the biasing resistors forum a voltage divider.
What I don't understand is why the 20k resistor is considered, (paralled wirh 5k) when the current (as far as ib is concerned) is actually flowing down thru the 5k into the base, then out thru the 10k to ground.
So, How does the current flowing down thru the 20k impact ib?
 

Ratch

Mar 10, 2013
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Thanks for the explanation but it doesn't answer my question.
I understand the Thevenin resistance calculation, but not necessarily why its used in this case. I also get that the biasing resistors forum a voltage divider.
What I don't understand is why the 20k resistor is considered, (paralled wirh 5k) when the current (as far as ib is concerned) is actually flowing down thru the 5k into the base, then out thru the 10k to ground.

Current present in the 5k resistor splits into two paths. One path is through the 20k resistor, and the other is through transistor base and into the 10k emitter resistor. Looking in from the base side, the emitter resistor appears to be 10k * (1+beta).
So, How does the current flowing down thru the 20k impact ib?
You want the current in the 20k resistor to be at least 10 time more than the base current ib, so that the changes in the base current ib will not change the voltage bias point.

Ratch
 

Tha fios agaibh

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Current present in the 5k resistor splits into two paths. One path is through the 20k resistor, and the other is through transistor base and into the 10k emitter resistor.
Right two paths, but only one into ib via the 5k. The other path in flowing down through the 20k so I don't see the 20k as part of the equation. Or, to put it another way; Current is not flowing the other way from ground up through to 20k and into the base.
I would expect base current to be;
ib=(8-VB)/5k.. The 20k resistor seems irrelevant as far as ib is concerned.
 

Ratch

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Right two paths, but only one into ib via the 5k. The other path in flowing down through the 20k so I don't see the 20k as part of the equation. Or, to put it another way; Current is not flowing the other way from ground up through to 20k and into the base.
I would expect base current to be;
ib=(8-VB)/5k.. The 20k resistor seems irrelevant as far as ib is concerned.

No, (8-VB)/10k is the emitter current. The base current is (8-VB)/(10k*(beta+1)). As mentioned before, the emitter resistor looks like 10k*(beta+1). You need the 20k resistor to be at least 10 times smaller than 10k*(beta+1) to stabilize the base voltage.

Ratch
 
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