# Trying to solve a circuit analysis problem

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24

In this problem I need to calculate the value of the emitter current (Ie). I know that the answer is 1,3 mA but I don't know how to calculate that. I don't know how to correctly use the β -> ∞ part and I don't understand what Vces is. My thoughts were that, because I know that β = Ic/Ib, we can assume that the value of Ib is really really small. We also know that Ie = Ib + Ic so because we assume that Ib is really small we can say that Ie = Ic. That is where I don't know how to proceed.

#### Ylli

Joined Nov 13, 2015
1,053
If Beta is approaching infinity, that would indicate there is little or no base current. With no base current the base voltage will be zero.

Vbe = 0.7 volts, so the emitter voltage will be -0.7 volts, and the voltage across the 1K emitter resistor will be 1.3 volts. 1.3 volts across 1K yields a current of 1.3 mA.

Just as a check, collector current will be emitter current - base current, and since base current is zero, the collector current also equals 1.3 mA. With 1.3 mA of current though the 3.6K collector resistor, we can calculat that the collector voltage is about 5.32 volts. Vce = 5.32 + 0.7 = 6.02 volts. The transistor is no where near saturation.

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24
If Beta is approaching infinity, that would indicate there is little or no base current. With no base current the base voltage will be zero.

Vbe = 0.7 volts, so the emitter voltage will be -0.7 volts, and the voltage across the 1K emitter resistor will be 1.3 volts. 1.3 volts across 1K yields a current of 1.3 mA.

Just as a check, collector current will be emitter current - base current, and since base current is zero, the collector current also equals 1.3 mA. With 1.3 mA of current though the 3.6K collector resistor, we can calculat that the collector voltage is about 5.32 volts. Vce = 5.32 + 0.7 = 6.02 volts. The transistor is no where near saturation.
Why is the emitter voltage -0.7 V and how do you know that voltage across the 1K resistor is 1.3 V ?

#### Ylli

Joined Nov 13, 2015
1,053
Base voltage is zero, and you are given Vbe = 0.7. So the emitter will be at -0.7 volts. With the emitter at -0.7 volts and an emitter source voltage of -2.0 volts, that leaves 1.3 volts to be dropped across the 1K emitter resistor.

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24
Base voltage is zero, and you are given Vbe = 0.7. So the emitter will be at -0.7 volts. With the emitter at -0.7 volts and an emitter source voltage of -2.0 volts, that leaves 1.3 volts to be dropped across the 1K emitter resistor.
I see. That clears things up a lot. Do you mind explaining me the part in your first comment where you wrote Vce = 5.32 + 0.7 = 6.02 V
Why did you sum these 2 voltages to get Vce? Don't we already know that Vce is 0.7 V? And why isn't the transistor saturated, as you wrote? In what situation would it be saturated and what those that mean?
Sorry if I'm a bit annoying I just want to understand this completely.

#### Ylli

Joined Nov 13, 2015
1,053
With the very high beta, there is virtually no base current, so the emitter current and the collector current will be the same.

With 1.3 mA of emitter current, we will also have 1.3 mA of collector current. That collector current flows through the 3.6K collector resistor. 1.3 mA flowing through a 3.6K resistor will drop 4.68 volts across that resistor (E = I*R). We have a collector source voltage of +10.0 volts, and a drop of 4.68 volts across the 3.6K resistor, leaving +5.32 volts on the collector (10 - 4.68 = 5.32).

We earlier determined that the voltage on the emitter is -0.7 volts, so with a collector voltage of 5.32 volts there will be a total voltage across the transistor (Vce) = 5.32 + 0.7 = 6.02 volts. (technically it is 5.32 -(-0.7), the difference between the two points, but that looks odd...)

Our 'data sheet' tells us that if the transistor was saturated (conducting the most it can) the voltage between collector and emitter would be about 0.2 volts. The actual Vce is much greater than that, so the transistor is not in saturation.

#### Ylli

Joined Nov 13, 2015
1,053
For a saturated condition, assuming ideal conditions and a Vces of 0.2 volts.

We are given the condition that there is no significant base current (beta near infinite), so there will be no voltage drop across the base resistor. The base voltage will be zero. And with Vbe = 0.7, so no matter what we do, the emitter voltage is going to be -0.7 volts.

If the emitter voltage is -0.7 and the Vces is 0.2 volts, that means the collector voltage would be -0.7 + 0.2 = -0.5 volts if the transistor was in saturation. If the collector voltage is at -0.5 volts, the voltage drop across the 3.6K collector resistor would need to be 10.0 + 0.5 = 10.5 volts.

To get 10.5 volts across that 3.6K resistor would require a collector current of (I = E/R) 2.92 mA. Remember that collector and emitter currents are the same. So you would need to supply an emitter current of 2.92 mA to put the transistor into saturation.

With a -2.0 volt emitter source voltage and an emitter voltage of -0.7 volts, you tell me what value of emitter resistor would be required to get put the transistor just into saturation.

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24
Alright so I made an equation like this:
-0.7 V - 2.92 mA * Re (emitter resistor) + 2 V = 0
From that I can calculate that Re would have to be 445.2 μΩ in order for the transistor to be saturated. Am I right?

#### Ylli

Joined Nov 13, 2015
1,053
-2.0 volts emitter source, -0.7 volts at the emitter means a 1.3 volt drop across the emitter resistor.
R = E/I
R = 1.3 volts / 2.92 mA = 0.4452 K ohms, or 445.2 ohms.

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24
Oh right, I miss calculated. Anyway thank you so much for your help I definitely couldn't have done it without you. Are you available if I maybe need some more help with similar problems?

#### Ylli

Joined Nov 13, 2015
1,053
I'm sure someone will be. But this does look like homework, and most here would like to see your best effort before actually giving you any answers or methods. Expect to need to do that in the future.

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24
I'm sure someone will be. But this does look like homework, and most here would like to see your best effort before actually giving you any answers or methods. Expect to need to do that in the future.
This isn't homework. This is a task from a high school electronics competition from 2 years ago. I'm trying to solve as many of them as I can and think about applying for the competition this year.

#### Ylli

Joined Nov 13, 2015
1,053
So it is self-imposed homework.

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24

#### MrAl

Joined Jun 17, 2014
8,479

In this problem I need to calculate the value of the emitter current (Ie). I know that the answer is 1,3 mA but I don't know how to calculate that. I don't know how to correctly use the β -> ∞ part and I don't understand what Vces is. My thoughts were that, because I know that β = Ic/Ib, we can assume that the value of Ib is really really small. We also know that Ie = Ib + Ic so because we assume that Ib is really small we can say that Ie = Ic. That is where I don't know how to proceed.

Hello there,

If you write the nodal equations (or any other method) you get this:
ib=0.7/(Rb-Re*B-Re)

where Re is the emitter resistor, Rb is the base resistor, B is the Beta, ib is the base current.
You then take the limit as B goes to infinity and you get:
ib=0

That is how you determine that the base current is zero mathematically.

So you just write regular equation(s) and then take the limit as B (Beta) goes to +infinity. That's not too hard to do.

There is a question that comes up however, and that is can the voltage drop from collector to emitter really be 0.2 volts under these conditions.
Calculate the collector voltage too and find out.

Last edited:

#### WBahn

Joined Mar 31, 2012
26,398
This isn't homework. This is a task from a high school electronics competition from 2 years ago. I'm trying to solve as many of them as I can and think about applying for the competition this year.
Now remove the constraint that beta is infinite. Make it finite (but unknown for now) and find the collector current as a function of beta.

You might then change things around and find the value of the collector resistor, as a function of beta, that places the transistor at the saturation point.

Then shake it up a bit more and find the value of the emitter resistor, again as a function of beta, that places the transistor at the saturation point.

Then change things again and set the emitter resistor equal to the collector resistor and, again as a function of beta, find the value that places it at the saturation point.

Each of these forces you to look at the same circuit from a different perspective and work with different aspects that you can control versus must remain fixed.

#### MrAl

Joined Jun 17, 2014
8,479
-2.0 volts emitter source, -0.7 volts at the emitter means a 1.3 volt drop across the emitter resistor.
R = E/I
R = 1.3 volts / 2.92 mA = 0.4452 K ohms, or 445.2 ohms.
Hi,

Funny, i get 445.7 Ohms.

Thread Starter

#### nikola1912

Joined Oct 25, 2017
24
Hello there,

If you write the nodal equations (or any other method) you get this:
ib=0.7/(Rb-Re*B-Re)

where Re is the emitter resistor, Rb is the base resistor, B is the Beta, ib is the base current.
You then take the limit as B goes to infinity and you get:
ib=0

That is how you determine that the base current is zero mathematically.

So you just write regular equation(s) and then take the limit as B (Beta) goes to +infinity. That's not too hard to do.

There is a question that comes up however, and that is can the voltage drop from collector to emitter really be 0.2 volts under these conditions.
Calculate the collector voltage too and find out.
No it cannot. The
Hello there,

If you write the nodal equations (or any other method) you get this:
ib=0.7/(Rb-Re*B-Re)

where Re is the emitter resistor, Rb is the base resistor, B is the Beta, ib is the base current.
You then take the limit as B goes to infinity and you get:
ib=0

That is how you determine that the base current is zero mathematically.

So you just write regular equation(s) and then take the limit as B (Beta) goes to +infinity. That's not too hard to do.

There is a question that comes up however, and that is can the voltage drop from collector to emitter really be 0.2 volts under these conditions.
Calculate the collector voltage too and find out.
Collector voltage is 5.32 V so Vce cannot be 0.2 V because emitter voltage is -0.7 V. What do we know because of this? That the transistor is in active mode ?

#### Jony130

Joined Feb 17, 2009
5,243
What do we know because of this? That the transistor is in active mode ?
Exactly this. Your BJT is in active region because Vce > 0.2V.

#### MrAl

Joined Jun 17, 2014
8,479
No it cannot. The

Collector voltage is 5.32 V so Vce cannot be 0.2 V because emitter voltage is -0.7 V. What do we know because of this? That the transistor is in active mode ?
Hi,

Please restate your point of view as it is not that clear from what you said so far.

Intuitively, if we have a Beta of say 100 and we get a base current of 1ma, then with a Beta of 1000 we would get a base current of 0.1ma, and with a Beta of 10000 a base current of 10ua, and etc., etc. So the higher the Beta the less base current it takes to get any particular finite current in the collector. It makes sense then that with a Beta that tends to infinity that the base current goes down to either zero or very very close to zero.
We can look closer at the math behind this if you like, but if you still feel differently about it now you can post your math to prove otherwise and i'll be sure to go over it carefully.