# Calculating CE BJT output resistance

#### LvW

Joined Jun 13, 2013
1,337
...And let me save us some trouble. You are absolutely right, what I said is before is confusing and what I say here is all hogwash, so please accept my apologies and just ignore it...
I am wondering what is the purpose of such a sentence.
Does it help to clarify misunderstandings? I have some doubts.
And - after reading the technical part of your last post I am pretty sure that there is a misunderstanding regarding the term "to turn off".

* If you read my posts again you will notice that I NEVER have used the phrase "turn off" but only "set V=0" because this is a clear term that cannot be misinterpreted.

* In contrast, in your post #16 and #20 you are using the phrase "turn off", which - for my opinion - needs clarification.

* As an example - let me quote one sentence contained in your last post:
"But how did you get that small signal equivalent circuit that doesn't have any DC supplies in it? By turning off the DC supplies!"

Turning off? What does this mean?
Disconnect? Switch-off? Disable? Power-down?
I think, nothing of it is true.
Instead, the dc power is, of course, used for calculating the operating point Q and the transconductance gm (which depends on Q). And after this has been done, a small-signal model can be drawn which simply does not show the dc supply (because of the nature of the small-signal model).
OK - you call this a "switch off"-condition, and thats the reason for the misunderstandings. Couldnt you imagine?
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You have a lot of experience in communicating with people from other (non-english speaking) nations - didnt you experience that such things happen from time to time?
Therefore, I think, there is really no cause to be ironically (your last sentence). Instead, its better to explain your opinion in greater detail - thus, decreasing the probability of misunderstandings.

EDIT: As another example for the confusion caused by the term "turn off" look at MrBuggys post#9 where he is asking if "disconnection" of a source is allowed.

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#### WBahn

Joined Mar 31, 2012
26,398
I'll certainly allow that "set V=0" is clearer than "turn off". However, the use of the terminology "turn off" is widespread, in part because it is generic and applies equally to both voltage and current sources. If we only use "set V=0" then that gets cemented into people's minds and when they encounter a current source, they "set V=0" because that is what they have heard over and over. The better terminology is "set the output to zero".

I would say that what I am primarily guilty of is not making clear what is meant by turn off. This is a result of having made it repeatedly clear many, many times that "turning off" a supply means setting it's output to zero and then failing to recognize that I am now talking to an audience that has not heard be say that over and over.

As we get to this material, usually first encountered with superposition, I explain very explicitly that, for a voltage supply, that means setting the output voltage to zero. It can have any current, but the voltage across it is zero. Well, this describes an ideal piece of wire, so "turning off" a voltage supply, from a circuit analysis standpoint, is equivalent to replacing it with a short circuit. In contrast, setting the output current of a current source to zero means that it can have whatever voltage across it but no current through it. Well, this describes an open circuit, so "turning off" a current source is equivalent to replacing the source with an open circuit. I state this many times in the early part of a class.

As for "turn off" leading to confusion about "if disconnection" of a source is allowed, a similar confusion exists with "set V=0" since, with a physical circuit, you do NOT set the voltage of the DC supplies to zero. In both cases, "turn off" and "set V=0" are circuit analysis techniques that can seldom be applied with a physical circuit.

#### LvW

Joined Jun 13, 2013
1,337
Hello WBahn, thanks for clarification.

As for "turn off" leading to confusion about "if disconnection" of a source is allowed, a similar confusion exists with "set V=0" since, with a physical circuit, you do NOT set the voltage of the DC supplies to zero. In both cases, "turn off" and "set V=0" are circuit analysis techniques that can seldom be applied with a physical circuit.
Yes, of course. As I have mentioned in my former post, in case of a small-signal model I never would say "set Vdc=0" or "turn-off V,dc" because both terms do not apply at all.
The dc supply simply does not appear in the drawing because it is no circuit diagram. It is just an equivalent system that contains some parts and some artificial blocks (transconductance or controlled sources) and applies for signals only. And the dc supply belongs not to the class of signals. Thats all.

#### anhnha

Joined Apr 19, 2012
886
But it does not HAVE to be zero -- that just makes the computation easier.

Let's show this by example.

I have a block box containing a circuit that has a Thevenin equivalent circuit consisting of Vth in series with Req. I want to find both Vth and Req but can't change anything inside the box nor am I allowed to open the terminals or to short the terminals. I can, however, get measurements on the terminal voltages and the terminal currents.

If your instructor, or anyone, maintains that the sources (other than your test source) MUST be zero, then they would have to conclude that it is impossible to determine Req for this black box circuit.

Is it impossible?

Of course not. There are two obvious ways to do it. Put a test source across the terminals (either voltage or current, we'll use a voltage source) and collect current data for two different values of the test source voltage.

I1 = (V1 - Vth)/Req
I2 = (V2 - Vth)/Req

I1*Req = (V1 - Vth)
I2*Req = (V2 - Vth)

Vth = V1 - I1*Req

I2*Req = (V2 - (V1 - I1*Req))

I2*Req = V2 - V1 + I1*Req

I2*Req - I1*Req = V2 - V1

(I2 - I1)*Req = V2 - V1

Req = (V2 - V1)/(I2 - I1)

Remember what I said about the resistance being the slope of the voltage-current curve?

But notice that

Req ≠ V2/I2 ≠ V1/I1

Similarly, you can find both Req and Vth by placing two different values resistors across the terminals and solving the two resulting equations for the two unknown parameters, Vth and Req.

See? Two unknowns, two data points needed. But they can be ANY two different data points.

By turning off the internal sources, we have simply created the situation in when we KNOW one data point without actually taking a measurement -- when know that V=0 when I=0 and vice-versa. So let's take just one measurement, V1 and the resulting I1, and then assume that I2=0 if V2=0 (or that V2=0 if I2=0). Thus, we DO have that

Req = V1/I1

And only a single measurement is needed -- but we can only find one of the two parameters, too.
This seems interesting but I wonder how this method works to calculate output impedance of the circuit in post #1 as it have up to two ports not only one.

#### WBahn

Joined Mar 31, 2012
26,398
This seems interesting but I wonder how this method works to calculate output impedance of the circuit in post #1 as it have up to two ports not only one.

It doesn't matter how many ports the circuit has, as long as the circuit is linear -- and that has to be true if you are going to zero all of the other supplies, too.

#### anhnha

Joined Apr 19, 2012
886
Not quite understand.
In your example, there is only one port. To find the impedance seen at that port we need to measure two times.
First time: V1, I1
Second time: V2, I2
Req = (V2 - V1)/(I2 - I1)

Now apply that method for the circuit below to find output impedance.
Example 1: Not zeroing input voltage source.
Apply a voltage source at the output port and do two measurements as above.
And we get:
Zout = (V2 - V1)/(I2 - I1)
Example 2: Zeroing input voltage source.

Apply a voltage source at the output port and do two measurements as above.
And we get:
Zout = (V4 - V3)/(I4 - I3)

What is the reason for choosing the method in Example 2? I don't see why it makes easier in this case. Maybe, I am missing something.

#### LvW

Joined Jun 13, 2013
1,337
What is the reason for choosing the method in Example 2? I don't see why it makes easier in this case. Maybe, I am missing something.
The anser was given, for example, already in post#8.

#### anhnha

Joined Apr 19, 2012
886
The anser was given, for example, already in post#8.
I am confused.

You said:
(post #8)
A resistor across a port is measured by connecting a voltage source across this port and measure the resulting current. Thus, it is clear hat the observed current must be caused by THIS voltage source only (and not by any other source in the circuit).
Therefore, the input signal must be zero.
WBahn:
(post #10)
But it does not HAVE to be zero -- that just makes the computation easier.

#### Jony130

Joined Feb 17, 2009
5,251
We use "Example 1" when we want to measure Zout, Zin in real amplifier on the benchtop.
We use "Example 2:" if we want to use circuit theory to find theoretical Zout.

#### anhnha

Joined Apr 19, 2012
886
We use "Example 1" when we want to measure Zout, Zin in real amplifier on the benchtop.
We use "Example 2:" if we want to use circuit theory to find theoretical Zout.
Are Zout from two this example approximately equal?

#### LvW

Joined Jun 13, 2013
1,337
I am confused.
Why? What is not clear in post#8 ?

* Without zeroing the input source you have to perform two mesurements with two different voltages at the otput (two equations for two unknown quantities) , and
* with Vin=0 only one single measurement is necessary (one equation for the ratio Vout/out)

#### WBahn

Joined Mar 31, 2012
26,398
Not quite understand.
In your example, there is only one port. To find the impedance seen at that port we need to measure two times.
First time: V1, I1
Second time: V2, I2
Req = (V2 - V1)/(I2 - I1)

Now apply that method for the circuit below to find output impedance.
Example 1: Not zeroing input voltage source.
Apply a voltage source at the output port and do two measurements as above.
And we get:
Zout = (V2 - V1)/(I2 - I1)
Example 2: Zeroing input voltage source.

Apply a voltage source at the output port and do two measurements as above.
And we get:
Zout = (V4 - V3)/(I4 - I3)

What is the reason for choosing the method in Example 2? I don't see why it makes easier in this case. Maybe, I am missing something.
That was addressed in posts 7 and 10. In the second method you only have to take one measurement because you know that if V3=0 then I3=0.

#### WBahn

Joined Mar 31, 2012
26,398
Are Zout from two this example approximately equal?
If the circuit is reasonably linear, they should be reasonably close.