# Calculating CE BJT output resistance

#### MrBuggy

Joined Nov 7, 2013
18
Why do we need to zero the input voltage to calculate the output resistance of this BJT CE configuration?

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#### WBahn

Joined Mar 31, 2012
26,398
Who says you do? Doing so just makes the answer trivially obvious by inspection.

#### The Electrician

Joined Oct 9, 2007
2,866
Why do we need to zero the input voltage to calculate the output resistance of this BJT CE configuration?
I've seen that image on the web before; can you provide a link to it?

Also, can you provide a link to a statement that you need to "zero the input voltage"?

#### LvW

Joined Jun 13, 2013
1,337
Mr. Buggy,

I think, the answer is obviuos - if you remember how the output resistance is defined and how you would measure it.

#### WBahn

Joined Mar 31, 2012
26,398
We haven't mentioned this yet and since it's your first post you may not know. Before we can give any detailed help, we like to see YOUR best shot at answering the question. That gives us a starting point from which to guide you toward the correct answer.

#### MrBuggy

Joined Nov 7, 2013
18
I've seen that image on the web before; can you provide a link to it?

Also, can you provide a link to a statement that you need to "zero the input voltage"?
This is from Sedra and Smith's Microelectronic circuits book. I will provide the img of the statement later when I get home.

I went to my professor this morning and he says this is how we do it. I do remember zero-ing voltage sources when trying to calculate the Thevenin resistance of a circuit. Thinking of it I never asked why, I just blindly zeroed the sources.

My best guess would be: voltage sources see other voltage sources as "short", and see other current sources as "open". But then again... I do not know why.

#### WBahn

Joined Mar 31, 2012
26,398
This is from Sedra and Smith's Microelectronic circuits book. I will provide the img of the statement later when I get home.

I went to my professor this morning and he says this is how we do it. I do remember zero-ing voltage sources when trying to calculate the Thevenin resistance of a circuit. Thinking of it I never asked why, I just blindly zeroed the sources.

My best guess would be: voltage sources see other voltage sources as "short", and see other current sources as "open". But then again... I do not know why.
Your instructor is doing you a disservice. To find the resistance seen between to ports, you need two data points since resistance is the slope of the relationship between V and I. But if you turn off all the sources, you KNOW that one of those points is V=0V, I=0A so you only have to get one more. If any of the sources are turned on, the it is all but guaranteed that the V vs I line does NOT go through the origin.

#### LvW

Joined Jun 13, 2013
1,337
I must confess that - due to language problems - I do not understand everything WBahn has explained in his last post.
Therefore, I try to do it with my own words (which I hopefully understand):

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.

#### MrBuggy

Joined Nov 7, 2013
18
I must confess that - due to language problems - I do not understand everything WBahn has explained in his last post.
Therefore, I try to do it with my own words (which I hopefully understand):

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.
Yes this makes sense, thank you.

I'm thinking of another interpretation here. We know that ideal voltage sources have zero internal resistances. So when we calculate for output impedances we can short the voltage sources. Is that correct?

In real life situations though, when calculating the output impedance of any circuits, should we disconnect the power source of the circuit? or should we take into account the internal resistance of the source?

#### WBahn

Joined Mar 31, 2012
26,398
I must confess that - due to language problems - I do not understand everything WBahn has explained in his last post.
Therefore, I try to do it with my own words (which I hopefully understand):

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.
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.

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#### The Electrician

Joined Oct 9, 2007
2,866
But it does not HAVE to be zero -- that just makes the computation easier.
It seems to me that this sentence would make more sense with the addition of the word in red, or have I misunderstood something?

#### WBahn

Joined Mar 31, 2012
26,398
It seems to me that this sentence would make more sense with the addition of the word in red, or have I misunderstood something?
Nope, you spotted a typo that really counts. Thanks! I've updated the original post.

#### LvW

Joined Jun 13, 2013
1,337
But it does not HAVE to be zero -- that just makes the computation easier.
.
OK - of course, you are right. There are many cases for which we have two or even more alternatives for measuring circuit properies. Making two measurements for eliminating the unwanted influence is a kind of "compensation" measurement.
However, as you know, the method with Vin=0 is commonly used - and this was the initial question.

#### LvW

Joined Jun 13, 2013
1,337
I'm thinking of another interpretation here. We know that ideal voltage sources have zero internal resistances. So when we calculate for output impedances we can short the voltage sources. Is that correct?
Please, be exact (to avoid misunderstandings). I suppose you speak about SIGNAL sources, correct? Of course, all DC supplies must remain switched-on.
In case the signal source has an internal resistance, the input node must remain terminated with this resistance (only Vsignal=0).

In real life situations though, when calculating the output impedance of any circuits, should we disconnect the power source of the circuit? or should we take into account the internal resistance of the source?
You are asking if the power source has to be disconnected? Of course, not.
And remember, the signal source has to be SHORTED (set to zero), NOT disconnected.

#### WBahn

Joined Mar 31, 2012
26,398
OK - of course, you are right. There are many cases for which we have two or even more alternatives for measuring circuit properies. Making two measurements for eliminating the unwanted influence is a kind of "compensation" measurement.
However, as you know, the method with Vin=0 is commonly used - and this was the initial question.
The initial question, very explicitly, was whether we NEED to zero in the input. This was again the main point in the followup in post #6 where the OP makes clear he isn't asking what the common method is, he is trying to understand WHY it is the common method and WHY zeroing the other sources is done. Otherwise, he would have been satisfied with his instructor's response that this is the way we do it.

I applaud his inquisitiveness.

#### WBahn

Joined Mar 31, 2012
26,398
Please, be exact (to avoid misunderstandings). I suppose you speak about SIGNAL sources, correct? Of course, all DC supplies must remain switched-on.
In case the signal source has an internal resistance, the input node must remain terminated with this resistance (only Vsignal=0).

You are asking if the power source has to be disconnected? Of course, not.
And remember, the signal source has to be SHORTED (set to zero), NOT disconnected.
And, Mr Buggy, please make the distinction between calculating and measuring. Calculating is where you have the schematic and pull out a pen, some paper and a calculator and get to work. Measuring is where you pull out the device, some measurement tools such as a function generator and oscilloscope, and get to work.

If you are calculating, you turn off ALL supplies. This is because turning them off only involves changing the schematic and we are assuming a linear system that doesn't mind having the supplies set to zero and/or shorted as far as the math is concerned. If you are measuring, then that is an option that is seldom available, and this is where what LvW is saying comes into play.

Also, it generally depends on if you are looking for the DC, or large-signal, resistance or, more likely, if you are looking for the AC, or small-signal resistance. If the small-signal resistance, it may also depend on what frequency range you are interested in.

#### LvW

Joined Jun 13, 2013
1,337
The initial question, very explicitly, was whether we NEED to zero in the input. This was again the main point in the followup in post #6 where the OP makes clear he isn't asking what the common method is, he is trying to understand WHY it is the common method and WHY zeroing the other sources is done.

* In post#8 I answered the OPs question and told him WHY the signal source needs to be zero in case of calculating at the output the ratio Vout/Iout. Anything wrong?
* More than that, in my post#13 I agreed with your remark that there is another (more involved alternative).
* WBahn, what do you want in addition? What is your problem ?

#### LvW

Joined Jun 13, 2013
1,337
WBahn, I am sorry but I must place some comments to your last post
If you are calculating, you turn off ALL supplies. This is because turning them off only involves changing the schematic and we are assuming a linear system that doesn't mind having the supplies set to zero and/or shorted as far as the math is concerned. If you are measuring, then that is an option that is seldom available, and this is where what LvW is saying comes into play.
For my opinion, this sounds a bit confusing.
a) If you calculate on the basis of the small-signal model there are no dc supplies at all that could be switched off. This model assumes fixed dc bias conditions and, thus, a certain operating point.
b) If you are calculating on the basis of the circuit diagram (which happens rather often), the dc supply must NOT be switched off, of course. Otherwise, you have no operating point at all.

Also, it generally depends on if you are looking for the... large-signal resistance....
May I ask: What is the definition of a "large signal resistance"?
I suppose you mean the resistance which would exist for signals extending the small-signal region, correct? According to the definition for large-signal operation this would involve circuit non-linearities. Right?
However, in this case, both voltage and current do NOT have the same (sinusoidal) shape anymore. Question: Are we allowed to compute the voltage-to-current ratio? How should we proceed (which values: peak, rms, mean value)?

#### WBahn

Joined Mar 31, 2012
26,398
* In post#8 I answered the OPs question and told him WHY the signal source needs to be zero in case of calculating at the output the ratio Vout/Iout. Anything wrong?
* More than that, in my post#13 I agreed with your remark that there is another (more involved alternative).
* WBahn, what do you want in addition? What is your problem ?
Fine. You're absolutely right. The initial question clearly was limited to whether Vin needs to be set to zero in the method where Vin=0. So Vin must be zero. How foolish of me for even suggesting otherwise.

#### WBahn

Joined Mar 31, 2012
26,398
WBahn, I am sorry but I must place some comments to your last post

For my opinion, this sounds a bit confusing.
a) If you calculate on the basis of the small-signal model there are no dc supplies at all that could be switched off. This model assumes fixed dc bias conditions and, thus, a certain operating point.
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!

You applied superposition to the original circuit diagram that has both DC and signal supplies involved. You first turned off the signal source and kept the DC supplies on. You analyzed the circuit to find the operating points of all the devices -- primarily the nonlinear components -- that you are going to replace with their small-signal equivalents.

You then turned off the DC supplies -- which is why there are no DC supplies in the resulting small-signal diagram, and also why any connection to a DC supply that is referenced to ground (or to another ground-referenced DC supply) is treated as a small signal ground connection -- and turned on the signal supplies after making the replacements of the non-linear components with their linear small signal equivalents (and, of course, the small signal equivalents don't HAVE to be linear, they just almost always are as that is usually the main goal of using them).

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.