Hi folks,

I must be thinking in the wrong way.

I know that output impedance is defined by first setting the small signal input voltage to zero in an amplifier (say a BJT common emitter) and then measuring the output resistance from the output.

But during the working of the amplifier, Vi will become different than zero obviously, so will the output impedance actually change ?

It seems to be that the output impedance of an amplifier is not really a constant but changes because Vi changes.

Can anyone please clarify this?

Oh boy.

 

When you say "measuring" are you talking about doing this with a physical circuit, or via calculation? If for a physical circuit (or simulation, for that matter), how are you measuring the output resistance?

If the circuit is linear, the output impedance won't change with input signal -- of course this is just another way of saying that if the output impededance changes with the input signal, then the circuit isn't linear.

In general, most well-design amplifier circuit circuits are reasonably linear over their intended operating range -- but another way of saying this is that the intended operating range is the range over which the amplifier circuit is reasonably linear.

Most amplifier circuits use transistors, which are highly nonlinear. So, yes, if you change in the input signal just about every parameter associated with the circuit changes because you have changed the operating point of a nonlinear device. But in well-designed circuits operating within their intended region of operation, the changes are small and/or tolerable.

 

When you say "measuring" are you talking about doing this with a physical circuit, or via calculation? If for a physical circuit (or simulation, for that matter), how are you measuring the output resistance?

If the circuit is linear, the output impedance won't change with input signal -- of course this is just another way of saying that if the output impededance changes with the input signal, then the circuit isn't linear.

In general, most well-design amplifier circuit circuits are reasonably linear over their intended operating range -- but another way of saying this is that the intended operating range is the range over which the amplifier circuit is reasonably linear.

Most amplifier circuits use transistors, which are highly nonlinear. So, yes, if you change in the input signal just about every parameter associated with the circuit changes because you have changed the operating point of a nonlinear device. But in well-designed circuits operating within their intended region of operation, the changes are small and/or tolerable.

I mean just calculating it in the circuit diagram. Not really measuring, sorry.

I found a formula for the output impedance for example and it's this:

Zo = Vs / ( Vs/Rc + Vi/re + (Vs - Vi)/Rf )


As you can see it depends on Vi, which is the input voltage small signal.

if I set Vi = 0, then the usual thevenin impedance results.

But of course Vi will not always be zero during the circuit functioning, so this implies that the impedance will change with Vi. I thought the impedance should be constant, but apparently it isn't...

My texbook defines the output impedance as the thevenin impedance. But really if it changes with Vi then this definition is only an approximation, and I only now discovered this.

 

Which of those terms dominate Zo? If it's the term with Vi in it, then it's probably not a "well-designed" circuit.

Is that formula that you "found" the formula that applies to your circuit?

 

Which of those terms dominate Zo? If it's the term with Vi in it, then it's probably not a "well-designed" circuit.

Is that formula that you "found" the formula that applies to your circuit?

Vi will not dominate as it's in the denominator, but as you can see, if one sets Vi = 0, then the output test voltage Vs cancels out and leaves one with an output impeance of 1 / (1/Rc + 1/Rf), which is the impedance of a simple textbook resistor feedback amplifier.

What I did is I applied a test voltage Vs to the output, and calculated the output current, and then just divided Vs by the current to find Zo.

But I decided to see what happens if I don't just set Vi to zero beforehand, so I left it in the formula, and so Zo depends on it, meaning it's a function of it. I did not know this fact before and I thought that Zo would be a constant.

I found from this that Zo is not really constant but varies with Vi. So I came here to ask if this is a general known phenomenon....My textbook always defines Zo as a constant through, and sometimes it doesnt depend on Vi, but sometimes it does, however the book always sets Vi to zero before defining Zo, so it looks as if Zo never depends on Vi. But just setting Vi = 0 doesn't mean Vi is zero in the circuit, so Zo can also be changing based on Vi. Do you know what I mean ?

 

In other words, why do they always define the output impedance with Vi = 0 ? What's the point of that if the output impedance sometimes depends on Vi ?

 

The input signal must be zero for you to measure the output impedance of an amplifier.

 

The input signal must be zero for you to measure the output impedance of an amplifier.

Why ?

 

Why ?

When you try to determine the current through the unknown output resistance (caused by a voltage source that is connected at the output node) there must be not another current caused by the input signal.

 

Another thing to keep in mind is that there are two different output resistances -- large signal and small signal. So how you figure it out depends on which one you are looking for.

 

When you try to determine the current through the unknown output resistance (caused by a voltage source that is connected at the output node) there must be not another current caused by the input signal.

I know but during the working of the circuit, the input voltage will be there, and won't that change the output impedance ?

What I do not understand is why the output impedance is defined with Vi set to zero given that it clarly can depend on Vi, since any Vi at the input will change the output current and with an output test voltage at the output, the output current will also change, changing the output impedance.

 

I think maybe this is just the usual thevenin resistance that we look for when defining output impedance. So my confusion is misguided. We short the input voltage according to thevenin's theorem so thats why be must. We then could simply calculate the remaining resistance, but since there is a depended source in the amp, that is impossible without using a small test voltage at the output...

Would this be correct?

 

See

 

See
View attachment 153955

The axes are not labelled......... What voltage are you varying? Vi ?

 

I initially set the current of the transistor 1mA as a voltage source V2.The voltage of the source V3 specifies the change in the bias mode (current and voltage of the collector) by the command ".step param A -35mV 35mV 0.1mV." A "is the parameter specifying the offset value (in the X axis its value.) A sharp drop in the output resistance occurs due to the"due to low voltage on the collector (transistor in saturation mode).

 

Anyone else would like to help me understand what's going on?

my suspicion is that the output resistance is the thevenin resistance.

 

Anyone else would like to help me understand what's going on?

my suspicion is that the output resistance is the thevenin resistance.

In post #3, you say "I found a formula for the output impedance for example and it's this:
Zo = Vs / ( Vs/Rc + Vi/re + (Vs - Vi)/Rf )"

Where did you find this formula? If on the web, give us a link to it. If from a book, take a picture of that page with the formula and post it here.

Post a schematic of the circuit that the formula applies to.

It's generally better to give more information pertaining to your question, rather than less.

 

See