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Impedance of active devices

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zachtheterrible

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I understand very well how there is impedance of capacitors and inductors, simply the resistance that it has to AC. But I can't quite grasp exactly what the input and output impedance of a transistor or amplifier is.

It seems in this book that I am reading, all of a sudden I am expected to understand this! Could somene help me understand this? thanx :lol:
 
zachtheterrible said:
I understand very well how there is impedance of capacitors and inductors, simply the resistance that it has to AC. But I can't quite grasp exactly what the input and output impedance of a transistor or amplifier is.

It seems in this book that I am reading, all of a sudden I am expected to understand this! Could somene help me understand this? thanx :lol:
One definition of impedance is the change in voltage divided by the change in current. If you know calculus, this can be written as Z=dv/di.
So, for instance, if you drive the base of a common emitter stage with 10uA dc, and you change tha current by 10nA, and this causes the base voltage to change by 25uV, then the input resistance (at this bias current) would be

Rin=25uV/10nA=2.5kohms.

This value changes as the base bias current changes.

If you do a similar measurement at high frequency where the input capacitance draws current, the input impedance would be a complex number, meaning it would have a magnitude and a phase angle.
 
i have a hard time actually understanding exactly what is going on with impedences also. i know that in transmitters the line and antena have to match. and i understan how the antena and line impedences are derived. the part that ii dont understand is the output impedence. lets just say for a simple example that i had a 741 op amp. what would determine the output impedence. is it something that is buit in or is it something that can be changed externally. is it actually an impedence on the output or is that just the impedence that is recomended on the output for optimal performance?
 
The example that made the idea of amplifier output impedance click for me was a simple push-pull two MOSFET setup. Imagine that amp is trying to drive its max output voltage into a load. In this case the top transistor will be all the way on (idealy). Even when all the way on the transistor still has Rdson resistance based on its gate voltage. Now you have Rdson in series with your load resistor. If the load resistor is smaller or about the same size as Rdson the amp can't reach the desired output voltage because of the internal resistance of the driving transistor.

This push pull setup is used for digital I/O pins on a uC. The datasheet often talks about the minimum digital '1' voltage when driving the maximum pin current. The reason this voltage spec isn't just the supply voltage is the output impedance as described above.
 
i know that in transmitters the line and antena have to match.

This is a different sort of idea than an amplifier's output impedance which I talked about above. With transmission lines and antennas you have to think about the AC impedance of wires themselves. When you have a simple wire it had capacitance, inductance and resistance. For low frequencies you can ignore the capacitance and inductance because they are small. But as you get to higher frequencies the inducance and capacitance starts to have an effect. You can have a cable that has 50Ohm impedance at RF frequencies but only has an Ohm or so impedance at DC. Its important that the the impedances match because the signal will bounce off any impedance change and create all sorts of weird effects.
 
One definition of impedance is the change in voltage divided by the change in current. If you know calculus, this can be written as Z=dv/di.
So, for instance, if you drive the base of a common emitter stage with 10uA dc, and you change tha current by 10nA, and this causes the base voltage to change by 25uV, then the input resistance (at this bias current) would be

Rin=25uV/10nA=2.5kohms.

This value changes as the base bias current changes.

If you do a similar measurement at high frequency where the input capacitance draws current, the input impedance would be a complex number, meaning it would have a magnitude and a phase angle.

i think that this is probably the exact thing that i need to understand. unfortunately i don't know calculus :cry:

bmcculla, that's a good example. What is Rdson?
So output impedance is that the output voltage cannot be the same as the source voltage?

Well then, what does input impedance mean? That the signal that actually gets processed cannot equal the amplitude of the original signal?

I think things are starting to click :lol:
 
Zach, I drew this simple one-tranny audio amp as an example. I'm driving it with a 1 millivolt AC signal, swept from 10Hz to 100kHz. Below is a plot of V(in)/I(in) (remember Z=V/I). Keep in mind that we are in the frequency domain here, not in the time domain like you see on a scope. Notice that the units on the vertical axis are in ohms. This is the input impedance of the amplifier.
At midband, it is about 6kohms. This will vary with the beta of the transistor, but will always be less than 8.8k (midband), because that's the equivalent resistance of the bias network (R4||R3). At low frequencies it goes up because of the 1uF input coupling cap. At high frequencies it goes down because of the input capacitance of the transistor.
This may still be baffling, but it's basically the same idea as an ohmmeter, only it's AC instead of DC. You apply a voltage and measure the current, or you apply a current and measure the voltage, then calculate V/I.
 

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Rdson is Resistance drain to source when the transistor is on.

So output impedance is that the output voltage cannot be the same as the source voltage?

That is the result of output impedance. Output and input impedances are convenient because they let you treat the amplifier as an ideal amplifoer with a resistor to ground on the input and a series resistor on the output.

Say your input is a simple resistor divider made from 2 10k resistors. If you have a nice high impedance input, like 1Mohm, there won't be any loading problems. This would be just like putting a 1Mohm resistor in parallel with the bottom 10k resistor - it won't have any real effect. However if you have a really low impedance input, like 100ohms, the parallel resistance will cause the input to change. The input voltage would be more like a voltage divider with 10kohms and 100ohms causing really poor readings of the input voltage.

This is why signals with a high output impedance need a high impedance amplifier to be amplified well.

The idea behind calculus is really simple. Z = dv/di just means for a really small change in V and I. This is the slope if the I, V curve for the amp at a specific point - like Z = (V2-V1)/(I2-I1) where the two points are close. All calculus does is make the two points infinitely close.
 
BMCULLA, THANK YOU SO MUCH!!! THAT MAKES SO MUCH SENSE! WOW!
from those few simple words, i get it!

so the ideal amplifier has infinite input impedance, and no output impedance, correct?
and a high input amplifier could amplify either a low impedance output or a high impedance output, correct?

What causes high and low impedance in an amplifier? is it because of the coupling capacitor? or does it have to do with the construction of the transistor? (lets keep it to single transistor amplifiers to keep things simple)

i still don't quite get the Z=v/i. i mean i get the calculation, but i don't quite get how to apply it to a circuit. sorry for my thickheadedness, i know you've already explained it, and im sure its a very good explanation, but . . . i dont get it :lol:
 
so the ideal amplifier has infinite input impedance, and no output impedance, correct?

When its a voltage amp yes.

a high input amplifier could amplify either a low impedance output or a high impedance output, correct?

Yes. The downside to really high input impedances is they pick up a lot more noise; they can detect tiny currents.

What causes high and low impedance in an amplifier? is it because of the coupling capacitor? or does it have to do with the construction of the transistor? (lets keep it to single transistor amplifiers to keep things simple)

Any input capacitors or resistors will affect the input impedance. There are also capacitors that are a product of the transistor construction. Also the bigger the transistors (more silicon) the lower the output resistance but you get more input capacitance (for MOSFETS at least: my schooling was a bit weak on BJT analysis).

R = V/I is the defining equasion for resistance. Resistors have a linear I-V curve - resistance is always the same no matter what voltage and current - so the slope is always the same. Nonlinear devices like transistors have an I-V curve that changes depend on curent and voltage. If you want to find the equivalent resistance at a specific point on the curve all you need to do is find the slope of the curve at that point. dv/di is just the calculus way to get the slope at a specific point on the I-V curve.
 
Hi Zach,
You can make any ordinary bipolar transistor have a higher input impedance by reducing its input bias current. You can also "bootstrap" its biasing resistor with its output signal that is the same as its input signal. Therefore with the same signal at each end of the biasing resistor, no current flows in it and its impedance is extremely high.
Of course, junction field effect transistors have a very high input impedance because their input is reverse-biased and doesn't draw current, and Mosfets don't even have a junction.

The output impedance of an opamp circuit is extremely low at low frequencies (but not very powerful) due to thier high voltage gain of a few hundred-thousand and negative feedback. If a load tries to pull an output up or down, the negative feedback causes the output to do the opposite so that the output voltage hardly changes. Ohm's law shows that a tiny voltage change divided by a fairly high current into the load equals a very low resistance.
 
Wow, thank you so very much everybody, you have made the mystery of impedance so much simpler! i'm not kidding you, if i hadn't found this forum about a year ago, i would've given up electronics forever because of this exact kind of thing.

I know that I don't fully understand impedance yet, but I have the knowledge that I need for now, just to get through this chapter. If I need more help (and I probably will) I will make sure to bring this thread back up.

thanx again :lol:
 
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