Impedence

[beenuseren]

Hi,

Can anyone explain this?

What is impedence matching?

Is it right that the impedence of a voltage source should be very less?
Where as the impedence of a signal source should be very high?

I am very confused. Please help me.

Thanks,

 

[FusionITR]

Impedence matching is when the output resistance is equal to the load resistance for maximum power transfer.

Pload = I^2RL

Pload is maximum when RL = Rs (Rs is the source resistance).

 

[Dr.EM]

For maximum power transfer, the output impedance of a stage should be equal to the input impedance of the next. For maximum voltage transfer (audio signlas etc) the output impedance of one stage should be 5-10 times lower than the input impedance of the next stage.

 

[RadioRon]

Impedance matching is used when one wants to pass the maximum amount of power (not the maximum voltage or the maximum current) from one stage to the next. You don't need to match impedances from source to load when you are trying to transfer a signal that is represented only by a voltage. In this case, it is best to have a low impedance source and a high impedance load so that you suffer the minimum amount of voltage drop in the source. This applies to the majority of experimenter circuits, generally speaking, including most logic circuits.

There are many times, however, when you do want to pass the maximum amount of power. For example, consider a stereo amplifier in your audio system. The transducer that creates the sound, that is the speaker, needs a lot of power to operate, and there really isn't any alternative to a speaker if you want to create a loud sound. Since it costs money to generate a lot of power, it pays to be efficient about it, hence our interest in getting as much power to the load as possible for a given amplifier circuit.

It is easiest to understand maximum power transfer by looking at the alternatives first. Let's say you made your load a very high resistance, while your source had a low output resistance. You would get a very good transfer of voltage, but not much current flow. Conversely, if you lowered your load resistance to a very low value, but left your source resistance higher, you would increase the current flow but your voltage delivered to the load would drop very low. Don't forget that power equals voltage times current so you want to find a way to have both of these be as high as possible. It just so happens that you get the maximum power when you deliver half of the available voltage and half the available current, which happens when your load resistance equals your source resistance. This is the point of maximum power transfer for resistive (non-reactive) sources and loads.

Note that I'm not using the word impedance here. This is important because it gets a bit more complicated when we say impedance. Impedance means the resistance plus reactance, a complex number, and reactance has a sign to it (that is, it can be negative or positive). I won't go through the math, but it is a fact that maximum power transfer occurs between a complex (meaning with resistance and reactance) source to a complex load when the resistance values are equal and when the reactance values are equal in magnitude but opposite in sign.

RF engineers get intimately familiar with this concept because at very high frequencies it is more convenient to measure power transfer than it is to try and measure instantaneous voltage or current at a particular point in the circuit. They learn quickly that impedance matching involves adding some capacitors and inductors between two stages to alter the impedance of a load so that it looks like the CONJUGATE of the source impedance. Conjugate is our big word meaning the same resistance and equal magnitude but opposite sign of reactance.

 

[RadioRon]

I wanted to add that the output resistance of an ideal voltage source is zero. The output resistance of an ideal current source is infinitely high. The output resistance of an RF signal source can be anything in theory but because RF engineers usually seek maximum power transfer it is customary and most convenient to set the output resistance to an industry standard, like 50 ohms for example. In this case, the output reactance must be zero, so the output impedance is also 50 ohms.


Also, to answer beenuseren's specific questions:
"Is it right that the impedence of a voltage source should be very less?" the answer is yes.

"Where as the impedence of a signal source should be very high?"
This depends on what you mean by signal source. These words are vague and describe anything that delivers a signal, so I am not sure what you mean by this. Could you explain the meaning of "signal source"?

 

[beenuseren]

Thanks everyone for the reply. I understood the concept well. Thanks so much Ron. You explained very well.

I have this another doubt. This is regarding BJT.
In CE configuration of a BJT, with the increase in base current the Collector-emitter voltage reduces. How is it possible? Can you explain me theoritically rather than mathematically?

Also when we compare the CE configuration with CC config, the input impedence of the CE is low and the o/p impedence is high. So is correct that the output voltage of CE config cannot be used as a voltage source?

Also in an Opamp the output impedence is low, which makes it ideal for a voltage source. Is it true?

thanks,
beenu

 

[stevez]

It appears that the question is not related to radio frequency but to add a bit of information on the subject I'll comment on radio frequency transmitters - at least what I think I understand.

My old ham radio transmitter (vacuum tube final amp) required that the load (transmission line/antenna) match the transmitter to some extent. Final tuneup of the controls on the transmitter (capacitors and coil) allowed a pretty good 'match' based on meter readings. This was done for a couple of reasons - maximum power transfer was one and to allow the amplifier to run at a higher power level. Another important reason was to keep the final amplifier running within it's design parameters - a severe mismatch would result in overheating and failure of the amplifier. I've described the results in non-technical terms. Maybe someone can comment on the 'electronics' or physics of what goes on in a mismatched transmitter.

Today's equipment (for radio amateurs anyway) is such that the final amplifiers are sensitive to significant mismatches. Built in protection circuitry senses mismatch conditions and the input power is reduced or shut off entirely. Comments provided by various amateurs suggest that the protection is far from reliable as they've ruined their final amps as a result of mismatch.

 

[audioguru]

Thanks everyone for the reply.

Correction to above. The output impedance of an audio amplifier is extremely low, so that it damps the resonance of a speaker and controls the motion of its cone very well. It is a voltage source with a high current capability and does not match the impedance of the speaker.

I have this another doubt. This is regarding BJT.
In CE configuration of a BJT, with the increase in base current the Collector-emitter voltage reduces. How is it possible? Can you explain me theoritically rather than mathematically?

An increase of base current causes an increase of collector to emitter current. The increased current in the collector resistor increases the voltage drop across it so the voltage between the collector and emitter reduces.

Also when we compare the CE configuration with CC config, the input impedence of the CE is low and the o/p impedence is high. So is correct that the output voltage of CE config cannot be used as a voltage source?

A high impedance is a current source, not a voltage source. When you reduce the collector resistor's value or reduce the load's value then the voltage at the collector becomes reduced. Changing the load of a low impedance voltage source wouldn't change the voltage.
The collector of a CE stage is commonly used as a voltage source when its load is a high impedance.

Also in an Opamp the output impedence is low, which makes it ideal for a voltage source. Is it true?

Yes, the output impedance is extremely low when the opamp has negative feedback, but its max current is limited to about only 20mA.

 

[RadioRon]

Thanks Audioguru, I stand corrected on my example of an audio amplifier.

To try and answer the question about how the collector current can be so much more than the base current, which is an essential part of the answer to why the collector voltage drops....(well he did ask for theory)
In a common emitter configuration using a bipolar transistor the physics go something like this:
"The (emitter) junction is biased in the forward direction and holes from the emitter region (P type material as in a PNP transistor for example) are injected into the N-base region, producing therein a concentration of holes substantially greater than normally present in the material. These holes travel across the base region toward the collector, attracting neighboring electrons, finally increasing the available supply of conducting electrons in the collector loop. As a result, the collector loop possesses lower resistance whenever the emitter circuit is in operation. In junction transistors this CHARGE TRANSPORT is by means of diffusion wherein the charges move from a region of high concentration to a region of lower concentration at the collector. The collector, biased in the opposite direction, acts as a sink for these holes, and is said to collect them."
(excerpt from Radio Handbook, 22nd Edition, William Orr, 1981, Howard W Sams & Co)



You can use the output of a common emitter stage as a voltage source if it has a relatively low output resistance when compared to the load that it is driving. There are no hard and fast limits as to what defines a voltage source so it depends on the relative difference between output resistance of driver versus the input resistance of the load. For example, it is easy to lower the output resistance of a common emitter stage by lowering the collector resistor value and then biasing the stage to suit. If you use a collector resistor of 470 ohms the output resistance of the stage will be about 470 ohms and if you then drive a load of 20Kohms, you can say that the driver is a voltage source.

 

[audioguru]

It sounds good, Ron.

 

[Russlk]

Quote:
Also when we compare the CE configuration with CC config, the input impedence of the CE is low and the o/p impedence is high. So is correct that the output voltage of CE config cannot be used as a voltage source?

The output impedance of the CE transistor at the collector is high, but a transistor with no load is not very useful. When there is a load connected, the output impedance is nearly equal to RL.

 

[beenuseren]

Hi,

I have a doubt.

This is with respect to transistor.

In CE configuration of a BJT, how does the increase in base current, reduces the width of depletion region in the CB junction? can anyone explain this theoritically?

thanks,

 

[beenuseren]

And also why are the polarities of the EB junction voltage drop and CB junction voltage are opposite in polarity?