Circuit Impedance-"Loading"

[Electrix]

Hi, just want to get a few things clear..

-------Why is it important that a circuit(say an amp) should have high input impedance and low output impedance.
(My intution says That a high impedance at input will draw less current from the power supply-which is be desirable.)

--------When does a circuit "load" another circuit ?

Awaiting your views.

Thanks.

 

[fsahmed]

Let us consider the amplifier as a two port system, dividing it into two segments, input & output.

At the input the amplifier will be the load & the input signal will be the source. Since the input signal is not very strong in power (otherwise it wouldnt need amplification in the first place), it isnecessary that the load doesnt take much toll on it, that is it does not draw too much current from the source. (Current drawing or loading is discussed ahead). Therefore the impedance of the input side of the amplifier must be high,sothat the current drawnis verylow(i=v/z).

ON the other hand, at the output,the amplifier circuit is the source, while the load is ... well, the load. In this case, the circuit should transfer maximum current to the load, having minimal voltage drop internally (though max. power transfer theorem suggests source impedace should be equal to the load impedance but that is another story). Hence the output impedance of the amplifer should be very low.

About loading, the current drawn by a circuit is given by ohms law, stating i=v/z, where z is the impedance of the circuit. But remember that current is the measure of flow of free electrons, suppose that the conductor of the source doesnt have enough free electrons to form the current that is required by the circuit according to the ohms law..... then the voltage at the source will appear to drop. This is the worst case of loading effect. Loading is the change in supply voltage levels when a load is connected to the supply (source) output. This occurs because if no load is connected, the impedance between the source terminals appears to be infinite,hence maximum voltage is dropped at the terminals.When a load of finite impedance is applied, comparable to the internal impedance of the supply the circuit results in a voltage divider network, showing less voltage across the load than that which was measured at no load condition.

 

[Electrix]

Cicuit Loading

About loading, the current drawn by a circuit is given by ohms law, stating i=v/z, where z is the impedance of the circuit. But remember that current is the measure of flow of free electrons, suppose that the conductor of the source doesnt have enough free electrons to form the current that is required by the circuit according to the ohms law..... then the voltage at the source will appear to drop. This is the worst case of loading effect. Loading is the change in supply voltage levels when a load is connected to the supply (source) output. This occurs because if no load is connected, the impedance between the source terminals appears to be infinite,hence maximum voltage is dropped at the terminals.When a load of finite impedance is applied, comparable to the internal impedance of the supply the circuit results in a voltage divider network, showing less voltage across the load than that which was measured at no load condition.

So a circuit with low O/P imped will load its output less ? Is there any other way that loading can be avoided ??

 

[Nigel Goodwin]

Re: Cicuit Loading

About loading, the current drawn by a circuit is given by ohms law, stating i=v/z, where z is the impedance of the circuit. But remember that current is the measure of flow of free electrons, suppose that the conductor of the source doesnt have enough free electrons to form the current that is required by the circuit according to the ohms law..... then the voltage at the source will appear to drop. This is the worst case of loading effect. Loading is the change in supply voltage levels when a load is connected to the supply (source) output. This occurs because if no load is connected, the impedance between the source terminals appears to be infinite,hence maximum voltage is dropped at the terminals.When a load of finite impedance is applied, comparable to the internal impedance of the supply the circuit results in a voltage divider network, showing less voltage across the load than that which was measured at no load condition.

So a circuit with low O/P imped will load its output less ? Is there any other way that loading can be avoided ??

I'm now confused :roll:

You seem to be making this VERY complicated!.

It's really very simple, the output impedance and the input impedance form a simple potential divider. The example below shows a CD player, output impedance shown by Rout, and an amplifier, input impedance shown by Rin.

If Rin = Rout, then the potential divider will attenuate the signal by half, which isn't what you want! - if Rin > 5 x Rout, then attenuation is small enough to be ignored. So it's common to have the input impedance at least five times the output impedance that feeds it.

You should be aware that this is for best 'VOLTAGE TRANSFER', which is what you usually want (certainly in audio equipment), for best 'POWER TRANSFER' the two values should be equal - you can easily confirm this with ohms law and power calculations.

 

[electronist]

Let me explain the concept in easy words,
suppose you have a wall socket that provides you with 230V mains.
Now suppose you connect a low resistance ( for example a conductor like a copper wire) across the line and neutral. What should happen?
Tremendous current will flow through the wire and blow off your fuse. This happened because the impedance of your ckt (i.e copper wire) was very low and hence you pulled a lot of current which blew off the lest sensitive part of your ckt (i.e. fuse). Now suppose you connect a high resistance ckt into your mains. since the resistance is high it will pull very little current from the mains and not load the mains (hence your fuse will not blow off). In this fashion you will always want that whatever load or ckt you have should have high input impedence so that you do not harm the ckt supplying the source ( voltage or signal.).

 

[zachtheterrible]

This is a very simplified amplifier (no biasing, coupling, etc). It is also a model: In real life, R1 and R2 would not be actual resistors.

R1 represents input impedance. If its too low, current will flow straight to ground, instead of going through the BE junction of Q1. If its extremely high, very little current will flow through R1, but more current will flow through the BE junction of Q1.

R2 represents output impedance. Obviously it should be low or not much current will get to the load.

So, an ideal amplifier has infinite input impedance and zero output impedance. Remember too that input and output impedance are not simply resistances, since the definition of impedance is resistance to an alternating current.

Hope this helped, it was the only way that i figured it out :lol: