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Need impedance matching explanation and verification

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faradayfan

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impedence zs

Hi everyone this is my first post. I am a electronics noob so please be patient with me.

I don't understand the concept of impedance matching. Why would you want to increase the impedance in the source, I would think that any increase in the impedance of the source is just wasted power. Maybe my problem is I am thinking in terms of DC current and resistance when adding resistance to the source does not make any sense. Why does increasing the source impedance to where it is equal to the load impedance increase the power available to the load? Less impedance (resistance) should mean more power available.
 

flatulent

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does increasing the ot impedance decrease power

It has to do with the voltage division caused by the source and load impedance forming a voltage divider. You get maximum power transfer if they are equal.

Also, ordinary sources have current limitations.
 

FvM

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impedance matching

In RF engineering, impedance matching does not involve adding resistance to the source rather than transforming the existing source impedance by a lossless network.

Wideband or pulse systems may use lossy networks to achieve frequency independant impedance matching when maximum output power isn't the primary objective.
 

doraemon

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Re: impedance matching

Hello!

It is possible to prove this as follows.
Suppose you have a source (with a certain impedance) and a load.
You can model this by associating serially:
- A pure voltage source (V)
- A source impedance (Zs)
- A load impedance (Zl)

By Ohm' law, you can say that the current in the circuit is:
I = V / (Zs + Zl) (equation A)
In all the following, XX2 means square of XX while 2XX means ... 2XX.

Now you are interested in the load, so you can calculate the power of
the load:

Pl = Zl x I2

If you use equation A to replace I:

Pl = Zl * (V / Zs + Zl)2

which becomes:

Pl = V2 / ((Zs2/Zl) + 2Zs + Zl)

If you want the power in the load to be maximal, then
(Zs2/Zl) + 2Zs + Zl should be minimal

If you use differential calculation: differentiate the above equation
by reference to Zl

d/dZl (Zs2 / Zl + 2Zs + Zl) = -(Zs2 / Zl2) + 1

When this differential is 0, it means that you have an extremum.
(Zs/Zl)2 = 1

This means that Zs is equal to +/- Zl

By further derivation, you can prove that the only solution that
gives a maximal power is that Zs = Zl

Careful: this does not mean that you have maximal efficiency (it's not the case).
It means that the max power that you can transfer is when both impedances
are equal in which case you get a 50% efficiency.


Beside this, you said : Why does increasing the source impedance ...
increase the power

An intuitive explanation of matching:
Example: you have a car battery (I mean not an electic car, a gas car).
What you usually want is to transmit a lot of power to the starter so that
the car has good chances to start quickly.
First experiment: you have a battery with an impedance of 10 mOhm.
You have 2 motors, one 10 ohm and one 5 ohm. I guess you agree that the
5 ohm device will be more powerful than the 10 ohm. In this case, lowering
the load impedance increases the power. I think this replies to your question
except that you don't increase the source impedance, you decrease the
load impedance.

Now another experiment: you have a starter with an impedance of 1 mOhm.
I guess you agree that the battery itself will burn about 90% of its power by
itself. In this case, raising the load power will increase the power.

This proves that between these 2 situations, there is a high point where the
load's impedance allows you to draw a maximum power. And the numerical
proof is above.

That's about it!

Dora.

faradayfan said:
Hi everyone this is my first post. I am a electronics noob so please be patient with me.

I don't understand the concept of impedance matching. Why would you want to increase the impedance in the source, I would think that any increase in the impedance of the source is just wasted power. Maybe my problem is I am thinking in terms of DC current and resistance when adding resistance to the source does not make any sense. Why does increasing the source impedance to where it is equal to the load impedance increase the power available to the load? Less impedance (resistance) should mean more power available.
 

akshay_d_2006

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impedance matching

Reffering to the 1st post. you can undertand it logically too.

Its quite simple.

-----[*source/output Impedence]------[#load/Input Imp]--------
| |
| |
| |
(V,in) (V,out)
| |
| |
| |
--------------------------------------------------------------------------

*here [abc..] is the resistance.
Now we must have the o/p impedence very low so that the voltage drop(voltage across[Source/O/p impedence] is less] and maximum voltage is available at input impedence(i/p of another appliance connected normally called as load)

If it is high, the voltage drp is high, thus the voltage available at output i.e INPUT OF OTHER DEVICE is less.

On the other hand INPUT IMPEDENCE (output in the diagram) shold be very high so that even if less current is flowing through it; it multiplied with high resistance gives you hegh viltage to runn the output device.

**UNDERSTAND THE DIAGRAM--
*The output impedence is the impedence of one thing (say BJT amplifier) seen from the output side ot that BJT.
#The input impedence is the impedence of other device (say loud speaker) connected after BJT. (i.e the impedence it the load here speaker)
So in diag. the o/p imp. is o/p of BJT amplifier and i/p is i/p of speaker.


Impedence matching is done where these conditions are not achieved. so as to get the mentioned abv conditions.

Added after 3 minutes:



-----[*source/output Impedence]------[#load/Input Imp]--------
|..........................................................................................|
|..........................................................................................|
|..........................................................................................|
(V,in)...................................................................................(V,out)
|..........................................................................................|
|..........................................................................................|
|..........................................................................................|
--------------------------------------------------------------------------

Added after 40 seconds:

In the diagram do not considet the dots...
 

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