Impedance question.

[caps]

To maximise voltage transfer ( as in a mic), input Z should be much higher than the output impedance of the source feeding it. And I think for efficient power transfer Z's should be matched.

Ok, this isnt enough for me. I want to know why. Thing is I dont understand how a higher impedance helps transfer voltage from the low Z feeding it. Is there some basic Ohm's law Im missing out on here or something? thanks.

 

[CJ]

If you load down the mic, it does not have enough gas in the tank to keep that voltage on the primary pumped up. No current behind the voltage. Like a 100 foot high dam with ten feet of shoreline.

 

[caps]

Thanx CJ, but LORD ! I am really one bottle short of a slab today....

Im thinking, BUT...

High Z = high Ohms, therefore the mic output (150 - 200 Ohm) sees , say 1.2K for example, higher RESISTANCE to the mic signal so less output. OBVIOUSLY wrong, but I cant get my head round it dam it . :grin:

 

[caps]

Ok.

Could it be that high input impedance , using E = I X R, gives us a bigger voltage drop across the primary? Hence a larger signal to go on and be amplified ?

Maybe there should be a "Newb" section on this forum. :gri

 

[pstamler]

Think of it this way: A microphone has an impedance; you can pretend it's a pure voltage source with an impedance in series with it. Let's say that impedance is 150 ohms.

Load it down with, say, a 1.5k load. You now have a pure voltage source feeding a voltage divider, composed of Rseries (150 ohms) and Rshunt (1500 ohms). [We will ignore questions of balanced lines, etc.; the numbers work out okay either way.) The equation for a voltage divider is:

Vout = Vin x Rshunt / (Rshunt + Rseries)

In the case of a 1500 ohm load (shunt) and a 150 ohm microphone (series) the total voltage output will be 1500/1650 x the imaginary pure voltage source, or 0.909 x the pure voltage.

Now change the load to 150 ohms. The situation is the same, and the equation is the same, but now it works out to 150/300 x the imaginary pure voltage source, or 0.5 x the pure voltage.

Combine those two numbers, and the microphone loaded down by 150 ohms produces 0.5/0.909 times as much voltage as the one loaded by 1500 ohms, or about 5.2dB less.

It's also possible that, if the microphone is a condenser, the lower impedance will draw more current out of the microphone's internal amplifier than the amplifier is designed to produce cleanly, and along with your output going down, your distortion will go up.

Peace,
Paul

 

[Brian Roth]

No one bothered to mention the fact that 'modern" mics are designed to provide a relatively low Z (say, 150 Ohms) towards the load (say, 1500 Ohms of greater).

That makes the input stage of a transformer-coupled input "interesting".

It needs to be "happy" with a 150 Ohm source impedance, yet provide a 1500+ Ohm input impedance.

Bri

 

[Emperor-TK]

Caps, here is a pretty good analogy that I saw on another website to help understand impedance matching. I can't remember where I saw it, otherwise I would just link to it. But let me try to paraphrase. I invite the more learned members to step in if I mess anything up :wink:

- A car battery and a car headlight are both low impedance devices.
- A AAA-battery and a flashlight bulb are both high impedance devices.
- Connecting the car battery to the car lamp gives the most amount of light that the the system can deliver. This is an efficient transfer of power, but the battery is loaded by the big lamp.
- Connecting the AAA battery to the flashlight bulb is similar to the above system in terms of efficiency and loading, but less power is delivered. Still, the power efficiency is maximized.
- Connecting the AAA battery to the car lamp doesn't work. The AAA battery is loaded down immensely by the lamp, struggles, farts, and dies.

And finnally, the analogy that represents how modern bridging impedance matching works....

- Connecting the car battery to the AAA flashlight works fine, but does not give the maximum power transfer that the souce is capable of. What we do get though, is a condition where the source is negligibly loaded by the lamp. In other words, the battery barely knows that the lamp is there and is capable of just being itself, without the load affecting it. It is a way of isolating the source from the load so that the source can act more independantly, regardless of changing load conditions. This way we can have sources and loads that were not specifically designed to interact together.

I think I got that right.

-Chris

 

[PRR]

> To maximise voltage transfer.., input Z should be much higher than the output impedance of the source feeding it. And I think for efficient power transfer Z's should be matched. Ok, this isnt enough for me. I want to know why

Simple extension of Ohm's Law. The source and load form a resistive divider. What happens for various ratios of source and load resistance?

Take the regulator off your car alternator, and spin the engine 6,000RPM. You will get about 100V no-load. If you short it, you will get about 100Amps. We can assume the alternator can be represented by a 100V source in series with a 1 ohm resistor.

This could be handy if you are out in the woods with your car, a voltmeter, an ammeter, and an electric heater that can be re-wired to any resistance.

What is the load that will show the highest reading on the voltmeter?

What is the load that will show the highest reading on the ammeter?

What is the load that will give the most heat in the electric heater?

To visualize it better, here's a picture of six identical sources with six different loads:

Important extra point: what is the power dissipated in the alternator for each load condition? What load gives the least heat in the alternator? the most heat in the alternator? This is not a factor in microphone loading but matters in line driving and becomes critical in speaker driving.

 

[Svart]

it helps to think in terms of current sourcing and current sinking. A device produces XXX amount of current (and also voltage so Ohm says) but is being met by resistence/impedence. Simply, it takes more current to drive more of a load. so a high current output( low impedence) device will drive a device that does not need high amounts of current(high impedence) much easier than one that sinks a larger amount of current.

so:

high current sources will drive devices that sink little current easily

low current sources won't drive devices that sink high current easily

high current sources will drive devices that sink high current


so really you are just looking at another side of the same law.. Ohm's law.