Output / Source Impedance, Input Impedance

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bjoneson

Well-known member
Joined
Mar 1, 2014
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170
Location
Oakland, IA
I think I might finally have my brain wrapped around all of this...

"Input Impedance" - the impedance as "seen" from the source feeding the input. The signal voltage must be developed across this impedance (load). It makes sense to me now, why most modern inputs are relatively high impedance. The higher the input impedance, the "easier" it is to develop the signal voltage across it (less current required). In modern circuits, the "signal" is the voltage, the driving current is much less relevant. So why not have an astronomical input impedance like 10M Ohms or something? The tradeoff is that the higher the impedance, the more noise is introduced (Johnson, etc...). So it seems a good practice is to have the minimal input impedance that the connected device can "comfortably" drive. All of this I, I now feel pretty confident with.

"Output Impedance / Source Impedance" - There's many, many articles / resources that define it. It's the impedance as "seen" from the outside looking back at the input. Where I've fallen down in understanding, is why does the source impedance matter? General rule of thumb is that it should be "low", but why? If we're just trying to develop a voltage across the "Input Impedance" of the receiving device, why does output impedance even matter? It's not been clear to me how it fits into the equation.

I finally think I got a grasp on it though. If I understand correctly, technically the output isn't just developed across the input impedance of the sending device. It's developed across *both* the output impedance and the input impedance in series. This creates a voltage divider with the "input" of the receiving device at the node between the source impedance and the input impedance. The ratio between the source and input impedance determines the signal loss (voltage drop).

Have I graduated from Impedance Apprentice, to Impedance Journeyman yet?
 
You are getting close. There's just a couple of points that need to be amplified.

It is often useful to draw a simple equivalent circuit of what you are talking about. This uses a combination of 'perfect' components to describe the behaviour of the imperfect real world. I have attached a simple equivalent circuit of and output driving an input.

The output consists of a perfect voltage generator (with zero source impedance)  with a resistor Rout connected in series to represent the output impedance. The input, assumed to be connected by a length of cable, is represented by the resistor Rin which is equal to its input impedance.

You can immediately see the input forms a potential divider with the output resistance. These days we are interested in getting the maximum voltage from the out put to the input. This mean we want Rout to be small and Rin to be relatively large. If Rout is 600 ohms and Rin is 10K then the voltage drop due to this potential divide is just 0.5dB. If Rout is 75 ohms, the drop reduces to 0.06dB so from this point of view there is not a lot to be gained by very low output impedances.

If you want to drive a pair of 600 ohm headphones or some older gear with a 600 ohm input impedance, then if Rout is 600 ohms you will lose 6dB of signal. Worse, you will lose 6dB of headroom. If Rout is 75 ohms, even driving a 600 ohm load will only drop the level by  1dB.

From the noise point of view, Rin and Rout are in parallel. This is because the perfect generator has zero source impedance. If its output is zero then Rin is connected to ground. So, even if Rin is 10Meg, If Rout is 600 ohms, then the resistance for noise purposes is 10Meg in parallel with 600 ohms which is close to 600 ohms. If Rout is 75 ohms then the noise resistance is pretty close to 75 ohms.

Real cables are not perfect and will have some capacitance across the signal wires (represented by Ccable in the diagram). This changes the simple pot divider into a low pass filter. The -3dB point will be determined by Rout and Ccable. In most situations this is not a problem but for long cable runs it can be. You might have 200pF per metre capacitance in a cable. 100 metres of this will be 20nF. The -3dB point of this with a 600 ohm source will be 13KHz. With a 75 ohm source, the -3dB point is over 200KHz.

Hope that helps.

Cheers

Ian
 

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ruffrecords said:
You are getting close. There's just a couple of points that need to be amplified.

It is often useful to draw a simple equivalent circuit of what you are talking about. This uses a combination of 'perfect' components to describe the behaviour of the imperfect real world. I have attached a simple equivalent circuit of and output driving an input.

The output consists of a perfect voltage generator (with zero source impedance)  with a resistor Rout connected in series to represent the output impedance. The input, assumed to be connected by a length of cable, is represented by the resistor Rin which is equal to its input impedance.

You can immediately see the input forms a potential divider with the output resistance. These days we are interested in getting the maximum voltage from the out put to the input. This mean we want Rout to be small and Rin to be relatively large. If Rout is 600 ohms and Rin is 10K then the voltage drop due to this potential divide is just 0.5dB. If Rout is 75 ohms, the drop reduces to 0.06dB so from this point of view there is not a lot to be gained by very low output impedances.

If you want to drive a pair of 600 ohm headphones or some older gear with a 600 ohm input impedance, then if Rout is 600 ohms you will lose 6dB of signal. Worse, you will lose 6dB of headroom. If Rout is 75 ohms, even driving a 600 ohm load will only drop the level by  1dB.

From the noise point of view, Rin and Rout are in parallel. This is because the perfect generator has zero source impedance. If its output is zero then Rin is connected to ground. So, even if Rin is 10Meg, If Rout is 600 ohms, then the resistance for noise purposes is 10Meg in parallel with 600 ohms which is close to 600 ohms. If Rout is 75 ohms then the noise resistance is pretty close to 75 ohms.

Real cables are not perfect and will have some capacitance across the signal wires (represented by Ccable in the diagram). This changes the simple pot divider into a low pass filter. The -3dB point will be determined by Rout and Ccable. In most situations this is not a problem but for long cable runs it can be. You might have 200pF per metre capacitance in a cable. 100 metres of this will be 20nF. The -3dB point of this with a 600 ohm source will be 13KHz. With a 75 ohm source, the -3dB point is over 200KHz.

Hope that helps.

Cheers

Ian

Ian, thanks so much for the thorough reply. I've started reading Doug Self's book "Small Signal Audio Design", and he expects the reader to have some "fundamental" knowledge. This has all helped put much of the text into context for me.

I might actually get a good night sleep tonight without nightmares of being chased by electrons! :)
 
The modern scheme of low output/high input impedance is called a "bridging" termination or interface and generally described as a 1:10 ratio, which also delivers a less than 1dB insertion loss (voltage).  Very old school terminations used to be the same impedance to insure maximum "power" transfer, these were the legacy 600:600 ohm interfaces.  Modern design has pretty much evolved beyond transformer interfaces using power transfer to instead maximize voltage transfer which is best at lower source and higher input Z.

As Ian noted there are reasons to not attempt zero source impedance (cable capacitance can cause circuit instability). While mostly a cosmetic concern, too high input impedance can seem noisy when nothing is plugged in. This is pretty much harmless, since it quiets down in use, but the customer is always right and kept happier with high but modest input impedances (tens of k ohm).

JR
 
bjoneson said:
I finally think I got a grasp on it though. If I understand correctly, technically the output isn't just developed across the input impedance of the sending device. It's developed across *both* the output impedance and the input impedance in series. This creates a voltage divider with the "input" of the receiving device at the node between the source impedance and the input impedance. The ratio between the source and input impedance determines the signal loss (voltage drop).
That's correct, and what everybody else said.
I would like to add to that that low impedance guarantees protection against unwanted interference.
-Electrostatic interference is created by capacitive coupling; decreasing the impedance of the connection* just creates a better voltage divider.

-Electroagnetic interference translates into a voltage that is applied almost perfectly equally on both legs. Since it is developped in the conductors, it's very low Z (a few 1/10th of ohm), so the actual source Z doesn't do much to counter it. In a typical unbalanced connection, the source Z has almost zero capacity of reducing interference. But it is relatively easy to demonstrate that in a balanced connection, the efficiency of Common Mode Reduction, measured by CMRR, is directly inverse-dependant on the output Z.




*Whatever is defining the connection impedance - it can be the source Z, the load Z and/or any arbitrary load. But since low output Z/high input Z also optimizes the level of desired signal, it's the preferred arangement.
 
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