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8th April 2010, 11:49 AM  #1 
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Join Date: Jan 2010
Location: Shanghai

determining input/output impedance
Guys,
Can anyone tell me how I determine the input impedance of my (power) amp, and teh output impedance of my preamp? cheers, K 
8th April 2010, 02:23 PM  #2 
diyAudio Member

Usually, the input impedance of a tube amp is the grid resistor. It is chosen to be lower than the grid capacitance while being as high as possible to avoid loading the preamp.
Typical values are between 50K and 100k for tube power amps, and 10K or more for solid state. Some tube amps with a pentode input were 250K to avoid loading the preamp. HTH Doug
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8th April 2010, 06:05 PM  #3 
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Join Date: Oct 2004
Location: Maui, Hawai'i, USA

Yes. The input of a power amp will either have a grid groundreferencing resistor between the signal hot and ground, or the track of a volume potentiometer (the grid being connected to the wiper) in the same position. The value of this resistor/pot is the input impedence of the circuit, close as makes any difference.
There are mathematical expressions with which to determine the approximate output impedence of the preamp, but the simplest way to determine is to test. Put a largish (470K or so) resistor across the signal output hot and ground, and a signal on the input. Adjust the volume control for, say, 0.5 or 0.1 volts AC on the output. Now replace the 470K resistor with a 10K or 20K pot, and dial down the resistance until the signal voltage is half the original value. Measure the resistance of the load; this is the Zout of the circuit. You can also do this by putting smaller and smaller value resistors across there, but it's laborious. How do I know this, hm? When you disconnect and connect the loads, turn off the circuit power. Aloha, Poinz AudioTropic Last edited by Poindexter; 8th April 2010 at 06:07 PM. 
8th April 2010, 06:27 PM  #4 
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Join Date: Jan 2010
Location: Jeffersonville, Indiana USA

output impedence
If the output of a tube device is not transformer coupled, then you can first order determine the output impedance by looking at the size of the plate resistor. For a transformer output coupled tube, you need to measure the impedence of the primary winding, which is not the same as the DC resistance.
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14th February 2015, 02:42 PM  #5 
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hello
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14th February 2015, 05:40 PM  #6 
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Join Date: Dec 2010
Location: Amesbury, MA

Another way to do the output impedance I think would be to place a resistor say 22k across the output of the preamp measure the voltage out at a specific volume level. At that same volume level measure the voltage after paralleling another 22k with the original 22k effectively giving you halve the resistance.
Zout=V1V2/((V2/R2)(V1/R1)) R1=22K and R2=11K 
15th February 2015, 01:26 PM  #7 
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I'm working on a power amp design for which there is no known output impedance. If you were to substitute an 8 ohm load in spice, then a 4 ohm load and do the above calculation would that work the same way?
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15th February 2015, 01:51 PM  #8 
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Join Date: Feb 2010

Yes you can.
In principle you should determine two pairs of voltages and currents with two different load resistances. Then you calculate Δu/Δi = Rout. Example: Adjust your device to supply 8V to 8 ohms. The current is then 1A. Then change the load to 4 ohms and measure the voltage accros the 4 ohms load. Assume it is dropped to 6 V. Now the current is 6V / 4 ohms = 1.5A. Then use the equation above (Δu/Δi = Rout) : Δu = 8V  6V = 2 V; Δi = 1.5 A1A = 0.5 A So Rout = 2V/0.5A = 4 ohms. Last edited by artosalo; 15th February 2015 at 01:58 PM. 
15th February 2015, 01:54 PM  #9 
diyAudio Member

Thanks!
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15th February 2015, 05:21 PM  #10 
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Join Date: May 2007

If you are measuring output impedance then using several resistors near the expected value are the best way to do it.
If you are calculating output impedance then often the best way is to calculate voltage gain A (i.e. voltage out/voltage in) into an open circuit, and transconductance gm (i.e. current out/voltage in) into a short circuit. Then Zout = A/gm. This is because open circuit and closed circuit can often simplify calculation. 
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