determining input/output impedance - diyAudio
 determining input/output impedance
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 8th April 2010, 11:49 AM #1 diyAudio Member   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 pre-amp? cheers, K
 8th April 2010, 02:23 PM #2 diyAudio Member   Join Date: Feb 2004 Location: Wheaton IL. Blog Entries: 30 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 __________________ Scienta sine ars nihil est - Science without Art is nothing. (Implies the converse as well) Mater tua criceta fuit, et pater tuo redoluit bacarum sambucus
 8th April 2010, 06:05 PM #3 diyAudio Member   Join Date: Oct 2004 Location: Maui, Hawai'i, USA Yes. The input of a power amp will either have a grid ground-referencing 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 diyAudio Member     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. __________________ Dynakit ST70, ST120, PAS2,Hammond H182(2 ea),H112,A100,10-82TC,Peavey CS800S,1.3K, SP2-XT's, T-300 HF Proj's, Steinway console, Herald RA88a mixer, Wurlitzer 4500, 4300
 14th February 2015, 02:42 PM #5 diyAudio Member     Join Date: Jul 2009 hello __________________ Artificial intelligence is no match for natural stupidity- Red Green
 14th February 2015, 05:40 PM #6 diyAudio Member   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=V1-V2/((V2/R2)-(V1/R1)) R1=22K and R2=11K
 15th February 2015, 01:26 PM #7 diyAudio Member     Join Date: Jul 2009 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? __________________ Artificial intelligence is no match for natural stupidity- Red Green
 15th February 2015, 01:51 PM #8 diyAudio Member     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 A-1A = 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     Join Date: Jul 2009 Thanks! __________________ Artificial intelligence is no match for natural stupidity- Red Green
 15th February 2015, 05:21 PM #10 diyAudio Member   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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