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
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
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
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
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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.
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.
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
Zout=V1-V2/((V2/R2)-(V1/R1)) R1=22K and R2=11K
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.
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.
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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.
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.
Another way to measure output impedance is to run the amplifier with no input signal, and drive a sinewave signal through the output terminals. Measure the AC current flow and the AC voltage developed across the output terminals, and use Ohm's law to determine the output impedance.
This is particularly simply in a Spice simulation. Just put an AC current source in series with the output load resistor (or in place of the output load resistor; it doesn't really matter), and divide the resulting AC voltage across the output terminals by the current provided by the source.
(In fact it's handy to put the current source in series with the output load resistor in the Spice simulation, because then when you want to go back to looking at other characteristics of the amplifier, you can just add a wire short-circuiting the current source in the Spice schematic.)
Chris
OK this is because I guess the stage before the pre amp sees the resistor in series with the signal and adds the resistor to to ground and sees a constant input impedance
Like in here Electra-Print.com The Ultrapath (I use this design)
Like in here Electra-Print.com The Ultrapath (I use this design)
In a stage like that, the circuit's small input capacitance has little effect in the audio band. So, the volume control can be considered unloaded, just equal to a simple fixed resistance.
Moving the wiper will have little effect on the input impedance or frequency response, unless the circuit's input capacitance is larger than around C = 1/ (2Pi x 40kHz x R/4), where R is the value of the volume control's resistance. This value of C will give a 1dB loss at 20kHz.
Moving the wiper will have little effect on the input impedance or frequency response, unless the circuit's input capacitance is larger than around C = 1/ (2Pi x 40kHz x R/4), where R is the value of the volume control's resistance. This value of C will give a 1dB loss at 20kHz.
How low?
Intelligently designed, not too very low.
If the load after the pot was infinite, there would be no drop of input impedance at all.
In a somewhat extreme case, 10k pot with 10k load, it drops to 5k input impedance.
Knowing that, the designer picks a coupling cap for the worst-case 5k input impedance, allowing response to extend another octave at very low knob settings. (Or doesn't: in cases where woofers are being blown, I have scaled for 40Hz roll-off at low volume but 80Hz when volume is maxed-out, lessening bass abuse.)
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