My Antique Sound Lab mg-s115 a kt-88 single ended designe
5w triode 15w pentode list it`s output impedience as 4,8ohm.
I am assuming 4 ohms in triode and 8 ohms in pentode mode?
I fed 3.55vac through a 8 ohm resistor into the output transformer and measured 1.76vac across the output transformer
so it does check out about 8 ohm. Did I do this right?
Thanks
5w triode 15w pentode list it`s output impedience as 4,8ohm.
I am assuming 4 ohms in triode and 8 ohms in pentode mode?
I fed 3.55vac through a 8 ohm resistor into the output transformer and measured 1.76vac across the output transformer
so it does check out about 8 ohm. Did I do this right?
Thanks
Woody - Sorry NO, try to think of it this way
From an AC signal point of view -
you have a "signal generator" with a hot side going to the positive speaker terminal via a "Source Resistance" which is the output Impedance of the Amp and a return (cold) side going to the negative speaker terminal.
Measure Output Voltage with no connection to the speaker terminals and Output set to give you approx 1 or 2 V AC.
There is no voltage drop across the "Source Resitance" since there is no "circuit" to carry current so you are measureing the "signal generator" voltage. ( call it V1)
Connect the 8 Ohm Resistor to the speaker terminals and measure the voltage again. (call it V2)
The difference (V1 - V2) gives you the voltage drop across the "Source Resitance" which IS the Output Impedance in this case.
Now knowing the voltage drop across an 8 Ohm load (V2) and the voltage drop across the output impedance (V1 - V2) you can work out The value of the Output Impedance from a simple ratio calculation.
Using above terms
Output Impedance = [(V1 -V2) / V2] x 8 Ohms
Sorry if this was "talking down" to you. I wanted to make it as clear as I could not only how to do but why you do it this way.
For those of you who are paranoid about running your amps without a load ( to measure V1) an alternate method is to measure voltage across an 8 Ohm resistor and then across a 4 Ohm resistor. Using the visualisation above you can work out the algebra for yourself.
Aside: running your amp without a load is OK as long as you keep output well below maximum power voltages (clipping level) - say no more than half.
Cheers,
Ian
From an AC signal point of view -
you have a "signal generator" with a hot side going to the positive speaker terminal via a "Source Resistance" which is the output Impedance of the Amp and a return (cold) side going to the negative speaker terminal.
Measure Output Voltage with no connection to the speaker terminals and Output set to give you approx 1 or 2 V AC.
There is no voltage drop across the "Source Resitance" since there is no "circuit" to carry current so you are measureing the "signal generator" voltage. ( call it V1)
Connect the 8 Ohm Resistor to the speaker terminals and measure the voltage again. (call it V2)
The difference (V1 - V2) gives you the voltage drop across the "Source Resitance" which IS the Output Impedance in this case.
Now knowing the voltage drop across an 8 Ohm load (V2) and the voltage drop across the output impedance (V1 - V2) you can work out The value of the Output Impedance from a simple ratio calculation.
Using above terms
Output Impedance = [(V1 -V2) / V2] x 8 Ohms
Sorry if this was "talking down" to you. I wanted to make it as clear as I could not only how to do but why you do it this way.
For those of you who are paranoid about running your amps without a load ( to measure V1) an alternate method is to measure voltage across an 8 Ohm resistor and then across a 4 Ohm resistor. Using the visualisation above you can work out the algebra for yourself.
Aside: running your amp without a load is OK as long as you keep output well below maximum power voltages (clipping level) - say no more than half.
Cheers,
Ian
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