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midmoe 20th January 2013 03:21 AM

impedance calculations
would a crude way to determine an unknown amplifier's output impedance be to hook up a variable resistance in series with an AC ammeter to the output,and input a 1K sign wave, adjust the resistor for max amps out, then read the resistance of the resistor? True, this would not be a superaccurate reading in a typical system, as not inductive, but should show a ballpark reading.....I'm building a circletron with 6- 6550 otputs in a monoblock. I listen to my bose 901's on my mcintosh 275 with passive crossovers to ev-300 18"ers as sub/ woofers. I'm fixing up the high end with 3 pioneer ribbon tweeters (ART-55D's) per side.....crossed off the 901's with a simple .5mfd per trio....I,m pretty happy with this but a friend gave me a tip about fluttermans and this led to finding out about the circletron. In my existing system I'm using old school 901's with the 8 ohm speakers, so I figure I can series them up giving 64 ohms, use them as my main drivers....hoping I can bias the 6550's up, and lower the plate voltage down, to the point the output impedance gets down close to this level...I am planning on using the same passive cross over, but autotransformering down the impedance from the woofer output to go to the 18"ers....using things on hand like standard output xfmrs. with a 70 volt line output as "input" and hooking up the woofers to the 8 ohm tap. This would leave the input of the xfmr hanging, only using the secondary, hence the "autotransformer" impedance matching idea. If this idea fails, I just bi-amp the 18's with the macintosh and use the circletron for the boses and tweeters. Does any one have experience on seriesing up the 901's speakers this the way I converted them to "800s"...I took out the front speaker, blocked up the hole, turned them around, flipped them on their sides, took off the grill cloth (which really helped the high end), hooked the ribbons on the I have 8 speakers, but seriesed they might be high enough for my circletron without resorting to 6AS7's or 6C33B/C russian tubes...and I figure, if it doesnt work out, I can use the 6550's in my macintosh. thanks..midmoe

cnpope 20th January 2013 05:47 AM

You could estimate the output impedance by driving the amplifier into an 8 Ohm resistive load, with a sinewave input, and measure the output voltage. Then, add an additional resistor in parallel, and measure the voltage again. Model the amplifier by an ideal constant voltage source V_0, in series with an "internal impedance" Z.

Before adding the parallel resistor, the voltage across the 8 Ohm resistor will be V1=V_0 8/(8+ Z). After adding the resistor, lets say of R Ohms, the voltage will be V2=V_0 Rp/( Rp + Z), where Rp=1/(1/8 + 1/R). From these two measurements, you can solve the equations to find Z.


Elvee 20th January 2013 07:16 AM

To be accurate, the two resistors method has to be performed at very small signal levels, otherwise it can be off by a large amount.
See discussion here, both for sim and reality:

DF96 20th January 2013 10:35 AM

It depends what you mean by 'output impedance'. Strictly this means what is also known as source impedance. To measure that you can apply a resistance and adjust for maximum power output (not maximum current). Or apply one or more resistances in turn and calculate from the voltage drop. To get maximum current is easy: apply a short circuit to the output.

Often people say 'output impedance' when they actually mean optimum load impedance. This is harder to measure, because first you have to decide what it is you are trying to optimise. For a power amplifier it is likely to be the best compromise between power and distortion, so first you have to decide what you mean by 'best compromise'. The quick way is to read the datasheet for the valves, or see what the circuit designer says.

midmoe 20th January 2013 03:08 PM

thanks, all...I should have been more specific in my impedance measuring question...the impedance I want to measure is from the circlotron I am building....with no output xfrm....using 6-6550's in parrallel push pull....if the guestimate I have is in the 100 ohm range, I would use a 500 ohm calibrated rheostat in series with an AC milliameter and then sweep down for current peak....and I'm presuming a "short" across the output (effectively hooking only the meter across) would not give max current in this instance....possibly so in a SS or conventional output xfmr type amp...good suggestion on keeping the power down for measurements....thanks..midmoe

cnpope 20th January 2013 03:40 PM

It is indeed necessary to decide what one wants to mean by "output impedance." To a large extent, this depends on why one is asking the question, I suppose.

Measuring it by adjusting the resistance of the output load for maximum power could be OK in some circumstances, but it is probably not too useful in the case of an amplifier with a low output impedance. For example, a solid state amplifier would commonly have an output impedance of perhaps 0.1 ohms or so, and a vacuum tube OTL could easily have an output impedance that is a smallish fraction of an ohm. It would probably be inadvisable to try driving into such a low impedance load, and it would be running the amplifier far from its normal and intended operating conditions.

For an OTL, the maximum output power is principally limited by the maximum current the output tubes can carry. Lowering the output impedance would, of itself, have little effect on this.

Of course, once the tube line-up has been chosen, and hence the maximum power output has been settled, then there could be advantages to trying to lower the output impedance. For example, to minimise the frequency-dependence of the output if the speaker impedance varies significantly with frequency. And for damping, etc.

From these points of view, measuring the change in output voltage when an extra resistor is paralleled with a "normal" load (e.g. 8 ohms), is probably a quite appropriate way of giving meaning to the concept of output impedance.

I'm not sure why one would need to restrict to making the measurements at "very small signal levels." I should have thought that as long as one stays within the power capabilities of the amplifier, the results should be relevant and meaningful, and also reasonably uniform, regardless of the signal level. If the results did depend significantly on the level, then I would tend to think that would mean there was something amiss with the design of the amplifier.

Even in the case under discussion in this thread, where the output impedance is going to be pretty high, it is probably still simpler to measure the output impedance by measuring the voltage drop when paralleling an extra resistor with a resistive load, rather than trying to adjust the load resistance for maximum power. (As DF96 said, you would need to adjust the resistance for maximum power, not maximum current.)


DF96 20th January 2013 03:41 PM

For almost any voltage generator the maximum current is obtained from a short circuit. An exception would be one with some form of foldback current limiting - unlikely in a valve audio output stage of almost any architecture?

As I said, if you want output impedance then you adjust the load for maximum power, not maximum current. This assumes that the source will not be damaged by such a load. If it will, then you have to apply bigger resistances and do some maths.

PlasticIsGood 20th January 2013 03:42 PM

Use one amp with signal generator at its input to drive, via an 8 ohm resistor, the output of the other, with its input grounded. Measure voltage across its output, and across the resistor so you can calculate the current. Repeat for appropriate frequencies and amplitudes. The rest of the OP is far too complicated to read.

midmoe 21st January 2013 03:19 AM

to "plastic is good"....are you into catalin radios? I thought that might be why your username is such....midmoe

tomchr 21st January 2013 03:30 AM

Usually shorting the output of an amp isn't recommended. It may be OK on a tube amp, but a sand amp will self-destruct if the output is shorted. Unless there's some protection circuitry to turn the output off.

I much prefer the method where two slightly different resistors are used and the output impedance derived from the different output voltages measured.


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