unkown output impedance measurement?

Henry, do you mean the true Thevenin O/P impedance of the unit as a "near voltage" source, i.e. very low indeed, less than, say, 0.5 ohm? Or do you mean what is the nominal speaker impedance that it is set up for, e.g. 4, 8 or 16 ohms? Cheers, Roger

Henry, do you mean the true Thevenin O/P impedance of the unit as a "near voltage" source, i.e. very low indeed, less than, say, 0.5 ohm? Or do you mean what is the nominal speaker impedance that it is set up for, e.g. 4, 8 or 16 ohms? Cheers, Roger

A low, known AC voltage is applied to the secondary winding, and the voltage is measured across the other. Be careful, because the voltage on the primary winding will be high enough to shock you.

The voltage on the primary over the applied voltage gives you the turns ratio. Square that number and you have the impedance ratio from primary to secondary. Since the transformer is in a push-pull application, the primary impedance is 'Plate to plate'.

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I would go for a set of watty resistors in series, tone source and a DVM on AC volts, with amp set at low output. If on some set of resistors of total value of R , the o/p voltage is greater than with value R+d and R-d , where d is about R/4 then R is the output impedance. If highest at R-d then go down, in steps, till it peaks, if R+d is highest then go higher for peak value

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In message , Charles writes

Just in case the amplifier is 'unhappy' with such a low load, it might be better to load the output with a resistor low enough to produce a less severe - but measurable - drop in output voltage (say by 10 or

20%). Then do a simple 'pot-down' calculation to get the output impedance.

--
Ian

That is how we measured the input impedance on the PRC77 The input was fed a 1 kHz tone, and a series resistance was added, till the modulation dropped to half.

If it is an old military radio it is likely 600 ohms. What is the brand and model, or the military ID number?

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be

and

Sorry, not so. This will only give you what I originally mentioned, i.e. equivalent generator source impedance (see Thevenin), and it will be very low - no use in selecting speakers. You need the O/P transformer (OPT) ratio. Make sure the receiver/amplifier is OFF. Use a filament transformer run off a variac to energise the OPT secondary (use the speaker terminals) - keep it low, say 2 to 4 VAC. Measure the OPT primary and secondary voltages at a few levels. Calculate each ratio and average them. Figure out the correct plate load for the O/P tube (not covered here) - it will likely be in the range 4 Kohms to 8 Kohms, call it Rp. Then the correct speaker load will be Rp/OPT ratio squared. Ex: Rp =3D 5000 ohms, OPT ratio found to be

36:1. Then, speaker should be 5000/(36)^2 =3D 5000/1296 =3D 3.86 ohms. So use a 4 ohm speaker. Cheers, Roger

Surely this IS the output impedance?

But what does the invariable negative feedback (from the OPT secondary to an earlier amplifier stage) do to the output impedance?

--
Ian

Roger

Sorry, not so. This will only give you what I originally mentioned, i.e. equivalent generator source impedance (see Thevenin), and it will be very low - no use in selecting speakers. You need the O/P transformer (OPT) ratio. Make sure the receiver/amplifier is OFF. Use a filament transformer run off a variac to energise the OPT secondary (use the speaker terminals) - keep it low, say 2 to 4 VAC. Measure the OPT primary and secondary voltages at a few levels. Calculate each ratio and average them. Figure out the correct plate load for the O/P tube (not covered here) - it will likely be in the range 4 Kohms to 8 Kohms, call it Rp. Then the correct speaker load will be Rp/OPT ratio squared. Ex: Rp = 5000 ohms, OPT ratio found to be

36:1. Then, speaker should be 5000/(36)^2 = 5000/1296 = 3.86 ohms. So use a 4 ohm speaker. Cheers, Roger

(snip)

Ian, it reduces the source output impedance of the amplifier as a voltage generator (increasing the damping factor) but does not affect the speaker impedance to be used. The speaker impedance determines the O/P tube plate (anode) load via the OPT ratio. Cheers, Roger

In message , Engineer writes

But the original question was "What the easiest ways to measure the output impedance of a push pull tube amplifier?" It didn't mention loudspeakers.

--
Ian

Henry mentioned a miltary radio with a high output impedance (but I don't know any with P-P output...) I have a BC348 and the OPT (from a single 6K6 O/P tube) is designed for 'phones only, both "high" and "low" impedance. The "low" impedance 'phone O/P is 300 ohms so using a regular LS is not possible. I put a typical radio OPT in this radio (I left the original 'phones OPT in place so it could be restored to original sometime in the next 100 years!) The new OPT and the nominal

8 ohms LS used puts a reasonably correct plate load on the 6K6, about 7K. Volume is quite good - I'd guess about 2 watts output (but not measured.) There is no NFB in the original radio but there is a little bit now as I routed the Rk decoupling cap to the top of the OPT secondary as a token NFB (properly phased.) Of course, invariably there is a lot of NFB on P-P audio amplifiers and, as noted above, it reduces the true output impedance of the amplifier as a voltage source, but it does NOT change the correct LS impedance as the latter defines the O/P tubes' plate-to-plate load. I don't know why Henry first asked about a "PP amplifier then "morphed to a "military radio" but it doesn't matter - it's all the same physics! Cheers, Roger

Neil Sutcliffe was kind enough to send me a manual on his GR 783A which is a more recent version of my GR 583A Output Impedance Meter. The o/p xfrmr on my Hmmarlund Super Pro 210X Type O is 10 ohms and I looked inside and saw no evidence that it was changed. I confirmed that the meter works by the same technique on my Ten Tec SP 325 which is 600 ohms and it measured 600 ohms. The GR 583A is a neat piece of gear, easy as pie to use.

If the output really is 600 ohms it sounds like it uses a cathode follower and was never designed to feed a speaker. 600 ohms is the old line driving impedance and headphones are available that will work fine from this at line level.

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    Dave Plowman        dave@davenoise.co.uk           London SW
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Dave The SP 310X has a pair of 6F6 triode connected driving a pushpull transformer with output of 15 watts. It was mainly a military Rx BC 794 etc butmine is a civilian model. No cathode follower. Any headphones work fine at 10 ohms.

I can't see rhyme or reason for such a high power output at that impedance.

If 10 ohm headphones work fine I doubt the output is 600 ohms.

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    Dave Plowman        dave@davenoise.co.uk           London SW
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