Yet another "what to do with those tubes?" topic

[rankot]

I have a 100V distribution transformer I took from an old distribution amp. I believe it has 8 ohm on secondary, but I am not sure how to check primary and secondary impedances. How to know power capability - to measure core size, as for power transformers, or there is another way to do it Please help!

[Rob Strand]

If you put an AC source on the secondary, then measure the voltage on the secondary and the voltage on the primary you can get the turns ratio:

So if its n:1 

n = Vprimary / Vsecondary

Then to calculate the primary impedance,

   Zp  = n^2 * Zs 

where Zs = secondary impedance = 8 ohms

Eg.  100V, 8 ohm:   
        Secondary test voltage 6Vrms in
        Primary Measured 27.4Vrms   
        Turns ratio calc:   n = 27.4 / 6 = 4.57
        Primary impedance calc:  8 * n ^ 2 = 8 * 4.57^2 = 167 ohms
        Power estimation calc:  Pest = 100^2 / 167 = 60W

Caveats:
- the test voltage must not saturate the transformer
- it's best to test at say 400Hz with a series capacitor to prevent DC loading on the amplifier.
- if you use 50Hz/60Hz mains you need to test at a voltage somewhat lower than the rating.
  If not IR drops in the transformer windings will cause large errors in the estimate for n.

When you get Pest you could compare it against the size of other transformer.  However,
the size has to do with the low frequency response.

Another thing to check is to measure the primary and secondary winding resistances.

You can also drive the primary and measure the secondary.

(Very roughly: A 60W transformer will be approx 1.2kg,  30W approx 0.8kg.)

[rankot]

Great Rob, thanks! I'll put 200mV source on one side and measure another, just to be sure not to make few kV.

[amptramp]

Ohm's law to the rescue.  If you have a distribution voltage rating and the tap you are using has a power rating (usually marked on the transformer), you use the equation:

V * V/P = R

If you have a 100 volt transformer, the 1 watt tap is 10,000 ohms, the 2 watt tap is 5000 ohms, the 0.5 watt tap is 20,000 ohms etc.

The transformer has to be able to pass the highest power it is rated for and the rating is the highest power tap.  The tube manual entry for an output tube will give you the impedance the tube wants to see, which may vary with voltage applied.  Tubes may have ideal impedances which are not available with your transformer but fortunately, it is not that critical when you are building an amp to just get it to work.

[rankot]

I have one, but no markings at all. I measured it's transformation ratio using 12AC few months ago, and concluded it must be some kind of 100V audio transformer, since it was connected at output of that old amp box and transformation ratio wasn't high, but stupid me didn't mark transformer leads when I removed it, thinking it will never be of any use... Now I want to make my first tube amp, and it seems that I can use it It has 4 leads on P side and 4 on S side, and when I put 220V on primary, secondary gives 52V (but I am not sure which of those leads I used to measure). So according to Rob's formulas, it has Zp of 143 ohm with 8 ohm Zs, and 70W of power (and yes, it is heavy). I will see what other leads do.

[anotherjim]

This page has some ready worked out PT as OT info...
http://locofonic.alphalink.com.au/valves.htm#avenue
I think it's cool that you might get something decent enough going with a PT that you already happen to have.

[rankot]

Great resource, thanks!

[Rob Strand]

It has 4 leads on P side and 4 on S side, and when I put 220V on primary, secondary gives 52V (but I am not sure which of those leads I used to measure). So according to Rob's formulas, it has Zp of 143 ohm with 8 ohm Zs, and 70W of power (and yes, it is heavy). I will see what other leads do.

If you are sure it was a 100V transformer then all the primaries probably connect together and all the secondaries probably connect together.   Do you know how to work out the "outsides" of tapped winding by measuring the voltages?

220V on the primary might be excessive if you tested at a low frequency.

The thing that is tricky is if there is multiple secondaries you don't know which one is 8 ohms.   Like if I saw 4 secondaries I'd be thinking  common, 4 ohms, 8 ohms, 16 ohms.   If you drive a test voltage V16 into 16 ohms and common then should be able to measure voltages: 
8 ohm tap to common V8 = V16 / sqrt(16/8) = V16 / 1.414
4 ohm tap to common V4 = V4 /  sqrt(16/4) = V16 / 2

[Forgot to add: you get the same conclusion if it's 32, 16, 8 or 8, 4, 2.]

If you have the old chassis or you can find a pic of the old amp you might be able to see labels 4, 8 16 ohms.

The thing that's harder to work out is which winding is 100V.  It's common that the outer windings are 100V, ie the ones that produce the highest primary output with a test voltage on the secondary.

Just to be clear the power estimation formula assumes the following:
- we know the transformer is 100V
- we know what the 100V winding is
- we know impedance rating on the secondary testing winding.

If all these are not known the power has to be estimated another way.  You can make an estimate from the size, weight, and winding resistances.  You can also do winding heat rise test (which can take several hours to heat up) but you have to assume the manufacturer has used the lowest temperature rating wire.

[PRR]

Heat is almost never a problem with audio transformers. Saturation, bass distortion, usually sets a lower limit than heat.

In this case we know one side can stand 100V of "audio" (bass limit not specified; and sometimes quite poor) and the other side has parasitic resistance which should be "small" compared to speaker inpedance. 0.5 to 1.0 Ohms DCR may be appropriate for an 8 Ohm connection. (Cost is usually more important than economy in that business.)

[Rob Strand]

parasitic resistance which should be "small" compared to speaker inpedance. 0.5 to 1.0 Ohms DCR may be appropriate for an 8 Ohm connection

For generic Tx's  0.8dB and 90% efficiency seems to be a good ball-park for  transformers < 100W.
That implies the total secondary referred transformer DC resistance is,

   Rt' = 0.1 * Rs  (ie. Rs = 10* Rt')

Where you can measure, Rt' as,

   Rt'  = RDC_secondary + RDC_primary (1/n)^2
   transformer n:1

It's a fairly loose estimate.  Transformers which have good bass, or are good quality, or are on the large end may very well have lower Rt'.     It only lets you estimate Rs not the power.

I suppose the bigger issues in using a 100V for a tube amp is the impedance match to the tube isn't going to be great and the core might saturate in a single ended stage due to the DC.  There's definitely not going to be any air-gap on the core of a 100V transformer.

[PRR]

> It only lets you estimate Rs not the power.

If you know the primary voltage (100V), the turns ratio (not yet discovered), and a rough estimate of a happy secondary load, then you have a rough estimate of Power.

[Rob Strand]

If you know the primary voltage (100V), the turns ratio (not yet discovered), and a rough estimate of a happy secondary load, then you have a rough estimate of Power.

It works fine for single winding primaries and secondaries.

It goes a bit haywire for multiple primaries and secondaries.

With multiple primaries and secondaries:

For the secondary side it works best for the highest impedance tap.   Do all measurements and calculations with the highest impedance secondary tap.   It usually works out better that way.   The lower impedance windings have lower DC resistance but it doesn't scale in proportion to the rated impedance because the mean-length turn varies (and the winding space in under utilized).

For the primary it's quite confusing to handle it correctly.   All the primary windings are rated at 100V.   The primary with the lowest number of turns gives you full power but the other windings deliberately drop the output power so you can "distribute power".  The insertion loss and efficiency workout best based on the outer primary winding (highest number of turns).  However a power calculation with 100V won't give anywhere near the right power with that winding.   So for many transformers, you need to work out the secondary impedances using the calculations in my previous post.   But you calculate the power using 100V on the winding with the least number of turns.

The above "normal" scenario is for generic transformers.   The insertion loss for the full power case is worse than the low power case.  Higher quality transformers might keep the insertion loss down for the higher power taps.    The uncertainty in how the transformer is designed means you can have large errors in the estimates.    One thing that saves the impedance estimate the true values are discrete values like 4, 8 etc.  That means you can choose the whole number impedance closest to the calculated estimate.

A different scenario for the primary is if it just has 100V and 70V taps that's a whole different problem.

[Edit: fixed a few typos.]

[Rob Strand]

Here's an example where the above method does work well if you use the primary outer windings to estimate the secondary impedance.  This transformer is a better design in that the insertion loss is low (close to 0.8dB) for the highest power configuration (ie. the lowest primary tap)
https://docs-apac.rs-online.com/webdocs/0326/0900766b80326f99.pdf

If you estimate the secondary impedance using the outer primary winding you get 5 ohms instead of 16 ohms (a large error). 

If you estimate the secondary impedance using the inner primary winding you get 15 ohms which is close to spot on with the expected 16 ohms; here the DC resistance for the 30W primary tap was estimated to be 27 ohms.

Be warned!

[PRR]

> Be warned!

Indeed.

I won't argue the typicality of that part. It is probably normal commercial grade.

Note that pins 13 to 18 are called "16 Ohms". And on the left we learn 13-18 reads 6.5mH.

6.5mH does not get up to 16 Ohms until 400Hz! Oh, it will couple an infinite generator to 16r just fine. But that generator working at 100Hz must supply 4X the reactive current as the real current that gets to the speaker. (And not saved by winding resistance: that doesn't equal reactance until 4Hz!)

Ah-- inductance measured with 1KHz. They should know better. An iron-core will show more inductance below a few hundred Hz (depending on lam thickness and resistivity). So it may be "OK" to 150Hz or lower.

Anyway: for a tube design the first choice is perhaps the highest impedance primary which is rated 10K impedance (1 Watt @ 100V). Or 2W which is 5K, or 5W which is 2. These may be decent matches for 25C5 class tubes at 150V of B+. While we should have a gapped core, my observation is that these things are so cheaply stacked that they are far from un-gapped. And since they turn up real cheap in odd places, we may forgive some bass error.

[Rob Strand]

6.5mH does not get up to 16 Ohms until 400Hz! Oh, it will couple an infinite generator to 16r just fine. But that generator working at 100Hz must supply 4X the reactive current as the real current that gets to the speaker. (And not saved by winding resistance: that doesn't equal reactance until 4Hz!)

That's a good point.   Previously I did check that inductance predicted the turns ratio and it seems OK in that at 1kHz we haven't hit the lossy region:  based on L, n=sqrt(Lp/Ls) = 24.2,  and based on V, n=vp/vs = 25.  So not bad agreement.

I didn't think about the bass response for the tube amp.  Like you correctly point out it's not going to be that great driven by a tube.    If you are lucky, those transformers give cutoffs specs for -1dB or -1.5dB or -3dB and usually end-up somewhere between 20Hz and 300Hz.   But that assumes a *low source impedance* not a tube.   With a low impedance drive on that transformer I get f(-3dB) = 7.9Hz and f(-1dB) = 30Hz for the 1W primary, and f(-3dB)=31Hz and f(-1dB) = 120Hz for the 30W primary.   Those figures look quite acceptable under normal conditions but they don't apply when driven by a tube drive.

On the good side there are people on the web using 100V transformers as tube output transformers.  So I guess you can get away with it to some degree.   However, they seem to be push-pull designs which avoids the gaping issue but means the transformer has to have a centre-tapped primary.

http://www.ozvalveamps.org/optrans.htm
http://home.alphalink.com.au/~cambie/6AN8amp/M1115.htm
http://home.alphalink.com.au/~cambie/6AN8amp/Grant_Wills_6CM5amp.htm

For the single-ended case I suppose you could trade output power for core saturation by using the primary taps with less turns.  You could also do something nasty like connect the tap to the +V rail drive the tube of one side of the winding then feed DC through the other side to cancel the field but then you might as well do a skewed push-pull.   None of this is ideal.   

I think I like the push-pull with the centre-tap.

[Rob Strand]

My apologies for dragging out this thread but PRR's comment about the low frequency response inspired me to check how much error occurs in the turns ratio measurements when you test at 50Hz.     What happends is the transformer inductance forms a voltage divider with the winding resistance and this makes the measured voltage lower and changes observed turns ratio.   That's without considering saturation which will make things worse.

You get different results depending on which winding you apply the test voltage to.

For the example transformer I gave the PDF for:

Test voltage                            Observed Voltage Loss factor
Input on secondary (16ohm)  0.9952
Input on primary (1W)           0.9874
Input on primary (30W)         0.8302

So as expected the least error occurs when you drive the outer windings.   
In this case driving the secondary and measuring the primary gives the least error.
On some transformers driving the primary and measuring the secondary will give a lower error as it depends on the relative winding resistances.