need help with a transformer primary

Started by Marcos - Munky, October 21, 2020, 09:04:35 PM

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Marcos - Munky

Not a stompbox related question, but a tube amp related one. So, a friend was out of town visiting his uncle, which used to repair some electronic stuff but got retired, and came back with a random transformer his uncle had on a corner. The only thing his uncle knows is it was taken from a tube amp, but he can't remember the model or if the transformer works, but he sent me so I can check if I have some use for it.

The transformer have 4 taps on primary and 3 secondaries. First thing I did was to check for continuity/resistance between each tap of each winding. All 3 secondary windings doesn't have continuity with each otehr and with the primary. So far, everything seems ok. Then I tested the primary.

All 4 taps have continuity with each other, with different resistances based on what pair of tap I measure. I randomly named them 1, 2, 3 and 4, looking at them from left to right. I got those measures:
- fixing tap 1, I got 8r for tap 2, 7.6r with for 3 and 13.4r for tap 4.
- fixing tap 2, I got 8r for tap 1, 1.4r for tap 3 and 7.4r for tap 4.
- fixing tap 3, I got 7.6r for tap 1, 1.4r for tap 2 and 6.9r for tap 4.
- fixing tap 4, I got 13.5r for tap 1, 7.4r for tap 2 and 6.9r for tap 3.

With tap 4 fixed, the taps seems to be something like 220-127-120-0 respectively, because the resistance rises as I measured the other taps. But if that's correct the same should happen if I fix tap 1 and measure the other ones, but the resistance drops if I move from tap 2 to tap 3. That's strange. The transformer could have two windings for the primary to be connected in parallel or series so I can choose the primary voltage, but if that was the case I guess I shoudn't have continuity/low resistance between all taps.

Then I connected a small transformer to one of the secondaries, so I could check voltage on the other windings and try to understand what's happening on the primary. I used a serial lamp as a safery measure, which didn't lit. I think this is a nice thing, but the transformer wasn't connected directly to the mains, so I won't take any hasty conclusions. Using a 13.74V small transformer, I got those voltages on primary:
- taps 1 and 2: 4.856V. taps 1 and 3: 4.117V. taps 1 and 4: 8.9V
- taps 2 and 3: 0.73V. taps 2 and 4: 4.095V.
- taps 3 and 4: 4.82V

Again, I can't conclude anything with those measurements. If I fix tap 1 and move from 2 to 3, there's a voltage drop. The same if I fix tap 4 and move from 3 to 2. So it's not a 0-120-127-220 primary. I can guess the secondary voltages if I assume one of them being the heather winding and one being the HT winding, but I got no clue on those primary pats.

Any guesses?

Rob Strand

#1
As far as the primaries go.   Your measurements of  "tap x to tap y" and "tap y to tap x" agree which is a good start.

No doubt all your measurements have some zero error from the multimeter.   That confuses things a bit because,
- measure winding with resistance A  you get readfing A + R0
- measure winding with resistance B  you get  reading  B + R0
However if you measure A + B in series you get reading (A + B) + R0.   If you add the two individual measurements you would estimate A + B + 2R0.      For three windings you get 3R0 error when you estimate from the sum.

It looks like your meter is adding R0 = 1 ohm but we will try to work it out instead of assuming it.


Anyway, the way I look at it is for any pair the *smallest* measurement is for a single winding.  So I get,

Smallest resistance pairs
tap          R raw
1 to 3     7.6 ohm
2 to 3     1.4 ohm
3 to 4     6.9 ohm

So what that's looking like is the windings aren't in series.
They all join at 3.

Presumed two windings at a time:

               R raw           R calc from single         Difference ohms
1 to 2     8.0 ohm          7.6 + 1.4 = 9                 1.0
2 to 4     7.4 ohm          1.4 + 6.9 = 8.3              0.9
1 to 4     13.4 to 13.5 ohm  7.6 + 6.9 = 14.5     0.9 to 1.0       ; [arithmetic error here but doesn't affect conclusions]

So the meter offset R0 seems to be between 1.0 and 0.9, average 0.95 ohms

We can correct the measurements for single windings

Smallest resistance pairs
tap          R raw        Rcorrected
1 to 3     7.6 ohm    6.65 ohm
2 to 3     1.4 ohm    0.45 ohm
3 to 4     6.9 ohm    5.95 ohm

And as a check the largest measurement 1 to 4 with 13.4 to 13.5 ohm, if we apply R0 correction the true resistance should be 12.5 ohm.  Now add 1 to 3 with 3 to 4,     6.65 + 5.95 = 12.6 ohm.

Check other pairs:
Check 1 to 2:     measured = 8 - 0.95 = 7.05 ohm,  calculated = 6.65 + 0.45  = 7.1 ohm
Check 2 to 4:     measured = 7.4 - 0.95 = 6.45 ohm, calculated  =  0.45 + 5.95 = 6.4 ohm

All the resistances check out to within 0.1 ohm.

Based on that it's sure looks like 3 is a common point for all the windings.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Rob Strand

#2
Quote- taps 1 and 2: 4.856V. taps 1 and 3: 4.117V. taps 1 and 4: 8.9V
- taps 2 and 3: 0.73V. taps 2 and 4: 4.095V.
- taps 3 and 4: 4.82V


If we look at all the voltages to 3:

Tap                 V
1 to 3           4.117V
2 to 3           0.73V
4 to 3           4.82V


Now look at the non "3" pairs

                   measured           Summing individual voltages
1 to 2         4.856V                  4.117 + 0.73 = 4.847V
1 to 4         8.9V                       4.82 + 4.117  = 8.937
2 to 4         4.095                      4.82 -  0.73   = 4.09V


So we can see
- winding 13  is in phase with winding 23.
- winding 13 is in phase with winding 43
- however winding 23 is out of phase with winding 43

I think that's right.

Draw it like this:

Winding:
13:            3 is dot
43:            4 is dot
32:             2 is dot

From that you can see why 2 to 4 voltage subtracts winding23.


We still don't know if 4 or 1 is the intended "common".

1) 1 common  (I expect this the intended connection)
Tap     Vtest              Vtap
3          4.117            102
2          4.856             120
4          8.9                 220

2) 4 common (expected to work but not intended)
Tap     Vtest             Vtap
2          4.095            101
3          4.82               119
1          8.9                  220


EDIT:
I chose 1 common based on the phase, however if you look at the total winding resistance for each combination it looks like 4 common has lower resistance.  That would imply 4 common as 4 is the start of the winding.   Please check.
[See reply #5: using 4 as common gives lower primary resistances for 100V and 120V, so it is likely 4 is the start of the winding and the intended primary 0V connection.]
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Marcos - Munky

Wow, thanks for such an in depth analysis! It does makes sense winding 23 is out of phase with the other ones, since the voltage drop. It's strange the transformer doesn't follow a 0-110-127-220 order, or 0-100-127-220 order, or any other crescent voltage order, but anyway.

Quote from: Rob Strand on October 22, 2020, 01:03:04 AM
I chose 1 common based on the phase, however if you look at the total winding resistance for each combination it looks like 4 common has lower resistance.  That would imply 4 common as 4 is the start of the winding.   Please check.
How can I check for that? I guess I should check which one is the heater winding, by applying voltage on one secondary and checking the voltage on the other ones. Then, based on the measured voltage, calculate the ratio for 6.3V and use that ratio for the primary. Is that correct?

Btw, when I applied voltage on one of the secondaries, the small transformer (which is a 12V one) gave me a 13.74V reading and I was able to do the previous measurements. From the voltages I got from the other secondaries, I was able to guess I applied voltage on the HV secondary. This morning I connected the same small transformer to another secondary (a lower voltage one), using the serial lamp again, but this time the voltage dropped to around 7V and the small transformer got a bit hot, then I turned off everything. Does this indicates a problem with the transformer, or it may be related to the fact this time the voltages on the transformer were higher than before and the small transformer felt the small resistances on the big transformer windings as "shorts"?

amptramp

Resistance readings do not necessarily correspond to voltage ratios because you can have inner and outer portions of a winding.  Right behind me I have a Philco 37-10 radio that gives the output transformer winding resistances as 370 ohm on one side and 330 ohm on the other.  The number of turns is identical but one side was an inner winding and the other was the outer.

You are correct to use voltage ratios.

Rob Strand

#5
QuoteIt's strange the transformer doesn't follow a 0-110-127-220 order, or 0-100-127-220.
Yes, you can't think too much about transformers they are all over the place.    I have seen taps on primaries out of phase before for multi-voltages.   No big deal here.   Like some transformers were 110V with an extra unconnected 10V *primary* winding.  If you don't use the 10V winding you get 110V, if you wire the 10V winding in-phase you get 120V, and if you wire the 10V out of phase you get 100V.

QuoteHow can I check for that?
I was just saying maybe you should check my results.  (I was doing something yesterday and I was adding bits to the post as I was doing it).

The 1 common vs 4 common issue isn't about the voltages.    The issue is *both* 1 common and 4 common produce the *same* input voltage taps.    In other words you get the right output voltage if use the primaries in either direction.   The reason is the taps are 0V, 100V, 120V, 220V, so powering between "120V" and "220V" is like renaming them as "100V" and "0V".   Perhaps it's easier to see with this table

Tap     Vtap                Vtap
1          0                      220
3          100                 120
2          120                 100
4          220                  0

This is where the resistance comes in.     Normally the 0V tap is the start of the winding.    That uses the turns closer to the bobbin which are shorter and have less resistance.   Less resistance means the transformer runs cooler.

Smallest resistance pairs  (from previous post)
tap           Rcorrected
1 to 3      6.65 ohm
2 to 3     0.45 ohm
3 to 4     5.95 ohm

            Using                                 Using
            1 as common                    4 as common
Tap     Vtap         Rtotal                 Vtap              Rtotal
1          0V              -                        220V             12.6
3          100V         6.65                  120V              5.95
2          120V         7.10                   100V             6.4
4          220V         12.6                    0V                 -


So you can see if you use 4 as common the total primary resistance is less for 100V (6.4 ohm vs 6.65 ohm) and 120V (5.95ohm vs 7.10 ohm).   To me that means the start of the winding is likely to be 4.

So what you can check is to see if the wire for 4 looks like it starts near the core and the wires for 1, 2, 3 are on the outside.   That would confirm it.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

blackieNYC

I don't know if this helps, but on a couple occasions I have used a signal generator 50 or 60 Hz sine wave to imitate the Ac voltages.  For a 120vac primary winding, I'll apply 1.20 Vac so the math is easier.  Seems to work in identifying secondaries.
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Marcos - Munky

Got it, Rob. Thanks again for the very detailed reply. I'm at work rn, so I'll check later if I can see which wire is closer to the core. But I agree it seems tap 4 is the start of the winding.

BlackieNYC, I don't have a signal generator unfortunally. But your idea is very nice. When I have to identify secondaries, I use this small 12V transformer to feed the primary, measure all voltages and do some math. It takes a bit of time, but at least I don't have to deal with high voltage as if I use the mains.

Marcos - Munky

I took a look, and I can't see which winding is closer to the core. The transformer don't have wires but terminals, everything else is sealed, so I can't really see the windings.

I did some real voltage measures today. Fed taps 4 and 3 with 128.5V from mains. Got 7.82V, 29.35V and 355.2V from the secondaries. Assuming voltage losses and that 7.82V droping to exact 6.3V, the other voltages will drop to 286V and 23.6V. Great, this high voltage is almost the voltage I need to power a Marshall "18W"-like amp I'm building!

Rob Strand

QuoteI did some real voltage measures today. Fed taps 4 and 3 with 128.5V from mains. Got 7.82V, 29.35V and 355.2V from the secondaries.

I got slightly higher voltage estimates.


     No load   No load                8% reg
     test           mains adjust        at 100VA

Vp     128.5           120                         120
Vs1     7.87             7.30                         6.76
Vs2     29.35           27.41                25.38
Vs3     355.2           331.7                307.1


At a rough guess the transformer is about 100VA. 

Transformer estimated at approx 100VA based on primary resistance at 120V
with some allowance for 120V not using all the primary winding space.
No idea what the secondary resistances are or the core size.

For less than 100VA load the output voltage will be between the No load and
8% reg voltages in the table.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Marcos - Munky

Here the mains voltage is 127V-ish. So the high voltage probably will be a bit higher than what I need, but there's the option to use a resistor to drop it a bit.

Core size is 3.2 cm X 3.2 cm. Secondary windings resistances are 0.4r, 11.7r and 38r, from lower to higher voltage windings. For the tubes I'll use, I calculated around 51.4 VA needed.

Rob Strand

#11
QuoteCore size is 3.2 cm X 3.2 cm. Secondary windings resistances are 0.4r, 11.7r and 38r, from lower to higher voltage windings. For the tubes I'll use, I calculated around 51.4 VA needed.

Upfront I think you might need to measure the DC resistance of the 6.3V winding more accurately, say using a 10mA current like we discussed in some of your older transformer threads.

I did some temperature rise checks.  For the 120V primary a reasonable upper limit would be a 90VA rating and 9.5% regulation.  Assuming the same regulation on each winding of 9.5% and a total of 90VA,

sec1        6.66V at 0.58A
sec2        25.0V at 70mA
sec3        303V at 280mA

For an 18W'er I'm assuming 3x12AX7 tubes and 2xEL84 tubes.   The heater current for both sections of the 12AX7 is 300mA and the EL84 is 760mA each.   That totals a 6.3V current  = 3*300mA + 2*760mA = 2.42A.   At this point I *didn't* factor in a tube rectifier, which could be another 1A.

The 2.4A requirement seems soooo much higher than the basic secondary 1 rating of 0.58A.

So I tried to redistribute the load currents using the fact that you only need 51.4VA which is much lower than the tx rating.   So we pull more out of the 6.3V winding and less out of the 300V winding.   To a reasonable approximation, if the total winding power dissipation in use is less then the total winding power dissipation at full load.

The main point to realize is if we pull more current out of one winding the regulation on that winding gets worse.

So doing a lot of messy calculations  I concluded that pulling 2.4A from the 6.3V rail very marginal:
- quite a lot more than the basic rating, even with the idea of using less current from 300V winding.
  Without testing the temperature rise on that winding it's hard to know if you would get a hot spot
  on that winding.
- the regulation on the 6.3V winding is 22.9% due to the increased load and resulting voltage on that winding drops to 5.94V,
  which is starting to be marginal
- if we reduce the load on the 300V winding, the overall heat dissipation on the transformer
  is 5.0W which is less than the maximum of 8.6W, and that's with the 2.4A load on the 6.3V

I suppose you would really need to do a load test.   Keep in mind loading one winding will show better regulation than when all windings are loaded.


I've attached my spreadsheet but it's not for humans.   

To deciper it:
- green cells are things which can be entered
- orange cells are key results.
- s = secondary, p = primary,  t = total
- nl = no load, fl = full load
- All the dashed '  parts is my way of scaling all the windings to a common voltage.
  I chose 220V.   Anything with a dash '  is for calculation purposes.
- Winding resistances allow for the transformer getting hot (this is the KRtemp factor of 1.2).

There's two parts:
- The first part with the orange results at the middle-right are the secondary VAs,
   Maximum Currents and *loaded* voltage.  Assuming all the windings have the same regulation.
- The second part with the orange result at the bottom right is a new set of calculation where
   I can enter each secondary current.  The entered secondary current are relative to 220V.  The
   true currents appear in the orange results.
   I have increased the 6.3V winding current and decreased the 300V winding current
   so the total VA matches your 51.4VA requirement.

[Click to enlarge]

bug: the "Is [A]" columns should be "Is' [A]" as they are relative to the 220V normalized voltage.

If you also wanted a tube rectifier with another 1A load on the 6.3V line I'd say that's definitely going too far.

If you can get an accurate measurement for the 6.3V winding resistance maybe that will change the calculations
for the better.  It's very easy to get wrong estimates when using a DMM to measure low resistance values.

(Edit: A minor point is I've included 1.91VA from the 25V winding, in reality it should be set to 0VA and the 300V
winding set to 35.17 +  1.91VA so it still makes up 51.4VA.   Not a big deal.  It wont change the numbers much.)
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Marcos - Munky

Thanks again for the very detailed reply, Rob!

Quote from: Rob Strand on October 27, 2020, 05:18:25 AM
Upfront I think you might need to measure the DC resistance of the 6.3V winding more accurately, say using a 10mA current like we discussed in some of your older transformer threads.
Ouch, my bad. You're correct, I really forgot on the 317 as a constant current source. I tried to do it, but I think I made a mistake. I used a 100r resistor (measured 110.3r), so I should get a 13mA current (11mA for the measured resistance). I connected the positive lead of my power supply to the 317, with the resistor/reg pin connected to one tap of the winding and the other tap connected to the negative lead. Measured a voltage of 8.89V and a current of 3.2mA (shouldn't be 11mA?) on the circuit, which gives me a resistance of 2.8K...

Quote from: Rob Strand on October 27, 2020, 05:18:25 AM
I did some temperature rise checks.  For the 120V primary a reasonable upper limit would be a 90VA rating and 9.5% regulation.  Assuming the same regulation on each winding of 9.5% and a total of 90VA,

sec1        6.66V at 0.58A
sec2        25.0V at 70mA
sec3        303V at 280mA

For an 18W'er I'm assuming 3x12AX7 tubes and 2xEL84 tubes.   The heater current for both sections of the 12AX7 is 300mA and the EL84 is 760mA each.   That totals a 6.3V current  = 3*300mA + 2*760mA = 2.42A.   At this point I *didn't* factor in a tube rectifier, which could be another 1A.
I'm using two 6N2P for preamp and phase inverter (340mA ± 35mA for heaters, 3.2mA for each triode) and two 6EM5 as the output tubes (800mA for the heater and 35mA for the pentode). No tube rectifier, but a full wave diode rectifier. So it'll be 135mA for the high voltage tap (83mA plus compensation of 62% because of the full wave rectification) and 2.35A for the heaters.

Quote from: Rob Strand on October 27, 2020, 05:18:25 AM
I suppose you would really need to do a load test.   Keep in mind loading one winding will show better regulation than when all windings are loaded.

I've attached my spreadsheet but it's not for humans.   
Agreed, I'll try to do it after I finish some stuff and clear my workbench. Also, agreed on the spreadsheet :icon_lol:

While the transformer is "closed", there's a bit of the heater wire visible, and I think I can measure it with a caliper. Is this measure useful?

Rob Strand

#13
QuoteMeasured a voltage of 8.89V and a current of 3.2mA (shouldn't be 11mA?) on the circuit, which gives me a resistance of 2.8K...
That seems a long way off.   [EDIT:  when you test you need to feed the output into a load, like a 1ohm or 10 ohm, resistor then measure the voltage across the resistor, or the current through the resistor.]

Perhaps try this simple test circuit,

So set Vin to about 10V DC,  R = 1k.  Nothing needs to be precise as we measure stuff later, for example 9V is fine.
Where it shows L you connect the transformer 6.3V winding.
The transformer is not powered during the test.

Measurements:
- Before you start measure the 1k resistor value with DMM
- measure the voltage across 1k resistor ; VR in the pic
- measure the voltage across the 6.3V winding; VL in the pic
  It is important to measure this right at the transformer terminals.

Calculations:
- I_test   = "Voltage across 1k"  /  "Measure resistance of 1k"
- Rs = "Voltage a across 6.3V winding" /  I

Rs is the resistance of the 6.3V winding.


So for example,

Measurements:
- 1k resistor measures 997 ohm
- V across 1k = 9.73V
- Voltage across 6.3V winding   = 3.7mV

Calculations:
- I_test = 9.73/997 = 9.76mA
- Rs = 3.7 mV  / 9.76mA = 0.379 ohms


You can see that the voltage across the transformer is quite small and needs some care to measure.
As a check you can repeat the test with say a 100ohm or 220ohm resistor.  It will get warm so you need a 1W or 5W.
The lower resistor increases to test current to 50mA to 100mA and that increases the voltage across the transformer
to 20mV to 40mV so you get more digits in the measurement.

Quoteand 2.35A for the heaters.
So the transformer might just make it.

QuoteWhile the transformer is "closed", there's a bit of the heater wire visible, and I think I can measure it with a caliper. Is this measure useful?
If it is easy to do, it's worth doing.   The rating of wires in the transformer depend on the heat so the true rating is much less than the "open" air ratings you find in wire tables.   Transformer designs often use current density A/mm^2 or A/cm^2.  Your 6.3V winding will have a high current density for sure.


Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Marcos - Munky

Quote from: Rob Strand on October 31, 2020, 05:43:13 PM
That seems a long way off.   [EDIT:  when you test you need to feed the output into a load, like a 1ohm or 10 ohm, resistor then measure the voltage across the resistor, or the current through the resistor.]
Yeah, I didn't used a load, just the transformer itself. That may be the issue.

Quote from: Rob Strand on October 31, 2020, 05:43:13 PM
Perhaps try this simple test circuit,
Ok, so I first used a 100r resistor. Measured the "zero" resistance (touching the probes together), then the resistor, then calculated the difference between the values for the real resistor value. Got a 98.9r resistor.

VR = 8.65 V, that makes i = 87.462 mA. VL = 6 mV, that makes RL = 0.0686r. Very low value. Then I moved to a 1k resistor to check it again (yeah, I did the reverse order).

Did the "zero" resistance measurement again, then checked for the real resistance value of the resistor. 875.2r, kinda off but anyway.

VR = 8.89 V (I didn't used a regulated power supply), that makes i = 10.158 mA. VL was measuring zero... then I noticed my meter have a mV reading. Ouch. Selected the mV reading, got VL = 0.71 mV. That makes RL = 0.0699r. Again, a very low value.

Then I went back to a 100r resistor to check it with the mV reading. Got a 97.5r resistor (a different one than the one used before), VR = 8.75 V. So i = 89.744 mA, so far it's close to the first calculated current. VL = 7.13 mV, so RL = 0.0795r. Hmm, kinda higher than the other values.

Does this seems correct?

Rob Strand

#15
QuoteYeah, I didn't used a load, just the transformer itself. That may be the issue.
I should have been clearer.  When you test the current source you need to check it with a resistor so you can measure the output current and also see if you can predict the value of a low ohm resistor with the test set-up.   No problem though, the resistor only method is perfectly OK,  all it means is you have to measure the VR and the resistance.   If you aren't doing this everyday that's no big deal.

QuoteRL = 0.0686r
RL = 0.0699r
RL = 0.0795r
Does this seems correct?
Hmm, kinda higher than the other values.
It looks quite reasonable.  The difference from lowest to highest is only about 0.01ohm which is 10 milliohm.     That might be because you didn't put the dmm probes right on the transformer wires.   You can easily get contact resistances of 30 milliohm if you measured on the "connecting wire" side of the contact same  -yeah pretty weird.

So the biggest conclusion out of all that is the winding resistance is waaaaay smaller than the 0.4 ohm you measured directly from the multimeter.    If we take a ball-park figure of 0.07 ohm from your measurements  that's a factor of 5.7.   That's why it's important to measure low resistance with the right set-up.    That has a massive impact on the current estimate for the 6.3V winding.     

Very roughly, before I got 0.58A  for 6.3V winding so now it's going to be more like 3.3A.    That's looking much more comfortable.   It does mean the 6.3V windings will have higher voltages than I estimated before.   I'll need to put your new measurement  into my crazy spreadsheet to get a better estimate but really it's not going to be a lot different to the 3.3A.  Also since you are only pulling 51VA from a 90VA transformer you have plenty of unused capacity for extra loading on individual windings.  It's actually looking pretty good.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Rob Strand

OK I got a chance to plug the new numbers in.
here's a summary of the main values.


Transformer Estimate Summary                                         
'MM's TX, V2.0,  4 Nov 2020                                                               
                                                               
                                                               
40 deg C rise, 8.57W total power dissipation (when hot)                                                               
nl = no load,  fl = full load                                                               
                                                when hot               
                                                reg1% = 100%*(Vnl – Vfl) / Vfl               
Vp [V]          Rp @ 20C     VA            reg%               
100                6.4                 80.1          10.7               
120                5.95               93.8           9.1               
220                12.6               107.6         8.0               
                                                               
Current rating estimates                                                               
120V primary only                                                               
All windings reg% =   9.1                               
Vs nl [V]         Rs @ 20C    VA secondary     V_fl [V]         Is_fl [A]
7.3                   0.07               19.2                    6.69                2.87
27.4                11.7                1.6                       25.1               0.06
331.7               38                  73.0                    303.9               0.24
                                                               
                total VA =              93.8                               
                                                               


For 35C rise, total power dissipation is about 7.5W.  Derate currents by about 6.9%.

                                                           
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Marcos - Munky

Quote from: Rob Strand on November 03, 2020, 07:45:39 PM
QuoteYeah, I didn't used a load, just the transformer itself. That may be the issue.
I should have been clearer.  When you test the current source you need to check it with a resistor so you can measure the output current and also see if you can predict the value of a low ohm resistor with the test set-up.   No problem though, the resistor only method is perfectly OK,  all it means is you have to measure the VR and the resistance.   If you aren't doing this everyday that's no big deal.
No, it was my bad. I did just a bit of reading on the 317 as a current source, went with what looked to be the simplest approach and didn't even thought on the load resistor. The resistor only method was also kinda simple, it took me more time to throw all values on excel and set up some simple equations than measuring the whole thing :icon_lol:

Quote from: Rob Strand on November 03, 2020, 08:40:17 PM
Vs nl [V]         Rs @ 20C    VA secondary     V_fl [V]         Is_fl [A]
7.3                   0.07               19.2                    6.69                2.87
27.4                11.7                1.6                       25.1               0.06
331.7               38                  73.0                    303.9               0.24
Now those are nice numbers. Since I'm drawing 2.35A from the heaters tap, I'm still below full load estimation. The high voltage tap will use 135mA, a bit more that half the estimate value. The 25V-ish tap won't be used at all. I'll still have to watch out for voltages higher than the expected, but a few resistors can solve this if needed.

Thanks again for all the help, Rob!

Rob Strand

QuoteNow those are nice numbers. Since I'm drawing 2.35A from the heaters tap, I'm still below full load estimation. The high voltage tap will use 135mA, a bit more that half the estimate value. The 25V-ish tap won't be used at all. I'll still have to watch out for voltages higher than the expected, but a few resistors can solve this if needed.

Thanks again for all the help, Rob!
No worries, it work out OK in the end.   It looks pretty close to a textbook design now.

Yes, worthwhile watching the 6.3V voltage.  My table is for full load on *all* windings.  Since you won't be loading down the 300V winding to full capacity we would expect the voltage on the 6.3V winding to rise a bit.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Marcos - Munky

I'll watch both voltages. It won't hurt to use a resistor to drop both voltages as a safety measure, then measure voltages under load and adjust the resistor value as necessary. Thanks again!