No load voltage readings with a toroid PT

2L man

Tele-Afflicted
Joined
Nov 23, 2020
Posts
1,969
Age
63
Location
Finland
Obviously filament secondary coil current increase primary coil "loss" too and then effect HV decreasing it more?
 

mountainhick

Tele-Holic
Joined
May 2, 2021
Posts
609
Location
Rocky Mountains
@joulupukki and @mountainhick

You don't necessarily need an app. You can use Merlin's method of estimating the source resistance and then use his chart to estimate the B+ and the ripple current based on the source/load ratio.

I'll walk through simulating it in LTspice first using the specs for the 50VA 05TC300 from their datasheet. I'm not going to do a complete tutorial on LTspice. I'm just going to go through a few models to show how we can simplify this down to Merlin's method.

Hopefully @andrewRneumann will check this for accuracy.

View attachment 1027717


The first thing we need to do is calculate is the turns ratio. Antek gives an input voltage of 120 Vrms and an unloaded output voltage of 310 Vrms.

Vpri = 120
Vsec = 310
Turns ratio = Vpri / Vsec

120 / 310 = 0.38709...

The impedance/inductance ratio is the turns ratio squared.

Inductance ratio = (Vpri/Vsec)^2

0.38709... ^ 2 = 0.14984...

In LTspice we'll create 4 inductors with mutual inductive coupling to simulate the two primary coils and the two secondary coils. We don't know what the actual inductance values are for those coils, and we don't really need to know for a simple model. We can just use an arbitrary high value of 100 henrys for the primary coil, and calculate a value for the secondary coils using the inductance ratio.

View attachment 1027734

Using the inductance ratio, we calculate a value for the secondaries:
100 / 0.14984... = 667.36111...

We create two 100H inductors, L1 & L2, to simulate the primaries, and two 667.311H inductors, L3 & L4, to simulate the secondaries. We give these mutual inductance with the K statement (K1 L1 L2 L3 L4 1). We also need to add the DC resistance for each coil. So we create resistors in series with the inductors using the values from the Antek datasheet (note that in LTspice you could simply add the series resistances to the inductor properties instead of creating separate resistors).

We add a voltage source for the 120Vrms 60Hz mains input. In LTspice this needs to be Vpeak, not Vrms.

Vpeak = Vrms x sqrt2

120 x 1.41421... = 169.70562...

Lastly, we need a load to simulate. Going by the datasheet example, 295Vrms output with a 60mA load. Using Ohm's law:

295 / .06 = 4916.66666...

Adding a 4916.67 ohm resistor for the load, and running a simulation (0ms to 100ms = 6 cycles):

View attachment 1027762

Vrms: 298V
Irms: 60.609mA

The voltage is off by about 1% compared to the datasheet, most likely because we are not modeling any losses.


Since we're probably going to be using the primaries and the secondaries in parallel, we don't really need to model all of the coils, so we can simplify the model by combining them. The turns ratio doesn't change. We just need to divide the series resistances in half.

View attachment 1027770

Running the simulation we get the exact same results:

View attachment 1027771

Vrms: 298V
Irms: 60.609mA


Now, moving on to Merlin's method found here:


In that article, he describes how to calculate the source resistance of a transformer as follows.

The source resistance is equal to the resistance of the secondary coil, plus the resistance of the primary coil when 'reflected' to the secondary:
Rs = Rsec + Rpri/(Vpri/Vsec)^2
Where:
Rsec = measured resistance of the secondary winding.
Rpri = measured resistance of the primary winding.
Vpri = RMS primary voltage (e.g., mains voltage).
Vsec = RMS secondary voltage measured with no load.



We already did part of this calculation, which is finding the impedance/inductance ratio.

120V / 310V = 0.38709...
0.38709... ^ 2 = 0.14984...

The DCR of each primary is 22 ohms. In parallel that's 11 ohms. We take the combined resistance of the primary and divide it by the impedance ratio, which gives us the reflected impedance of the primary.

11 ohms / 0.14984... = 73.40972... ohms

Now we add that value in series to the resistance of the secondary. Since each secondary coil is 235 ohms, that's a parallel resistance of 117.5 ohms.

117.5 ohms + 73.40972 ohms = 190.90972 ohms

We can get rid of the inductors in our model, and simply use a series resistance of 190.91 with a voltage source. The voltage source will be the unloaded secondary voltage. In LTspice, this needs to be Vpeak (310 x sqrt2).

View attachment 1027795

Running the simulation we get practically the same result, accounting for rounding errors.

View attachment 1027800

Vrms: 298.01V
Irms: 60.612mA



So far, all we've done is simulate an AC load to show how we can model just a transformer. That's all I have time for at the moment. I'll post an update with rectification and filtering, and we'll compare it to Merlin's source/load ratio chart and calculations.

First read through I follow you. Clear and understandable. Thanks! A lot of it is already familiar, but your presentation closes a couple gaps. It'll take me a bit more of my own effort to engrain.

Always grateful for your contributions Phryigian!
 

joulupukki

Tele-Holic
Joined
Nov 26, 2020
Posts
687
Location
Utah
@Phrygian77

In your example, you hypothesize a 60 ma load. How specifically do you come up with that total in a real world case? The spec sheet combined max current ratings for the tube plates and screens? Or any derating?
+1 to this question. My best guess is that it's the average DC current drawn by the amp as described by The Valve Wizard but would you also include the heater lines in that calculation or ignore them since they are not the main secondary lines we're talking about?

Edit: my best guess is that we ignore the heater part of the circuit/transformer since I don't see any of that in the calculations so far. Seems like for heaters, as long as the transformer provides enough current (and the right voltage) for the tubes you're using, should be good to go.
 
Last edited:

mountainhick

Tele-Holic
Joined
May 2, 2021
Posts
609
Location
Rocky Mountains
+1 to this question. My best guess is that it's the average DC current drawn by the amp as described by The Valve Wizard but would you also include the heater lines in that calculation or ignore them since they are not the main secondary lines we're talking about?

Edit: my best guess is that we ignore the heater part of the circuit/transformer since I don't see any of that in the calculations so far. Seems like for heaters, as long as the transformer provides enough current (and the right voltage) for the tubes you're using, should be good to go.

You're right, Phrygian's model is only the High Voltage. Heaters are simple, add the amps for each triode pentode and tetrode on the spec sheets and refer to the filament current rating of the PT
 

Phrygian77

Poster Extraordinaire
Gold Supporter
Joined
Apr 30, 2016
Posts
6,354
Location
Crawfordville, FL
@Phrygian77

In your example, you hypothesize a 60 ma load. How specifically do you come up with that total in a real world case? The spec sheet combined max current ratings for the tube plates and screens? Or any derating?


I was using the AC load reference in the Antek datasheet for the 05TC300. I thought that I said that in my post. In fact, I said that the model was off by 1% compared to the datasheet. Look again at the Antek datasheet (where I circled it in red in my post). It shows at 0A (no load) 310 VAC RMS output, and right below that in the table, 295 VAC RMS output at 0.06A. My point was to show that the estimated source resistance and output was close to real word (or at least as specified by Antek).
 
Last edited:

joulupukki

Tele-Holic
Joined
Nov 26, 2020
Posts
687
Location
Utah
I think maybe what @mountainhick and I are wondering is now that we can see how the values are calculated for the PT, how does one check to see if a particular transformer will “work” for a given amplifier circuit with specific tubes and output transformer? For example, the RR763V (the 6V6 Blackvibe) that uses:

1x 12AX7
1x 12AT7
2x 6V6

Is it a matter of adding up, like mountainhick suggested, the max screen and plate currents of all the tubes? Or is there something more than that?

I also don’t understand how to know that the PT and the OT will “work” well together. Sure, I can just trust pre-published schematics, but in this case, how do I estimate/calculate which, if any combo, of PT & OT from Antek would work (for any amp circuit and not just the RR763V).

Knowing how all that works and is calculated would be one of the big breakthroughs I feel that would make a big difference in understanding in amp building.
 

mountainhick

Tele-Holic
Joined
May 2, 2021
Posts
609
Location
Rocky Mountains
I was using the AC load reference in the Antek datasheet for the 05TC300. I thought that I said that in my post. In fact, I said that the model was off by 1% compared to the datasheet. Look again at the Antek datasheet (where I circled it in red in my post). It shows at 0A (no load) 310 VAC RMS output, and right below that in the table, 295 VAC RMS output at 0.06A. My point was to show that the estimated source resistance and output was close to real word (or at least as specified by Antek).
Thanks, Tail between legs.
 

Phrygian77

Poster Extraordinaire
Gold Supporter
Joined
Apr 30, 2016
Posts
6,354
Location
Crawfordville, FL
Getting back to Merlin's method, for a rough initial estimate at idle, we'll ignore the preamp and screens, and just use the idle plate current. So, let's say we want a pair of 14W 6V6s biased at 65%:

2 x 14 W x .65 = 18.2 W

Since the 05TC300 is rated at 300 Vrms (even though we already know it will be less than this), we'll assume 300 x sqrt2 (300 x 1.4142) = 424.26 Vdc.

Using ohms law, voltage squared divided by power gives us resistance.

424.26^2 / 18.2 = 9889.92...

Our estimated source resistance (Rs) for the 05TC300 (secondaries in parallel) was 190.91 ohms, and our estimated load resistance (Rl) is 9889.92 ohms.

Rs = 190.91
Rl = 9889.92

Rs / Rl = 0.0193...

This is the source to load ratio, and this is what Merlin's graphs are based on.

psu2.jpg

Ignore the green dotted lines, which are from an example on Merlin's website.

The source to load ratio (Rs/Rl) is represented on the horizontal plane. From left to right, it is a decade from .01 to .1 (increments of .01), .1 to 1 (increments of .1), and so on.

We're missing one component here, which is the source AC frequency (60Hz) times the input filter capacitance times the load resistance (f*C*Rl). We'll use 40uF in this example, which is 0.00004F.

60 x .00004 x 9889.92 ≈ 24

This number just represents the amount of filtering relative to the load, and as you can see from Merlin's graph, there's not a lot of difference here between 2 and 1000.

We calculated our source to ratio as ≈0.0193, but we'll just call it .02, since that's the first grey line from the left in the decade.

The vertical plane is the DC voltage as a percentage of the unloaded RMS secondary voltage (in increments of 5%). So, for the 05TC300, we know that the unloaded voltage is 310 Vrms. Going up the graph to our f*C*Rl number, between 2 and 1000, 2 is about 122% and 1000 is about 125%. We'll just say that our 24 is 123%.

310 x 1.23 = 381.3

Our estimated B+ is approximately 381 Vdc. That's a pretty big drop from 426.26 Vdc, which is due in part to the source resistance. To quote Merlin ...

Maybe this would cause us to re-think the amplifier design, or maybe we would consider it OK.

His next graph represents the ripple current to load current ratio. Ripple current is the product of rectification and filtering. The RMS ripple current is the full load that will be seen by the transformer.

psu3.jpg


For our example, the ripple current will roughly 1.9 times the load current.

Since we didn't calculate the DC load current directly:

381.3 Vdc / 9889.92 ohms = .03855...
38.55 mA x 1.9 ≈ 73.25 mA
 
Last edited:

andrewRneumann

Friend of Leo's
Joined
Mar 22, 2020
Posts
2,171
Location
Cincinnati, OH, USA
Nicely done! Below are some of my notes.

60 x .00004 x 9889.92 ≈ 24

This number just represents the amount of filtering relative to the load, and as you can see from Merlin's graph, there's not a lot of difference here between 2 and 1000.

Yes, what this tells me is that there is not much to be gained by increasing the reservoir capacitance above 40uF. Maybe you could eek out few more DC volts. But say you had Rs/Rl = 0.03--there is no difference between fxCxRl of 2 and 1000. If you want more voltage in that situation, increasing the reservoir capacitance isn't the way to do it.

310 x 1.23 = 381.3

Our estimated B+ is approximately 381 Vdc. That's a pretty big drop from 426.26 Vdc, which is due in part to the source resistance. To quote Merlin ...

Maybe this would cause us to re-think the amplifier design, or maybe we would consider it OK.

As will almost all graphical methods, this number should be used in a 2nd iteration to see if you are converging on a solution. Go back to the beginning, using a B+ of 381V, recalculate the Rs/Rl and check the result. How close is it to 381V on the second time through? (I don't think it is too hard to add in the current from the preamp and screens. Why not make it as accurate as possible given the information we have? Some preamp tubes use a lot more current than a 12AX7.)

For our example, the ripple current will roughly 1.9 times the load current.

38.55 mA x 1.9 ≈ 73.25 mA

Ok, I don't want to steal your thunder, so I'll mix metaphors and tee this one up: what do we do with the 73.25 mA number?
 

Phrygian77

Poster Extraordinaire
Gold Supporter
Joined
Apr 30, 2016
Posts
6,354
Location
Crawfordville, FL
@andrewRneumann I almost didn't post that because, as I got into it, I realized there maybe questions that I couldn't answer about it. Merlin's graphical method seems to give slightly lower B+ numbers than I get out of LTspice. I'm wondering now how Merlin came up with them to begin with.

The RMS ripple current relates to the unloaded RMS voltage of the transformer. We don't ingnore the transformer's voltage drop. It's part of the equation. That's one thing that threw me off when I first started tinkering with linear supplies in LTspice. The numbers weren't adding up, and then I had one of those "duh!" moments.

310 x .07325 ≈ 23.7 watts
 

Phrygian77

Poster Extraordinaire
Gold Supporter
Joined
Apr 30, 2016
Posts
6,354
Location
Crawfordville, FL
I should also note, as you pointed out in a roundabout way, the estimated B+ with the initial estimated load resistance isn't close to our target of 2 x 14 W x 65%.
 

andrewRneumann

Friend of Leo's
Joined
Mar 22, 2020
Posts
2,171
Location
Cincinnati, OH, USA
@andrewRneumann I almost didn't post that because, as I got into it, I realized there maybe questions that I couldn't answer about it. Merlin's graphical method seems to give slightly lower B+ numbers than I get out of LTspice. I'm wondering now how Merlin came up with them to begin with.

Agreed. I get B+ about 10V less with Merlin then I get with Falstad (which includes 1.4V of diode drops).

I get "better" results using this graph (it should work with full-wave bridge as is):

1663385993787.png


Em is max input voltage (438Vpeak in your example). Rs/R is Rs/Rl. wRC is fCRl times 2pi (6.28).

I wonder if Merlin's charts are little off? Maybe just by 2 or 3%.
 

Phrygian77

Poster Extraordinaire
Gold Supporter
Joined
Apr 30, 2016
Posts
6,354
Location
Crawfordville, FL
Anyway, I think you'd be really pushing the limit of the 50VA. Compare it to the Hammond 290XX (Blues Jr.) with an Rs of 78.6 ohms rated at 260V at 170mA, or even the Hammond 290ABX (Vibro Champ XD) with an Rs of 157.3 ohms and rated at 293.5V at 100mA.
 

2L man

Tele-Afflicted
Joined
Nov 23, 2020
Posts
1,969
Age
63
Location
Finland
Does Merlin live GB? Possibly 50Hz mains make some of the difference to yours when discharge time is up to 20% longer?
 

Phrygian77

Poster Extraordinaire
Gold Supporter
Joined
Apr 30, 2016
Posts
6,354
Location
Crawfordville, FL
Does Merlin live GB? Possibly 50Hz mains make some of the difference to yours when discharge time is up to 20% longer?

f*C*Rl should account for that. I've been trying to think of a better why to describe that. On the graphs, it essentially represents represents the total AC load.
 

King Fan

Poster Extraordinaire
Ad Free Member
Joined
Jan 1, 2013
Posts
9,047
Location
Salt Lake City
Cool thread, you guys. I feel like the freshman looking for his pre-calc 101 class stumbling into a graduate seminar on Real and Complex Analysis. But the drawings on the board are beautiful... :)
 

andrewRneumann

Friend of Leo's
Joined
Mar 22, 2020
Posts
2,171
Location
Cincinnati, OH, USA
f*C*Rl should account for that. I've been trying to think of a better why to describe that. On the graphs, it essentially represents represents the total AC load.

I believe f*C*Rl represents the ratio between the Vavg (DC component) vs. Vripple (AC component) at the reservoir capacitor. It is a unitless factor that describes the "shape" of the voltage signal. If you increase the frequency of recharges (f), the capacitance (C), or reduce the load (by increasing load resistance, Rl), the ratio of Vavg to Vripple goes up. If you want a proof with formulas, I will try to oblige, or you can just take my word for it. ;)
 

joulupukki

Tele-Holic
Joined
Nov 26, 2020
Posts
687
Location
Utah
Cool thread, you guys. I feel like the freshman looking for his pre-calc 101 class stumbling into a graduate seminar on Real and Complex Analysis. But the drawings on the board are beautiful... :)
I feel like the guy at the back of the class who's hoping the professor doesn't call on to answer questions. ;)

Had a really busy weekend, but hope to do more studying of this course material during the week. Thank your @Phrygian77 and @andrewRneumann for the good material so far.
 




Top