Current flowing in an amp-power transformer center tapped bobbin - Watts

[percyhornickel]

Hi, I don`t know if this is the right place to post something about amps power transformers but I need somebody to help me or make clear some things I am not that sure.

After many fx circuits I would like to build a 5F1 or 5E3 Fender style amp someday.

I have been studing about power transformers calculations but I have seem some videos and have read some docs taking just the half of voltages of the secondaries bobbins (just the ones with center tap) in order to calculate the bobbin watts.

Every cicle (in a certer tapped bobbin) use half of voltages at time, so it dissipates the half of the watts and there is less current flowing every cicle, this is why is possible to use a thinner cable and use less space.

This is the only way I think I could recreate a HAMMOND 290EX (example) according with the datasheet measures and bobbins currents.

I made a nice excel file to calculate the transformers (there are many parameters on it) but as I said, the only way it could fits inside the EI-76.2 lamination with core 2.54 X 3.5 cm.

Is this right or I am wrong in some point?

P.H.

[Rob Strand]

There's a lot more to it than you think.  I'll go through the details below.  You might struggle with it.   You need a really clear head to get all the details right.  I could have even screwed up myself.    My only check is the result at the bottom seems to match the analysis and tables.

I have been studing about power transformers calculations but I have seem some videos and have read some docs taking just the half of voltages of the secondaries bobbins (just the ones with center tap) in order to calculate the bobbin watts.

Every cicle (in a certer tapped bobbin) use half of voltages at time, so it dissipates the half of the watts and there is less current flowing every cicle, this is why is possible to use a thinner cable and use less space

I suspect you are calculating things incorrectly.

For the same size transformer a full-wave rectifier will always produce less DC power than a bridge rectifier.   A full-wave rectifier only utilizes one half of the secondary winding on each cycle whereas a bridge rectifier using the entire winding on each cycle.

What's important for a transformer is the power dissipation in the windings.   You have Ip_ac^2*Rp in the primary for both full-wave and bridge where Rp is the primary resistance.   For the secondary we set-up the comparison as follows:   Suppose we have a dual secondary winding transformer.   We can wire the secondary as a full-wave.   For the bridge rectifier case we would use both windings in parallel and that will produce the same DC voltage as the full-wave case.  So here's where the difference comes in:   

The resistance of each *half* the secondary winding is Rs.   

I'm using Iout_ac to represent the rms current regardless of waveshape.

For the full-wave case each winding carry full current but only for half the time  Each secondary will dissipate (1/2)Iout_ac^2*Rs, so a total of Iout_ac^2*Rs.   

For the bridge rectifier case each winding only carries half the output current due to the parallel connection but carries current all the time.  Each secondary will dissipate (Iout_ac/2)^2*Rs, so a total of Iout_ac^2*Rs/2.

For the same output current the secondary of a full-wave rectifier dissipates twice the power as a bridge rectifier. 
So for the same secondary dissipation,

   Iout_ac_br^2*Rs/2 = Iout_ac_fw^2*Rs

So fullwave the can only produce 1/sqrt(2) = 0.71 the output current of a bridge rectifier.

The previous paragraph matches the secondary power dissipation.  However, when the primary current is reduce in the Full-wave case the dissipation in the primary is also reduced.  Instead of matching the secondary dissipation we can match the total transformer dissipation.   

For a typical transformer design the primary dissipation is about the same as the total secondary dissipation for an AC resistive load.  The bridge case (no cap) is similar to the AC load.

   Pdis_total = 2*(Iout_ac_br^2*Rs/2) = Iout_ac_br^2*Rs

but for the full-wave,

   Pdis_total = Iout_ac_fw^2*Rs + (Iout_ac_br^2*Rs/2)*(Iout_ac_fw^2/Iout_ac_br^2)
                   = 1.5*Iout_ac_fw^2*Rs

So for the same total transformer dissipation,

   Pdis_total = Iout_ac_br^2*Rs = 1.5*Iout_ac_fw^2*Rs
=>   Iout_ac_fw = Iout_ac_br/sqrt(1.5)
=>   Iout_ac_fw = 0.82*Iout_ac_br

As expected this predicts we can get a little more output (15%) from the full-wave than just matching the secondary power dissipation.   The true maximum output is somewhere between the two values and depends on how well heat is distributed throughout the winding.

Now lets look at the Hammond table,

https://www.hammfg.com/electronics/transformers/rectifier

We have the second from the bottom entry of the right column for a bridge,

      IDC_br = 0.62 * IAC_br

The factor of 0.62 respresents the increased winding losses of a capacitive load due to the pulsed current waveform compared to a sinsusoidal current waveform.   My Iout_ac is the rms current of the pulsed waveform.

The bottom entry on the left column corresponds to the full-wave,

    IDC_fw = 1.0 * IAC_fw

However, as shown the currents cannot be compared with the bridge case. [take a pause to think about that.]
The bridge circuit produces twice the output voltage of the full-wave.  Different winding voltages are required in each case.
That's why I constructed my argument with the same transformer with a parallel winding for the bridge case.
For the same DC output voltage the VAC of the fullwave case needs to be twice the VAC of the bridge case.
So for the same transformer size the IAC rating for the full-wave case is only half that of the IAC for
the bridge case.  So we have,

   IDC_fw = 1.0 * (IAC_br / 2) = 0.5 * IAC_br

In other words based on the Hammond table, for same size transformer and same DC output voltage,
the DC output current of the full-wave circuit is 0.5/0.62 = 0.81 times the output current of the bridge circuit.

If we now compare the results from the Hammond table with the dissipation analysis done earlier we can see the factor of 0.81 from the Hammond table matches the 0.82 from the dissipation analysis.   So the Hammond ratings table is based on the total transformer dissipation being the equal in the bridge and full-wave cases.

[R.G.]

I did an ad-hoc beginning on transformer basics including some first-steps calculations here:
https://ampgarage.com/forum/viewtopic.php?t=37347
Give it a look. I'll be doing extensions to that as time permits. The forum, The Amp Garage, might be a good place for you to go scope out, given the interest you've expressed.

[Rob Strand]

Here's a summary of results for a spice simulation which confirms the 0.82 factor.
The factor from the simulation is 0.84.

Note also the DC "derating factor" for the bridge rectifier with cap is 0.56
instead of 0.62 from the Hammond table.

It's quite normal for these factors to vary depending on specifics (like the size of the filter cap).

[merlinb]

290EX is rated for 275mA DC. The current in each half of its winding will be
Irms =  Idc / (PF * sqrt2)

Where PF is the power factor. This is typically around 0.6, giving an RMS current in each half of the winding of:
Irms = 0.275 / (0.6 * 1.41) = 0.32A rms

However, the apparent power in the whole core is not simply Vrms*Irms, it is about 1/(sqrt2) time less because of the way the current is shared.
660V * 0.32A / sqrt2 = 150VA

Which is the same as:
Vdc * Idc / (PF * sqrt2)

[percyhornickel]

Rob, R.G., merlinb...   Thank you for all the this information!!! As soon as possible I will try read deeply and take notes..  Saddly I´ll be in my work out of town until the next week (I work 7 x 7).I think I will need a week to read all this data and sure it will help me a lot.

I did a quick review:

- Rob you were right, I was not considerating the king of rectification it would use after the P.T., I will use a 5Y3 tube just like the 5F1 schematic does. Let me read a little bit better your entire post as soon as I come back.
I can not open the spice image you sent from here (proxy permisses at work), I will watch it soon.
I made a 5F1 ltspice simulation taking the hammond power transformers data and vacuum tubes models.


- R.G. I just watched a bit the ad-hoc beginning on transformer basics, and yes this is a great great resume, in my excel sheet I am considerating the material flux density, % core power loss, frecuency, etc..  ..to calculate turns for volts. Leter I will send a copy if you wish.

- merlinb, you said 290EX is rated for 275mA DC, how or where did you get that inf?, I just see the datasheet 100mA for the 275V/325V winding, 2.25A for the 6.3V landing and 3A for the 5V winding. I am still trying to understand this.

[merlinb]

- merlinb, you said 290EX is rated for 275mA DC, how or where did you get that inf?, I just see the datasheet 100mA for the 275V/325V winding, 2.25A for the 6.3V landing and 3A for the 5V winding. I am still trying to understand this.

Huh, are we talking about a different transformer? I see this info:
https://www.hammfg.com/part/290EX

[percyhornickel]

There is a 290AX and 290EX !!!!!, I didn´t note that!!

.....290AX is the one rated 100mA !!!!

290EX is 275mA!!!!!! you are right!!

[percyhornickel]

I did an ad-hoc beginning on transformer basics including some first-steps calculations here:
https://ampgarage.com/forum/viewtopic.php?t=37347
Give it a look. I'll be doing extensions to that as time permits. The forum, The Amp Garage, might be a good place for you to go scope out, given the interest you've expressed.

R.G. about the turns ratio I am totally agree when you said : (according hammond 169VS datasheet)

"The turns ratio is kicked up from 1:1 to something like 114.35/120.6.",

I understand that part, after this you wrote:

"..the hidden ideal transformer primary does not get 115Vac, it gets 115 - 0.647 = 114.353Vac(rms) as an input. This has to make an output of 120.6Vac(rms) on the secondary, so the voltage/turns ratio on the hidden ideal transformer has to be 120.6/114.35 = 1.054629 "

I don´t understant why turns ratio is inverse this time (secundary/primary)..  ...I´m trying to understand while I´m traslating to spanish..  ..maybe there´s something that I don´t see.

[Rob Strand]

I don´t understant why turns ratio is inverse this time (secundary/primary)..  ...I´m trying to understand while I´m traslating to spanish..  ..maybe there´s something that I don´t see

It's just a typo, it should be  114.35/120.6 like the earlier paragraph.   The effect of the magnetizing inductance and the voltage drops across the winding resistances always work against you so for a target 1:1 voltage ratio *under load* the turns ratio has to boost the output so the secondary turns has to be higher than the primary turns.

FWIW, typically the magnetization current is shown as an inductance.  However, in practice I've found there's a point where it's better to be modeled as a loss resistance.  Iron loss needs to be represented by a resistance in parallel with the magnetizing inductance.   For small transformers say less than 20VA the inductance dominates the magnetizing current,  at around 30VA the magnetizing current is share equally between the loss resistance and the magnetizing inductance, then for larger transformers a loss resistance works better.

The way you can see the effect is when you gradually load down the transformer and measure the output voltage.    If you try to match measured voltage with that predicted by the circuit model you find you need to get the right balance of magnetizing inductance and loss resistance in the model.   (Due to more complications even that model might not predict the primary current accurately.)

These are additional evil details.   The main idea RG is getting across is the turns ratio needs to be different than the textbook values when you consider magnetization and winding resistance drops under load.

[percyhornickel]

Ok Rob, I see all the transformers maths are so for be a linear result or an exact formula for everything.

The model R.G. is using has the main parameters I can see in the datasheet, windings resistence, current without load, etc..   ..very useful to make an good approximation about the parameters to consider.

One thing I do not understand is how is calculated the primary inductance (619 mH) from the values showed in the second image, is not that clear to me.

 ..." The magnetizing current is modelled as a primary inductance with a value that nets through the correct number of zoots.
And that's how I calculated Lpri = 619mH. With 115Vrms across a 619mH inductance, 0.49A of inductor current flows "....

[R.G.]

One thing I do not understand is how is calculated the primary inductance (619 mH) from the values showed in the second image, is not that clear to me.

 ..." The magnetizing current is modelled as a primary inductance with a value that nets through the correct number of zoots.
And that's how I calculated Lpri = 619mH. With 115Vrms across a 619mH inductance, 0.49A of inductor current flows "....

If you think about it a little, disconnecting the secondaries on a transformer makes it into just an inductor. The primary winding becomes an inductor fed by the AC voltage on the winding. So you know the voltage feeding the "inductor", and the current through it.

This lets you know the impedance that is limiting current flow - it's just the ratio of the voltage feeding the primary, divided by the current. Ohm's Law still works for inductive and capacitive reactance.

The impedance of any inductor is Z = 2 * pi * F * L;

so L = Z /(2 * pi * F).

[percyhornickel]

One thing I do not understand is how is calculated the primary inductance (619 mH) from the values showed in the second image, is not that clear to me.

 ..." The magnetizing current is modelled as a primary inductance with a value that nets through the correct number of zoots.
And that's how I calculated Lpri = 619mH. With 115Vrms across a 619mH inductance, 0.49A of inductor current flows "....

If you think about it a little, disconnecting the secondaries on a transformer makes it into just an inductor. The primary winding becomes an inductor fed by the AC voltage on the winding. So you know the voltage feeding the "inductor", and the current through it.

This lets you know the impedance that is limiting current flow - it's just the ratio of the voltage feeding the primary, divided by the current. Ohm's Law still works for inductive and capacitive reactance.

The impedance of any inductor is Z = 2 * pi * F * L;

so L = Z /(2 * pi * F).


[/quote

One thing I do not understand is how is calculated the primary inductance (619 mH) from the values showed in the second image, is not that clear to me.

 ..." The magnetizing current is modelled as a primary inductance with a value that nets through the correct number of zoots.
And that's how I calculated Lpri = 619mH. With 115Vrms across a 619mH inductance, 0.49A of inductor current flows "....

If you think about it a little, disconnecting the secondaries on a transformer makes it into just an inductor. The primary winding becomes an inductor fed by the AC voltage on the winding. So you know the voltage feeding the "inductor", and the current through it.

This lets you know the impedance that is limiting current flow - it's just the ratio of the voltage feeding the primary, divided by the current. Ohm's Law still works for inductive and capacitive reactance.

The impedance of any inductor is Z = 2 * pi * F * L;

so L = Z /(2 * pi * F).

Thanks R.G.!!!!!,  I am not that familiar with inductors, just an simple electronic self-leanrning guy (petroleum engineer).

[R.G.]

Kudos to you for being willing to dig into unfamiliar territory!

As Rob correctly points out, there are many more (and more evil!) details in transformer operation, and other ways to model what is going on inside.

The simple ideal transformer model was concocted as a way to get pretty-good answers to transformer questions. It gets reasonable answers quickly, and in its fuller variations has other parts added on to model the finer details.

Fancier versions of the ideal transformer model add on non-linearity to the primary inductance, non-linear core losses, eddy current losses in the iron and copper, leakage fluxes, inter-element capacitances, and non-lumped versions of all of these to get closer to what the windings and core are really doing.

It's a rabbit hole suitable for a lot of cold winter evenings! 

[percyhornickel]

Hey I have another question, trying to replicate/calculate the Hammond 290AX PT I see the secondary voltages in the DATASHEET:

Volt     Amp     Watts
702,03    0,10    70,20
    6,84    2,25    15,39
    5,42    3,00    16,25

POWER: 101.84 WATTS

If we calculate the core watts the transormer can manage according the datasheet:

             X(cm)    Z (cm)
CORE    3,20    3,25  =    10,40 (cm²) 
       
POWER 108,24 WATTS

How the requiered watts are higher than the watts the P.T. can manage?

Need help in here please!!

[Rob Strand]

Volt     Amp     Watts
702,03    0,10    70,20
    6,84    2,25    15,39
    5,42    3,00    16,25

POWER: 101.84 WATTS

Transformer ratings are based on the *loaded* voltages.   The idea is that represents the output VA under full load.

See below for calculations.  I get about 94VA.

If we calculate the core watts the transormer can manage according the datasheet:

             X(cm)    Z (cm)
CORE    3,20    3,25  =    10,40 (cm²) 
       
POWER 108,24 WATTS

Using the core center leg area is not an accurate way to predict the transformer VA.  It does not factor in the maximum flux, allowed winding resistances, core loss or allowed winding temperature. These formulas are rough at best.   There are a lot of assumptions which may or may not be met.

More accurate calculations are fairly complicated.

Dimension C in the datasheet is 3" (76mm) and dimension D is 1.28" (33mm).
From dimension C, the center leg = 76/3 = 25.3mm
From dimension D, stack = 33mm
So I'm getting an even smaller area than you.

The DC resistances seem consistent for a transformer of that size.
The specified weight of 1.33kg is also consistent with a transformer of that size.

Under normal circumstances with old-school winding insulation I would expect a transformer of this size to be rated at 60VA.  In order to get 94VA the only way around that is to allow the temperature rise of the windings to be higher.

Here's some details calculations

https://www.hammfg.com/part/290AX
https://www.hammfg.com/files/parts/pdf/290AX.pdf?v=1709681208


Data from specifications:

Winding      Vnom   Irated   VNL   R @ 20degC
Primary:   120V   -   -   3.927 ohm
Secondary 1:   650V   100mA   702.3V   277.1 ohm
Secondary 2:   6.3V   2.25A   6.841V   0.108 ohm
Secondary 3:   5.0V   3A   5.416V   0.091 ohm

Total weight 1.33kg


1) Nominal power rating

Transformer voltages are at full load conditions.
Transformer VA is are the output VA.
Output VA is the loaded voltage at full load current.

If we assume the nominal voltages are at full load:

VA = 650*0.1 + 6.3*2.25 + 5.0*3.0 = 94.2 VA


2) Predicted loaded output voltages and VA

The primary resistance cause a load on any secondary
to cause a voltage drop on all secondaries.

Primary currents for each secondary
Ip1 = (702.3/120)*0.1  = 0.5853A
Ip2 = (6.841/120)*2.25 = 0.128A
Ip3 = (5.416/120)*3    = 0.135A
total primary Ip_total = 0.848A

2a) Cold: 20degC

Internal primary voltage:
Drop across primary = Ip_total * Rp = 0.848*3.927 = 3.33V
Vp_internal = 120 - 3.33 = 116.7V

Calculated loaded voltage:

The primary voltage drop means we need to scale the output
voltages by (116.7/120) to account for the primary voltage drop.

Under load we then have for each winding,

   VS_FL =  (116.7/120)*VS_NL - IS*RS

      Loaded secondary voltage
Secondary 1:   (116.7/120)*702.3 - 0.1*277.1 = 655V
Secondary 2:   (116.7/120)*6.841 - 2.25*0.108 = 6.41V
Secondary 3:   (116.7/120)*5.416 - 3.0*0.091 = 4.99V

Total output VA = 0.1*655 + 2.25*6.41 + 3.0*4.99 = 94.9 VA

2b) Hot: say 80degC

The resistances will increase by a factor of 1.24.
The higher resistance causes more voltage drop.

      Rcold      Rhot
Primary:   3.927 ohm   4.869 ohm
Secondary 1:   277.1 ohm   343.6 ohm
Secondary 2:   0.108 ohm   0.134 ohm
Secondary 3:   0.091 ohm   0.113 ohm

Internal primary voltage:
Drop across primary = Ip_total * Rp = 0.848*4.869 = 4.13V
Vp_internal = 120 - 4.13 = 115.9V

      Loaded secondary voltage
Secondary 1:   (115.9/120)*702.3 - 0.1*343.6  = 644V
Secondary 2:   (115.9/120)*6.841 - 2.25*0.134 = 5.96V
Secondary 3:   (115.9/120)*5.416 - 3.0*0.113 =  4.89V

Total output VA = 0.1*644+2.25*5.96+3.0*4.89 = 92.5VA

The predicted output voltages look reasonably close to the spec.

[merlinb]

If we calculate the core watts the transormer can manage according the datasheet:
?
How the requiered watts are higher than the watts the P.T. can manage?
Need help in here please!

How did actually you calculate the core watts?
The short answer is, transformers are made from physical materials and the power calculations are not precise; they are design approximations.

[percyhornickel]

Sorry I took my time to read once and again and take notes. Rob that was a espectacular explanation, Just like the R.G.´s above!. I will print both to read at work (I am starting my work-week tomorrow).

My power transformer calculations were based on no load voltages (which seems to be around 8% over nominal voltages, at least on Hammond´s datasheets I have seem till now).

Thank you all for the information!!!!, I really appreciate the time you took to write all this lines.

P.H.