I want to concentrate on the basics and a few cause and effect relationships.

I will be updating this power supply page as the questions roll in.

Like my attenuator page I am choosing quantity over quality at the
moment.

I'll work on the formatting (quality) as I find the time.

Added 12/23/00 Updated 12/23/00

Lets first make a simple tube power supply schematic. The values for the transformer are similar, but not exactly the same, as a HV transformer I have.

Lets say with a 120 V RMS input we measure 315 V RMS on the secondary (output side) with the windings unloaded. That puts the primary to secondary turns ratio at 120:315 = 0.381. The secondary to primary turns ratio is 315:120 = 2.625.

It is important that the turns ratio for the model be measured with the transformer unloaded. If the loaded voltage is used and is 10% lower than the loaded voltage, the model can be wrong by more than 10%. One reason behind this is that the DCR of the primary is reflected to the secondary with the square of the turns ratio.

If the secondary resistance measures 210 ohms and the primary resistance measures 16.7 ohms the RSEC_MODEL =

RSEC_MODEL = 210 + 16.7 * (315/120)^2 = 210 + 115 = 325 ohms (lets ignore the diode resistance.)The peak voltage will be:

V_peak = sqrt( 2 ) * 315 = 445.47 (I'm using 446V peak in my models because I want to.)

Note: Pspice uses the peak voltage of a sinewave for its sinewave sources. It is easy to get the peak, average and RMS confused in Pspice. Check your work at least twice.

Note: I purposely left the leakage inductance out of the model even though the power transformer's leakage inductance can be important.

First lets just look at the freebie change. Lets vary C2 and verify that the only thing that changes is the peak to peak output ripple. We can see that V( OUT_C ) and RMS (I( V1 )) are not changing as C2 is changed. As expected the peak to peak output ripple is changing with C2.

C2 does not affect the ripple current in the transformer because the 270 ohm resistor hides C2's impedance from the diodes. If the 270 ohm resistor was too small, C2 would affect the ripple current in the transformer.

Note:

Before I would add 470 uF to C2, I would split R3 (270 ohms) into two 130 ohm resistors and use a CRCR filter of 130 ohms, 220 uF, 130 ohms, 220 uF.Now let's try a bunch of different values for C1. As C1 changes, I am assuming that the ESR of C1 changes in impedance proportionately. If C1 doubles, the ESR halves.When do you split capacitance and when do you just add capacitance?If a lower AC output impedance from the power supply is your prime concern, just add more capacitance to C2. If you are chasing noise, more than you are chasing bass tightness, split the capacitor and resistor. Programs like Pspice and PSUDII Duncan Amps PSU designer II can help you make a decision, but your ears, not mine, are the final judge. The electrical models and measurements don't tell then entire story; otherwise, why would we still have stereos that measure great but sound. . . lacking?

We can see as we vary C1 with RSEC_MODEL = 323 ohms, the output voltage, V( OUT_C), does not change that much nor does the RMS current in V1 (the transformer secondary). The peak to peak ripple on the output drops as we increase C1. WHERE is the doom and gloom I predicted about the RMS current rising as C1 is increased?

Let's take a closer look at the RMS current in the secondary. You can see that the RMS current in the secondary is getting slightly worse as C1 increases.

Well that does not look to bad to me, does it to you? (It shouldn't.) Notice that at 100 uF we have about 77 ma of RMS secondary current flowing with a 60 mA load. And this is only 1/2 of the current in the transformer. This discrepancy is accounted for in the power factor specification of the equipment. Power factor is the true power in watts over the apparent power in VA:

= ( 1/T * integral of (V(t) * I(t)) dt for t = 0 to T where T is the time for one cycle)/ (True RMS Volts * True RMS Amps )We have (315 V RMS + 315 V RMS) * 77 mA RMS = 48.5 VA of power being transferred through the transformer to get 350V DC * 60 mA DC = 21 watts. This is a power factor of 21 watts/ 48.5 VA = 0.433 = 1/2.31. The transformer must be designed to transmit 2.31 times the actual heating power (watts) going through it. (Yes, I know I lumped R3's loss (60 mA^2 * 270 ohm = 0.97 W) in the transformer.)

= Heating power/ (V_rms * I_rms)

= Watts/ (V_rms * I_rms)

= Watts/ VA

DO NOT use cos( phase angle) for power factor. That only works into
linear resistive, linear capacitive and linear inductive loads. Once we
add a swinging choke (in the AC line), a diode, a SCR, an electronic
switch etc., we are no longer ** linear**.

Now what?

Why did I read all this?Lets install a better transformer, one that has better regulation! If I cheat a little bit and set the RMS current in the transformer to 77 mA, we see the regulation of the modeled transformer is not the best, but it is not shabby either.Where are we going!

What happens if we change Rsec_model, which is the same as changing the regulation of the transformer?Regulation = ( 325 ohms * 77 mA RMS ) / 315 V RMS = 25V/ 315V RMS = 7.9 %.

WOW! Just about everything changed but the peak to peak ripple! This is the doom and gloom I was talking about.

Notice how the RMS ripple current is about twice the load current with the DCR equal to about 15 ohms. This would make our transformer power rating 2 * 315 * 120 mA = 75.6 VA!

For a 21 Watt supply we would need a 76 watt minimum transformer if the effective DCR was 15 ohms! And that does not include any load current for filaments! However, because of economics, a 315 V RMS output transformer with a mythical 15 ohms effective DCR will be rated for 100's of watts. There are several advantages for buying a transformer with good regulation that I will discuss later. I'm not done abusing this power supply yet.

Lets be extra mean to the power supply and set the DCR to 100 ohms (an obtainable number) and re-sweep C1:

With a 100 ohm secondary, even at 10 uF output capacitance we are above the 77 mA transformer current that we had with the 325 ohm effective secondary resistance with 100 uF C1 capacitance. Fortunately, a transformer with a 100 ohm effective resistance will usually have a higher power rating than one with 325 ohms. So don't worry too much if all you are changing is the power transformer.

Note: I ran C1 up past 1000 uF because I knew someone a few years ago that put insanely dangerous amounts of capacitance in the B+ of his tube amp. I'll add a regulator before I add that much capacitance. With 1000s of uF at 500V, I'd be afraid that if a component failed it would explode with enough force to put a hole in the wall of the house and injure the neighbors. I'd want individual high voltage fuses in each of the capacitors in the bank he that set up.

Lets now go back to the original supply values and sweep the load current from zero to 60 mA to check the load regulation at DC.

At near no load the voltage on OUT_C is 441V. The voltage on OUT_C is 350 V at 60 mA. If we use these currents to calculate the output impedance (which is the wrong way to do it) a 60 mA swing gives a 91V swing for an output impedance of

91 V/ 60 mA = 1517 ohmsIf we do it the right way and change the load current about 10% and look at the output impedance we get

(350.19 V - 342.11 V)/(68.10 mA - 60.19 mA) = 7.97V / 7.91 mA = 1007 ohms effective output impedance.I ran a sweep with RSEC_MODEL = 325 ohms and C1 = 100 uF and got the following for the DC regulation:

RSEC_MODEL = 325, C1 = 47 uF, C2 = 47 uF

(350.23 V - 342.45 V) / ( 67.99 mA - 60.27 mA) = 7.774 V/ 7.722 mA = 1007 ohms. A bigger C1 does not help the DC (or low frequency) regulation. A bigger C1 will help some with the noise and higher frequency regulation.I then ran a sweep with RSEC_MODEL = 100 ohms and C1 = 47 uF and got this:

(392.14 V - 387.46 V ) / ( 68.10 mA - 60.27 mA ) = 4.680 V/ 7.829 mA = 597.8 ohms effective output impedance.Below is what the output voltage and RMS transformer current look like with and equivalent secondary resistance of 100 ohms (RSEC_MODEL). The regulation is much better and the output voltage is closer to the peak of the sinewave.

The 1/2 secondary transformer current was 93 mA RMS.

Lowering the DCR of the transformer helped the DC regulation.

Where is all this voltage going as the load current is increased?. . . It is being lost in the DCR resistance of the transformer.

ALL THE CHARGE that is drawn out of the power supply capacitor between 60 Hz cycles is replenished at the crest of the sinewave. Drawing all the current at the crest of the sinewave distorts the sinewave. This distortion can couple noise into the filament windings of the tubes both on the same transformer and to a lesser extent to any equipment on the same wall plug outlet. With transformers with better regulation, the voltage distortion is less. This leads to less distortion on the filament voltage which leads to less grunge in the music.

What can we do to eliminate some of the noise that is coupled into the filaments? Add high voltage snubbers and filament snubbers. Even thought the sinewave will still be distorted, the snubbers help. The high voltage snubber helps prevents any ringing from the diode recovery from getting into filaments. The filament snubber helps reduce any internal and external noise coupling in on filament line. If the filament snubber is only addressing the upper audio band, that is usually good enough. As the frequency drops, we become less sensitive to the noise AND usually the filament couples less of it into the amp's output.

The other choice to keep rectification noise off the filament is to use a separate transformer for the filament supplies.

How does this relate to the Schade's curves? The Schade's curves for power supplies put all of this in a nice graphical form. But you need to know RSEC_MODEL to use the curves.

Added 12/25/00 Updated 12/25/00

The original schematic:

Add two diodes and an inductor to make it a choke input power supply:

Lets take a look at the inductor current and the output voltage:

Notice that the RMS current in the secondary 1/2 winding has dropped to 69 mA RMS. The drop is because the current is being drawn all the time, not just at the crest of the sine wave.

With an inductive input power supply, you should to keep the inductor value above L critical to get all the benefits of inductive input (lower RMS in winding, decent load regulation, low output ripple.)

R_load = DC V at first capacitor/ ( I load DC + I bleeder )Lets sweep the inductor value below and above L_critical and see that as Lc drops below L_critical, the output voltage starts to rise.

L critical = R_load/ 1130 for 60 Hz = R_load/ 942 for 50 Hz

L critical = (512 V/ .060A)/ 1130 = 8533 ohms/ 1130 = 7.55 Henry (Twice L_critical is recommended.)V_out = sqrt( 2) * 2/PI * V_rms

V_out_ideal = 0.900 * 630 V rms = 567.2Note that there is:V_out = V_out_ideal - drops = 567.2 - 71.28 - 2 * 0.7 V = 494.5 V

0.060 A * (270 ohms + 270 ohms + 2 * 325 ohms) =

0.060 A * 1190 ohms = 71.28 Volts drop in all the resistors.

I ran a sweep with 60 mA and 68 mA load current so we can calculate the DC output impedance:

(495.2 V - 485.8 V)/ ( 0.068 A - 0.060 A) = 9.3825 V/ 0.008 A = 1178 ohms.What is this rise in output ripple at 100 H in the above graph? It is ringing of the LC filter tank as seen below for Lc = 10 and 100 henries:

This is with in the modeling error of 1190 ohms of the actual DCR of all the components. This is an important note: the DC output impedance is equal to the total of all the DC resistances in the B+ power path.

Damping the LC tank is a complex topic I'll get to later

Synchronous rectifiers (active diodes) for a 60H HV output could be fun to build and design. Figuring out the shutdown protocol will be the hard part.

.

First version Dec 23 2000 Last change 12.23.00.

( New 2024 index page.) _( Old 2003 index page.)