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**jazbo8** · 08-09-2014, 09:42 AM

Hammond 270CX
PS Design Guide

**ACHIEVEIT** · 08-09-2014, 04:13 PM

Thanks a lot for the reply jazbo8. So in my case where i use a full wave diode bridge rectifier followed by a reservoir cap, according to the link the dc voltage under load is 0.9 times the secondary ac voltage. This comes in disagreement with merlin's approach where for the same situation the dc voltage under load is proposed to be 1.2 times the ac secondary. Could you help me understand why hammond's approach is correct and how the ratio is calculated?

Thanks a lot

**R.G.** · 08-09-2014, 05:18 PM

It's not a simple matter of ratios. It depends on the relative size of the filter cap and the load, and also the internal wiring resistance of the transformer. They're all interrelated.

First: run a sine wave into a series diode followed by a filter cap. The diode conducts whenever the voltage on the sine side exceeds the voltage on the capacitor side by one diode drop. For silicon, let's use 0.7V for that. So the sine through the diode charges up the cap until it reaches its peak voltage, then starts to decline. The cap holds the charge voltage, so it stays at the peak voltage it got.

The peak of a sine wave is 1.414 (square root of two, actually) times the RMS value. So if you are feeding in 10Vrms AC, the peak of the AC in is 14.14V, and the cap charges to 14.14-0.7 = 13.44V, and the diode current quits flowing there; the diode turns off when the sine has dropped about 0.45V, which is below the diode's forward voltage minimum conduction threshold.

If there is no load, the story is over. The sine comes up again one cycle later, but the cap is still full, so even if the diode "conducts" because it has enough forward voltage, no current flows. If you add a resistor across the cap, things change.

With a resistor, the cap voltage runs down. It's actually an exponential R-C run-down, but for small changes in voltage we can approximate the resistor by a constant current load and say it runs down at a constant rate. Let's say the example is loaded by a 13.44 ohm resistor to make the numbers easy. So the cap runs down at a rate of 1A.

For capacitors, I = C* dv/dt, or dv = I*dt/C for situations where dt and I are small enough compared to C to make the constant-current approximation true. This is true for all the power supplies you're going to be working with. "dv" is the ripple voltage, the amount the cap runs down between chargings from the AC wave through the diode. "dt" is the time between peaks of the AC wave. For 60Hz, dt is 16.666... mS

So for any cap C and load I, you can calculate an approximation to the ripple voltage from an input voltage of Vac.

Notice that the load current directly affects the ripple voltage, and that the output "DC" voltage is an average DC level with a sawtooth ripple imposed on it. Following stages of filtering will remove most of the ripple (we hope!) so the DC out is the average of the instantaneous DC plus ripple. In practice, this is the no-load DC minus half the ripple, so the DC out "sags" by half the ripple voltage as load goes up.

We can double the usefulness of any given cap by using a diode bridge instead of a single diode. The diode bridge lets us use both polarities of the AC wave instead of only one, and thus it cuts dt in half. The cost for doing this is only one diode drop on BOTH AC half cycles. So we lose another 0.7V, but gain twice the effectiveness of the cap, or half the ripple voltage for the same cap.

So far, so good, but you already knew all that whether you knew you knew it or not. Your question was about the transformer rating.
You wanted 440Vdc at ... um... five tubes of load? OK, you called it 20ma. Let's do a quick and dirty first.

If we use a bridge rectifier, dt is 8.333... mS, we'll assume ripple is zero and come back to that. I = 0.02, and we need to solve for Vrms on the secondary. So Vpeak from the secondary is Vout plus 1.4V or 441.4V. The RMS is then 441.4/1.414 = 312.16Vrms. This is a minimum.

We know that ripple exists, and we want to allow for it. Ripple is something you simply choose. You say "I want my ripple to be less than X" and then pick a cap that makes this come true. Generally, something less than 5% ripple is a good target, and often much less. So let's say 5%, or 440/20 = 22V. Then C = I*dt/dv = (0.02*.0083)/22 = 0.000007572 = 7.5uF That's even reasonable. A 10uF drops the ripple to dv = (I*dt)/C = 16.6V. 100uF would make this 1.7V. Kewl.

But the issue remains of what ripple and AC line tolerances do to you. Unless you're going to pick a target voltage, design to that, then let both transformer internal resistances and AC line variations go where they will, you have some calculating to do.

AC power line variation is a biggie. In the USA today, the voltage is nominally 120Vac, but varies somewhat. Exactly how much is open to question, and the variation varies from place to place. It's usually quoted as +/-10% or 108Vac to 132Vac. UGH. In actuality, it is usually on the high side of 120V, often 120 to 128VAc. That's what's coming out of the wall. Your transformer will expect some wall voltage; probably 120Vac for new-made Hammond. So your DC, which depends directly on the AC wall socket voltage, will vary +/-10% or whatever the wall socket really does.

On top of that there is the issue of what the transformer "regulation" is. The internal wire resistance of the transformer causes the AC output on the secondary to sag. The manufacturer quotes this as the no-load to full-load voltage drop and calls it "regulation" although it is only a sag. Hammond may or may quote this. There are some subtle ways to calculate this.

But for the moment, you're already at a decision point. Do you want rock solid 440.00Vdc, or do you want about 440Vdc and it's OK if it wanders up and down some? If you're OK if it wanders, you're done. Pick a transformer with about 320Vac to 330Vac secondary at about 100ma or more rating, and build it. If you want rock solid 440.00Vdc, you have to regulate it after the first filter cap and make the first filter cap DC voltage higher by enough to ensure that under lowest AC line and maximum ripple the output instantaneous DC is enough to run your regulator.

**martin manning** · 08-09-2014, 07:28 PM

Thanks for taking the time to post this R.G.!

**ACHIEVEIT** · 08-09-2014, 07:35 PM

Dear R.G. i can't thank you enough for the detailed analysis you provided me. I will definitely print it and keep it in archive (in case the internet collapses

). Let me see if i got it straight:

First of all i found a mistake in my initial calculation. The target dc voltage is ~310V, not 440v as i wrote. The reason is i didn't notice a dropping resistor in the initial schematic (i am building a rack preamp based on the preamp of a 100w tube amp). So, again hoping my calculations are correct, the dc target is ~310v.

I live in europe, so for a bridge rectifier it should be dt=10ms.
I have 5 12ax7 tubes and according to merlin the average current draw per triode is 1mA, so 10mA in total, although at times the consumption can rise, so let's say 20mA.
With a target ripple of 1% that means that the Vripple=3.1V.
So, for 10mA, C=32.25uF and for 20mA, C=64,51uF. With that in mind should i go with a 68uF or would it be considered over-filtering and a 33uF is sufficient?

I guess at this point i wouldn't mind voltage wandering a bit, so is it ok to look for a 225 to 235 vac / 100ma secondary? (Out of curiosity, if i didn't need a winding for the heaters would this mean that i could in theory connect the bridge rectifier directly to the mains since the voltage here is 230vac?)

Thanks again for your help!

**R.G.** · 08-09-2014, 08:25 PM

Originally posted by ACHIEVEIT View Post

Let me see if i got it straight:

First of all i found a mistake in my initial calculation. T... So, again hoping my calculations are correct, the dc target is ~310v.

I live in europe, so for a bridge rectifier it should be dt=10ms.
I have 5 12ax7 tubes and according to merlin the average current draw per triode is 1mA, so 10mA in total, although at times the consumption can rise, so let's say 20mA.
With a target ripple of 1% that means that the Vripple=3.1V.
So, for 10mA, C=32.25uF and for 20mA, C=64,51uF. With that in mind should i go with a 68uF or would it be considered over-filtering and a 33uF is sufficient?

Let's see: 310Vdc + 1.4V for diodes, divided by 1.414 = 220Vac rms. You want 1% ripple, which is 3.1V, and you want a max current of 20ma, so C = I*dt/dv = 0.02*.01/3.1 = 64.51uF. Yep, looks good.

As to whether it's over- or under-filtering, that gets into some practical subtleties. Getting to 1% ripple is hard on caps and other things for a number of reasons, largely because the diode-cap setup lets the cap only charge for a fraction of the AC power half-cycle. In that few tens of microseconds, all the energy needed for supporting the whole rest of the half cycle in the load has to flow into the cap. Then the diode turns off and the cap, newly zapped up to full, supports the load for the rest of the half cycle.

On the incoming side of the cap, the power is a series of very high pulse currents, the average of which has to equal the smooth flow of DC going out of the output side of the cap. The size of the ripple voltage is inversely related to the size of those pulses. 1% ripple means, literally, that the cap must do all of its charging in the top 1% in terms of voltage in the incoming sine wave. In practice, you quickly get to a place where the resistance of the transformer windings, the wires to the rectifiers and the caps, the ESR of the caps, and all the other imponderables matter more than your design work.

1% is a very small ripple, and there are often more cost effective ways to get to that than doing all in one cap right off the rectifiers. This is the original of the "pi" filter setup in older tube amps: cap to ground, series resistor, cap to ground, series resistor...

The first filter cap sets the peak currents from the rectifiers. In circuits with tube rectifiers, there will be a maximum capacitance specified because there is a peak current limit on tube rectifiers which kills them faster if exceeded. So tube amp designers adopted the practice of using a moderate cap for the first filter after the rectifiers, then a resistor, then another cap to ground.

If you juggle things just right, the resistor/ second cap gives you a big ripple reduction, more ripple reduction than DC lost. It goes like this:
The ripple from the first filter cap is some amount; call it 10% to be gross. The DC voltage lost in the series resistor is Idc*Rseries. The AC losses to the ripple depend on the ratio of the series resistor and the load in parallel to the reactance of the second cap. Since this is always parallel with the load, the loss to ripple is always larger than the loss to the DC, and you can make it even more so.

In your case, you want 310V-ish. Let's say you make your DC on the first filter cap be 325V, and use your 20ma load to wipe off 15V of DC. The series resistor is then 15V/0.02 = 750 ohms. If you want to reduce your ripple by 10:1, you can put a cap in parallel with the load that has an impedance of 75 ohms or less at the lowest ripple frequency, that being twice your line frequency. 2x line frequency is 200Hz, so you want a cap with a reactance no larger than 75 ohms at 200Hz. Since Xc = 1/(2*pi*F*C), we can rework that into
C = 1/(2*pi*F*75 ohms) = 10.6uF, assuming I punched the calculator right.

Notice that to get 10% ripple in the first cap after the rectifiers, you would have used something like 6.4uF, and you'd probably have used 10uF. So a 10uF first cap, a 750 ohm resistor, and another 10uF cap gives you 1% ripple (I'm approximating heavily here, but it's about right) instead of buying one 64 to 100uF cap and really stressing your transformer, rectifiers and the ripple current on your first cap. You've paid for this reduction in ripple with a minor reduction in DC voltage and some waste heat in the resistor.

Using an active "resistor" in terms of a power transistor or MOSFET being driven to only let through enough current to hold the right voltage on the second cap - that is, a regulator - can easily make the loss not 10:1 from first cap to second, but 100 to one, and give you rock solid DC voltage at the second cap too. This can be as simple as a MOSFET, a couple of resistors, and a stack of zeners to get the right DC reference voltage.

See how the road to hell is littered with minor improvements?

I guess at this point i wouldn't mind voltage wandering a bit, so is it ok to look for a 225 to 235 vac / 100ma secondary? (Out of curiosity, if i didn't need a winding for the heaters would this mean that i could in theory connect the bridge rectifier directly to the mains since the voltage here is 230vac?)

Hmmm. 230Vac*1.414 = 325V Kewl. Works.

... as a sudden death machine. It would be great, and some very old tube amps did that. But the "ground" on the (-) side of your first filter cap is connected to the AC power line alternately on both sides every AC cycle through a diode in the bridge rectifier. That makes your amplifier "ground" be connected to the power line, and someone touching your power supply's "ground" and a safety grounded equipment someplace else be a load directly across the AC power line, and therefore dead or dying.

This is very, very, very unsafe. Don't think about this any more than to notice how dangerous it is to you and others, and to appreciate how wonderful isolation transformers are.

And that brings me around to my standard warning about wiring mains voltages. Don't do this at all unless you already know how to do it safely, with no questions and not having to ask about it or look it up. Musical gear is fun, but it's not worth dying for, and that's not a joke or an exaggeration. Unless you're sure you already know how to do the wiring safely, get on-the-spot help from someone who's already trained to do it right. One of the worst scenarios is doing electrical wiring ...almost... right, and having it work OK for years until your child or wife picks up the guitar or microphone.

**jazbo8** · 08-09-2014, 08:44 PM

Originally posted by ACHIEVEIT View Post

Thanks a lot for the reply jazbo8. So in my case where i use a full wave diode bridge rectifier followed by a reservoir cap, according to the link the dc voltage under load is 0.9 times the secondary ac voltage. This comes in disagreement with merlin's approach where for the same situation the dc voltage under load is proposed to be 1.2 times the ac secondary. Could you help me understand why hammond's approach is correct and how the ratio is calculated?

Thanks a lot

R.G. already provided you with a very detailed explaination, so there isn't much to add. As to the appearant discrenpency between Hammond and Merlin's figures, please note that Hammond is referring to the average un-loaded output voltage, while Merlin's "rule of thumb" refers to the RMS loaded output voltage, they are two different ways to describe the output voltage.

As a refresher, the commonly used figures for un-loaded output voltage of the bridge rectifiers are:

Vavg = 2*Vpeak/Pi
Vrms = Vpeak/sqrt(2)

Hope this helps.

**Gregg** · 08-09-2014, 09:32 PM

To cut the long story short you need 350-360V/20-40mA for the high voltage and 6V3/2A (or 12V/1A). If it's a high gain preamp (with 440V it seems like a Mesa DR preamp to me) I always use DC regulated heaters so I order my PTs with a 13-15V/1A winding for the preamp tubes. You should definitely check TT's line of toroidal PTs that can fit in 1U rack unit as well. This one for example:

Ringkern Netztrafo 56VA - Tube-Town GmbH

or this one (you can use a voltage doubler and lower the voltage to your taste, also run first 2 tubes DC regulated, the rest on AC):

Ringkern Netztrafo 47VA - Tube-Town GmbH

**R.G.** · 08-09-2014, 11:18 PM

One of those subtle gotchas is that because of the very high peak currents in a solid state rectified, bridge-rectifier, low-ripple setup, the pulses of current cause the transformer to heat more than it would with a resistive load. Unless the manufacturer's specs show data for output current with a bridge rectifier and DC loading, you need to divide the specified current capability by 1.6 to 1.8.

This is because the heating in the transformer windings, which is what determines how much power it can put out without burning up, is determined by the RMS current in the rectifiers, and that's 1.6 to 1.8 times the DC average current out of the first filter cap. Sure enough, the bigger the cap and the smaller the ripple, the higher this number gets. A 1% ripple capacitor size pushes the RMS current from the transformer towards 1.8.

So yeah, if you need 20ma out on a long term basis, pick a transformer that is rated at 40ma rms or more - which is how the trannie makers usually specify their transformers, even if they know full well that you can't pull that much average DC out out of them.

And getting good-quality toroids that can fit flat inside a 1U rack case is a really good idea.

**J M Fahey** · 08-10-2014, 12:13 AM

Excellent analysis but don�t over think it.

Assuming voltage drop under load of, say, 15/20% is fine for power amps, which tax their supplies heavily and to boot have variable load, unless they are trues Class A , very unlikely.

So you�ll have more +B voltage to start with?
Good !!! , now you can easily pad it down and mainly filter it to what your preamp actually needs.

And unless it�s a 6 Ch (or more) PA mixer , HV consumption will be nil.

**nashvillebill** · 08-10-2014, 04:20 AM

to the original poster--have you downloaded Duncan's Power Supply Designer PSUD2 (It's totally FREE)

Change the load to a constant current load, and modify the rest of the power supply as required, and it will do the heavy lifting for you.

**ACHIEVEIT** · 08-10-2014, 03:33 PM

Thanks a lot R.G., jazbo8, Gregg, J M Fahey and nashvillebill for your input and thanks again R.G. for the detailed analysis of the phenomena in general - your answers are a precious asset to me

Unfortunately, i don't have access to a pc to use the PSUiiDesigner - i am on vacation, so i use a tablet to post. I am trying to understand how things work so as to order the pt asap. I was thinking about ordering it from a local experienced winder. I didn't know about the tube town pts - they sure look nice although i am not certain the 6v3 winding is center-tapped? The local winder gave me an even lower estimate than tt's, so i think i ll go with him since i will spare the shipping + i will get the center tap.

So, with the target dc being ~310v, say that the pt ht secondary is 245v. After the rectification, the voltage should be (in loaded state and considering the 1v4 voltage drop of the diodes) around 330v.

By choosing a 10uF reservoir cap, the voltage ripple is Vripple=t*I/C = 0.01*0.02/(10*10^-6) = 20v, which, percentage-wise, is ~6% of 330v.
Then by placing an RC filter with R=750 Ω, the voltage drop on the resistor is Vdrop=750*0,02= 15v and the power consumption is P=I^2 *R = 0.3W (so even a 0.5W seems sufficient although i would go with a 1W). With that in mind, the voltage of the second capacitor would be ~315V. So to further reduce the ripple by 10:1, Xc=75Ω, so C=1/(2*π*f*Xc)= ~10.61uF. So by placing the RC filter, i further reduce the ripple by a factor of 10 while ending up with ~315v and keeping the components happy and relatively stress-free.
Regarding the current of 20mA, 20mA*1.8=36mA is the minimum pt rating, so let's say above 40mA.
Have i understood anything wrong?

Do i need to ask the pt winder what the ht winding resistance will be in order to get a correct psuii emulation?

Thanks again!

P.s. Thanks R.G. for insisting on the safety. Soon after i posted my previous mesaage i thought about what i wrote and it was wrong, but thanks a lot for taking the time to explain it thoroughly.

**jazbo8** · 08-10-2014, 03:51 PM

I don't know how experienced you are with building amps, but if you are planning to build a SLO-like hi-gain rack preamp, perhaps a kit like this can save you a lot time and effort, especially for the chassis and the faceplate, and of course a proven layout/design.

**ACHIEVEIT** · 08-10-2014, 03:53 PM

Also, to see if i got this part right again, the dc voltage after the reservoir cap is calculated as:
The Vpeak of the ht winding is Vpk=245*sqrt(2)=346.48v
Taking the diode voltage drop into account, the voltage after the diode bridge is V=346.48-1.4= 345.08v, which is the no-load dc.
For a 10uF reservoir cap with I=20mA and 50hz mains, Vripple=20v
So, the dc voltage after the reservoir cap should be Vdc=345.08 - (20/2) = 335.08v instead of 330v i wrote above. Does this sound right?
Thanks

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