Regulated vs unregulated psu

Hi guys

I would like to know what are the advantage of both design. I know that for a unregulated We need to know the maximum load and manage to select the resistors and caps accordingly, both design need to have a power transformer with enought juice but other than that, what are the caracteristics that leads you to use one or the other. For the regulated one, their is plenty of informations within the regulators specs sheet which ease the works a little.

For premium low noise designs, a clean regulated DC supply can reduce the amount of hum in the noise floor coming in via op amp power supply pins due to PSRR.

JR

Thanks. So in the case of a opamp audio design, this is the way to go for best noise and clean signal ratio. For tube design, regulated psu, too? Which regulator?

The question makes no sense until you say WHAT you are powering, how sensitive it is to absolute voltage.

You don't regulate your incandescent light-bulbs, but they are more sensitive (brightness, life) to voltage than most electronics.

Many audio circuits work about the same over a w-i-d-e range of supply voltage. TL072 has near the same specs (except output swing) from 10V to 36V. A 12AX7 stage will work from 100V to 400V with little change of gain, proportional change of maximum output, and proportional change of THD at a given signal level.

Personally, I like un-regulated. Simpler. Less to go wrong. With simple math (and some understanding of parasitics) you can design the residual buzz/hiss as low as you want, perhaps lower than any regulator.

OTOH if your audio chip blows-up at 36V, and you need 30V to meet your required output level, then on ordinary wall-outlets you probably must regulate (or change requirements or chips).

If you take precision DC references from your power rails, as many synthesizers did, you may need obsessive voltage control. This is perhaps bad design, but on many-many module synths sometimes it was the only sane path.

Thanks prr. I prefer to draw unregulated psu for tube design but sometimes, i found it a little difficult to find the current load to design exactly the psu for a given voltage. For exemple, if i need to désigna psu for, i dont know, a gyraf g9, i have 2 tubes in paralelle to feed, so current Will sum together. Is there a way to easily calculate the current draw in each tube without referencing to the curves sheets of the tubes to draw the load Line, etc.

Regulators have less ripple generally but they are more complicated and sensitive. They break easier and require more focus on controlling the heat they generate, generally. I just built a big all regulated design for a tube project and it was a lot of fun but I went against common interest with it. Regulators are great for phantom power, relays, and solid state stuff at lower voltages. It can still work great for HT and heaters but it's less advised.

Non regulated circuits last longer and have less problems.

Tube power supplies.. those are lots of fun but I bet you can imagine the difference there.

> a way to easily calculate the current draw in each tube without referencing to the curves sheets

Some under-paid junior engineer worked hard on those data-curves. They are readily available even today. Why would you not want to use this rich data?

You can't just look at the tubes. 12AT7 can be run over 10mA, but in audio stages is rarely worked over 2mA. (Run rich for good gain at radio frequencies; run lean for high gain at audio freqs.)

In resistor-coupled stages you can get some clue from the plate resistors. If the supply is 300V and the plate resistor is 100K, it CAN'T pass more than 3mA (Ohms Law). In fact it could only pass 3mA if the tube were dead-short and doing nothing. More likely it is a "fair fight" between the resistor and tube to split the supply voltage (more or less according to signal). So 300V and 100K suggests 1.5mA. However the popular Fender 12AX7 stage sets the plate nearer 70% of supply voltage, so 300V implies 0.9mA.

And do not worry too much. A "300V" tube like 12AX7, with 100K on top, can easily stand a 420V supply, yet will make clean sound (for lower signal levels) with less than 200V supply. SO just build it and smoke-test it. Then if you are sure you want different supply voltages, modify the dropping resistors.

I think it is worth making the distinction between class A and class B circuits. In a class B circuit (most op amps) the supply current is signal level dependent. If you don't regulate the supply then as the signal level increases the power supply droops. This is used to deliberate effect i guitar amplifiers. For this reason, regulated supplies are commonly used for op amps.

In class A circuits, the total power draw is constant. In a properly decoupled class A amplifier this means the current draw is almost constant. If it isn't then the decoupling is not doing its job. This means it is not necessary to regulate the supply. On the other hand, class A circuits tend to have a poorer power supply rejection ration (PSRR) than class B circuits (this is not a rule but it is often the case). Op amps have a typical  PSRR of 80dB or more but a class A amp will  struggle to achieve 30dB. This means power supply ripple is less of an issue than voltage regulation for op amps. It means that power supply ripple is more of an issue than regulation for class A tube amps.


Many years ago, a smart guy named Scroggie realised that cascaded RC circuits reduced ripple much more than a single RC of the same total RC. So three circuits of R and C reduce ripple a lot better than one of 3R and 3C. It is not too hard to obtain 30dB attenuation per stage giving 90dB for a three stage RC circuit

Scroggie's original article is here:

https://www.dropbox.com/s/uq3qnct44cgkc9k/Scroggiefilters.pdf?dl=0

Cheers

Ian

Thanks for all these replies. Awesome source of knowledge here.

Prr: i always reference to the curve data but i was simply asking if their was other method of finding the current draw.  i guess in some case, the grid current must be taken in count? This particular one seems a little obsur since their is not so much information about it. Since the grid voltage is varying, how do We determine it's current?

Thanks Ian, now i clearly see the distinction between both.

I recently got Morgan Jones valve amplifiers tome 4 so i'll make a hard read of it and if i remember well, their is a psu section. I should find a lot of informations their, too.

Thanks again guys

Deepdark said:

Prr: i always reference to the curve data but i was simply asking if their was other method of finding the current draw.

Click to expand...

If you designed the circuit yourself then presumably you selected the idle currents yourself. You don't need to 'find them out', you defined them in the first place.! 

On the other hand, if you're copying someone else's design then it ought to quote some voltages and resistor values, from which you can work out what current is flowing (Ohm's law, innit).

Thanks Merlin. Yes when I design myself, I already work out those values, but, for now, I was looking at a schematic from Jensen of a cascode tube pre. I'd like to give it a try so I'll manage to find the current flowing and design the psu accordingly

But to be honnest, I once read somewhere that for tube psu designing, we had to take account of the grid current. This one in particular struggle me because I can't find valuable information about it. I know the voltage at the grid will vary, which is driving the tube. But I would like to know how to manage to find it's current value

Deepdark said:

But to be honnest, I once read somewhere that for tube psu designing, we had to take account of the grid current.

Click to expand...

Not usually... unless you mean the screen-grid current of pentodes/tetrodes?

> Since the grid voltage is varying, how do We determine it's current?

Grid voltage varies around its idle value at an _audio_ rate.

The power filtering should _absorb_ all audio (and line-hum) rate variations.

So the average current in a Class A stage, as seen by the power supply, will BE the idle current.

(Ian rightly points out that AB stages need further study. For 741/TL072 type opamps, carrying speech/music, not heavily loaded, the dynamic signal current may not matter compared to idle current. Beefy opamps driving heavy loads, thrifty opamps driving moderate loads, it may matter. FULL power test-tones can drive dynamic current above idle current long enough to sag an R-C power filter.)

> I once read  ..... take account of the grid current.

The internet is a million monkeys at a million typewriters; you can read ANYthing eventually.

In audio we "NEVER!" have signal-grid current. It just makes distortion and we hate that.

Yes, exceptions. Some of the big Bogen, McIntosh, and Ampeg/Fender power amps pushed lightly into signal-grid current on peaks. All tube guitar amps will try to push grid-current on peaks, but with R-C coupled grids the least grid current charges-down the coupling capacitor and we get grid-blocking instead. Several very-low-voltage schemes bleed a trace of G1 current to get useful plate current with very low supply voltage. And there are special-process tubes which run HUGE G1 current, signal at G2, to get a modest plate current from a 12V car.

In Pentodes, screen grid, G2 current, does happen. For any class-A small-signal stage this will be essentially constant, any audio-rate variation smoothed-out by the screen capacitor. In power amps, and especially over-driven power amps, G2 current will rise. You will never go wrong by using the general parameters in the tube data sheet. You will rarely fall very wrong if you copy a known-good commercial product. While Fender worked those poor 6V6 far beyond factory ratings, history has shown it works well enough for users who get paid to play and can change their own tubes as needed (very different from grandma's radio, or a radar above the arctic circle).

> if i need to design a psu for, i dont know, a gyraf g9, i have 2 tubes in parallel to feed

That happens to be easy, once you find the Philips ECC82 datasheet.

First stage has 2.2K cathode resistor returned to grid, and 47K and 47K (it is a bit odd, but for bias purpose both 47K are "in plate").

ECC82 datasheet page 3 shows various resistor value sets for various supply voltages. Mid-page is 2.2K cathode and 100K plate (close enough to 47K+47K). *Anode Current is explicitly given!* And if you do a V/I calculation, you will see that this tube with 100K+2.2K bias "acts like" a 151K to 153K resistor over the range from 100V to 400V. This is generally true-enough for any triode (and most pentodes) if you keep all the cathode/plate/screen resistors the same and vary the supply voltage: you can approximate the stage as "a resistor" and do simple Resistor Network calculations to get your dropping resistors.

You may also note that Voltage Gain does not change over the supply range from 100V to 400V. (Actually it does, and I wonder if this junior engineer skimped his notes, but the change of gain is not large, never enuff to make-or-break an audio amplifier.)

(Max) Output Voltage does change, and slightly faster than the change of supply voltage. Generally we only care about this at the last stage, and maybe the stage before this.

The G9's 2nd stage uses a cathode resistor, 680r, not on the chart. We might peep the similar 47K+1.2K conditions and guess "higher". 1.2K/680r is about double, does current double? If it did, plate voltage would fall very low, which means low current. It happens to work more like square-root of change of cathode resistor. So I would guess current near 4.3mA.

I had doubts about that, so I worked it out "the right way" on the curves. At the zero-grid curve the current in 680r must be zero. On the 2V grid curve the current must be 2V/680r or 2.9mA; etc for the 6V grid line. Connect the dots you found. The op-point *must* lay along this line. Now plot the 47K plate resistor. Assume it starts from 240V so it must also intersect 240V/47K= 5.1mA, and this is a straight line. Where the cathode line and the plate line intercept IS the op-point, 3.5mA.

The G9's final stage has no plate resistors and a low cathode resistor. You have to work it on the curves. (Being "curves", there is no trivially simple way to get accurate results; though within limits there are rules-of-thumb to get a rough-guess quickly.) By symmetry we know V2A V2B split the 245V equally, 122.5V top and bottom. The 470r also takes a few volts so guess 120V. No plate resistor so the line is vertical. (Technically you would plot 470 Ohms up from 245V, but in this case the difference is too small for my eyes or any practical purpose.) 6mA.

1.63mA+3.5mA+6mA= 11.13mA at 245V supply.

For other supply voltages over a reasonable range (maybe 150V-350V) the whole circuit could be approximated by a 245V/11.13mA= 22K resistance.

So far the curves have not lain near the maximum power dissipation. At high supply voltages they might. A glance at the last-stage shows that 220V per triode (440V total supply) puts you at the rated max 2.75W dissipation. But for any cathode at a high voltage you must also check the Heater-cathode voltage rating. On ECC82 this is 180V. If you actually ran 440V supply, the upper cathode would be near 220V, so the heater supply could not be grounded (as drawn). Cutting back to 360V even 300V total supply would be wise. Alternatively elevate the heater supply on at least 40V (but not more than 178V).

wow PRR. That is a really nice and well explained theorics. It explain me a lot at once hahaha. Of course, we have to understand the principal Under the topology, but so far I get what you wrote there.

Just 1 or 2 things to be sure I get it.  When you say "First stage has 2.2K cathode resistor returned to grid, and 47K and 47K", you are referencing to the 2* 47K plate résistors (R9 and R25) or R9 and R10??

The last stage, which looks like a SRPP stage, feed the bottom plate from the upper cathode, so we might think we run the tube symmetrically, so half the supply voltage. Is that it? Upper cathode pass through R30 (1k5) so I guess the voltage will drop a little. Is it what you were referencing when you said "The 470r also takes a few volts so guess 120V" is it a typo or there is no drop across the 1K5 but through R29 (470r).

Last thing to be sure I'm ok, what do you mean by "(Technically you would plot 470 Ohms up from 245V, but in this case the difference is too small for my eyes or any practical purpose.) " ?? I understand that we place the line vertically when there is no anode load resistor, (such as in mu-follower or srpp stage) but we run at 1/2 supply voltage, so we should put the line at 120V, as you did it. Why 245V??

Thanks again for your time PRR. These things really passionate me and my goal is to learn as much as possible

and in another note, just to validate if I'm ok.

Looking at the cascode mic pre schematic fiound on Jensen website, we have a first stage which is, of course, a cascode followed by what looks to be a srpp stage. So, the cascode stage: We have an anode resistor load of 100K. HT is 250V and after the resistor, we have 150V, so I can guess that the voltage drop across Ra is 100v/100k = 1ma. Is it the total current draw of the stage or the lower triode has to be taken in account, too? Lower anode voltage is around 82V and upper grid is 80v (around 1/3 HT) and lower cathode resistor is 1K5. I tried to draw it but it doesn't really work. I draw the load line taken in account the HT 250v and the Ra 100k which gave me 2.5ma. I then draw a vertical line from my lower anode voltage, 82v up to the 0 curve and it cross the load line. So I guess it's ok. But then, if a draw back the maths to find it's 1K5, at 1ma it gave me 0.15v (near 0v curve) which do not seems to work. From there, I'm sure I'm missing something really stupid, or I just care about something I shouldn't haha. 

> you are referencing to the 2* 47K plate résistors (R9 and R25) or R9 and R10??

R9 is on the *second* stage, which I compute separately.

R10, one end goes only to two capacitors, does not have any DC current.

The first stage has R9 on top and R7 under the cathode. For bias purposes, these are both in series in the plate-loop.

> Upper cathode pass through R30 (1k5)

Ah-ha. I missed that. That changes things. It is an odd detail. It complicates match-book analysis. I guess the first thing I would try is to assume that the Rk reflects through the tube as Mu to give the effective playe resistance. Mu is 17. 17*1,500 is 25.5K; 17* 470 is 8K. To these numbers add the "internal resistance", around 8K. 25K+8K= 33K; 8K+8K= 16K. Assume that B+ is split like a resistor divider of 33K+16K. Call it 2:1. From 240V that would be 160V on top tube and 80V on bottom tube. Plot 1.5K cathode line on the curves. It looks like both tubes will run near 4mA. Since they are in series with no DC path out of the mid-point, they MUST run the same current, about 4mA (not 6 as I figured from flawed reading of the plan).

Hehe that was Quick and looking at the cascode one, do i miss something or my first itteration was right?

> looking at the cascode

Well, you are looking at it, but have not given the link, so I am not looking at it. I do know how to Google, but it is your question.

damn, I forgot to link the schematic