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Help with a choke for first-time DIY...

mr mojo

2005-06-06 3:58 pm
Howdy folks, almost done with my first scratch DIY-PP 7591s, dual 5ar4s, 2 0A3s for screen regulation of each channel and a pair of 6AN8s in a cathodyne arrangement for gain/phase splitting.

My question is where and whether or not to use a choke. IIRC I've got 2 80uf lytics in series off the cathodes of the 5AR4s sharing 2 180k 10w wirewound resistors dropping voltage from the cathodes of the 5AR4s to the 0A3s. From there it's on to the output xfrmrs. After going through the 0A3s it's capacitor/resistor filtering to the 6AN8s.

Would I use the choke in place of the 180K dropping resistors? What would I gain by doing so? If I were to use a choke in place of the resistors, how do I figure out henries, ohms and amps needed?

Since they were'nt a part of the origninal layout there's not a lot of room, but I figure I've got roughly a 3"x3"x3" cube of space on each side of the 5AR4s and 0A3s so I'm also wondering if I even have space for them now.

Any and all help would be greatly appreciated!

OK....when using choke loading...the mid band Gain will roughly be the mu of the valve.... mu changes according to operating conditions..so you figure the actual mu ..... for simplicity figure mu=21 and plate resistance = 4700 ohms.....
Low frequency Gain is based on the inductance of the choke...
WHen the Inductive reactance EQUALS the plate resistance of the valve...you have you -3dB point....
So (6.28 x fr x L) = 4700 is the -3dB point...
I usually want my -3dB point much lower than 20Hz...to maintain phase as best as possible.....
I usually go down to 3Hz...but that will not be realistic in this case..
So try using 6.7 Hz..... then figuring for L you get 112 Henries...
The DC current for the inductor will be the idle quiescent current..For this tube it should be choosen with some headroom so figure 20mA....
For high frequency response you will have to deal with the Capacitance.... You form a LCR tank circuit.. You know your L and you know your R (plate resistance)....then you figure a C that will be appropriately dampened by the plate resistance...then wind a choke to meet this Capacitance or less...this is usually done by putting lots of thick insulation between each layer...can be calculated..

Oh...Boy!!! I am slow... I just figured out what you were really asking in your question....
When using a LC filter..this is refered to as a DOUBLE POLE filter and therefore has a -40dB per decade roll-off.....
So you are in USA...So the 60Hz mains becomes 120Hz ripple output from the supply voltage...
The resonant frequency of double pole filter is 1/ (6.28) (LC)^1/2

OK...for example..if you set the resonant frequency to 12Hz..then you will be down -40dB at 120Hz.... This would be a 20H choke with a 8.8uF cap...... This is not the only solution...as long as the product of the LC remain a constant then you can use different combination of L and C , within reason..... For example 10H choke with 17.6uF ....So you would actually use a 20uF....
This example of being -40dB down at 120Hz is the minimum I would use...any more than this would be great....
The DC resistance of the choke could be a problem in current limiting the OA3... SO you may need to use a resistor on just one of the filter stages and choke on the other one in series..


mr mojo

2005-06-06 3:58 pm

Please don't take this as snide-but I have no idea what you just said!

Gotta keep in mind I know just enough to be dangerous. I've restored a couple Scotts, I've got a lot of the basics of tube circuits, I can read a schematic and translate to a physical design but beyond that I'm awfully green. Part of the drive to DIY was to learn-which I have-and to create something unique, but as I get closer to finishing my amp the vast gaps in my understanding become clearer and clearer.


It seems you and cerrem are on the same page here as far as the current to the 0A3s. Looks like for now it's best to stick with the schematic. I guess since it was originally created by the engineers at Westinghouse I'd do best to stick to it for this build.

I sure appreciate the help from you both! I've got a feeling this won't be the last amp I build-maybe on the next I can dig a little deeper on designing the power supply myself.

Thanks again,

If you're talking about the current set resistor immedeately before your VR tubes, you could also use a constant current source. This would perform the same function as the resistor as well as providing superior PSU noise rejection. If you're talking about a resistor further upstream from your VR tubes, then you could substitute a choke.

Bas Horneman

diyAudio Moderator Emeritus
2002-04-03 12:03 am
The Netherlands
When using a LC filter..this is refered to as a DOUBLE POLE filter and therefore has a -40dB per decade roll-off.....
Hi Chris,

Could you explain how to use that calculation?

1/ (6.28) (LC)^1/2

for instance where does the 6.28 come from? And what would that calculation look like when using 50Hz?


I tried to feed this little calculator (3 pole is smallest) http://www.wa4dsy.net/filter/hp_lp_filter.html

but if I enter your 12Hz I get uH...instead of Henries and pF :xeye: plus no frequency the filter is fed with.


Paid Member
2003-06-12 7:04 pm
Maine USA
> where does the 6.28 come from?

Basic trigonometry. Converts radius to circumference. Nevermind why.

> this little calculator (3 pole is smallest)

Cute. Actually gives "correct" answers if you can convert pF to uF in your head. Correct but wrong because it ignores the relative costs of chokes and caps (for 50 years, caps have been MUCH cheaper than chokes for the same filtering effect). In fact we want very low impedances in and out, which suggest big caps. An "optimum" wave-filter is not necessarily the best power filter.

Your first cap off the rectifier must be sized to catch energy from the input sine peaks. 1,000uFd per Amp is a good trial value for low-volt work. More is better, up to somewhere past 10,000uFd/Amp where rectifier surge current can be insane. This rule comes from 60Hz work but is VERY rough and will still give you a place to start for 50Hz work.

Two channels of 7591 will eat ~200mA or 0.2A. 0.2*1,000= 200uFd.

200uFd, especially at over 450V, is rather costly. And with a vacuum rectifier, we may be over the peak current*time limit for the bottle (often 10uFd-40uFd for the bottles used in hi-fi).

Duncan's Power Supply Calculator shows why. At 400V 200mA we have only 6V peak-peak ripple, which is "small" compared to 400V. If money is a problem (it always is), a better rule of thumb is to start from 5% RMS ripple. For 60Hz this suggests that the load resistance times the capacitor should be about 40,000ΩuFd. For our 400V/200mA= 2,000 ohm load: 40,000/2,000= 20uFd. For 50Hz, use a constant of 48,000: 48,000/2,000= 24uFd.

This is good commercial practice. It may not be good audio practice. Guitar amps sometimes use "small" caps for dynamic effect (pluck is modulated by droop, sustain is brought out as power falls). Hi-fi often likes more solid rails with lower ripple (and crosstalk). Given the low price of Japanese bulk uFd, 40uFd to 100uFd is reasonable, 470uFd not incredibly insane for silicon rectifiers.

Mojo's 2*80uFd= 40uFd is a reasonable value. It will show about 2.5% RMS ripple. 2.5%*400V= 10V RMS, about 30V peak-peak. This is actually 30V dips from the 400V maximum available, so it means the output wave will show ripple-clipping at about 80% of the power possible with a perfect 400V supply. Even when not clipping, the 10V RMS ripple may cause buzz in the output. Push-pull pentodes are fairly resistant to power rail buzz: we know many classic amps worked with the values Mojo has.

Buzz-reduction for small volt-amp stages is usually done with resistor-capacitor dropping.

If we need lower ripple on the output, we can try an R-C, but the R wastes power. An ideal L-C filter won't waste power.

Pick an L and C value from thin air. Say 10H and 40uFd.

The frequency we want to reduce is the ripple frequency. For 50Hz, it is 100Hz.

Compute the reactance of the L and C at 100Hz:

10H*100Hz*6.28= 6,280Ω ~= 6K

1/(40uFd*6.28*100Hz)= 39.8Ω ~= 40Ω

Since the difference in reactance is large, we can use the approximate result:

Reduction in 100Hz buzz = 40/6K = 0.0067 or 1/150

So for 400V, 200mA, 40uFd -> 10H -> 40uFd:

10V/150= 0.066V RMS buzz

(Duncan's PSD gives 0.019V p-p, about 0.067V RMS, using low-R parts.)

We have computed only the first partial of the ripple, the 100Hz. "Buzz" includes higher partials. At the first cap they are all strong. But after an L-C filter, they fall off at 12dB per octave. This usually makes higher partials negligible: if the 100Hz is too low to hear, and higher falls off fast, the higher partials are too low to hear even with Fletcher-Munson.

Bas Horneman

diyAudio Moderator Emeritus
2002-04-03 12:03 am
The Netherlands
Thanks a lot for the comprehensive answer PRR...I appreciate your time!

So far so good :
10H*100Hz*6.28= 6,280R ~= 6K

1/(40uFd*6.28*100Hz)= 39.8R ~= 40R

Since the difference in reactance is large, we can use the approximate result:

Reduction in 100Hz buzz = 40/6K = 0.0067 or 1/150

But this part...where does the 10V come from? is that the 400V / 40R or is this from your earlier "It will show about 2.5% RMS ripple. 2.5%*400V= 10V RMS" so the 2,5% is that just a rule of thumb?

So for 400V, 200mA, 40uFd -> 10H -> 40uFd:

10V/150= 0.066V RMS buzz