Bias Circuit

[surfsup]

Hi guys, in this circuit:



we have Vac going through the first R (unknown value), then rectified, then thru the 15k, then the pot+47k

My question is how does one calculate the voltage drop across each resistor to determine the voltage at the Bias Tap between the pot and 15k?

Assuming 220k resistor for the first resistor, we have a total resistance RANGE of:

Pot @ max: 220+15+50+47 = 332k
Pot @ min: 220+15+47 = 282k

If I was using a 300-0-300 PT, I would have:

I=V/R=(300*.707)/332 or (300*.707)/282
Irange = 0.75mA - 0.64mA

Using this current I can solve for a V drop across each resistor. The problem is:

1) It doesn't match the duncanamps PSU Designer results
2) The first resistor is on the AC side of the diode and the voltage drop is different

Can anyone enlighten me on this a bit more?

[PRR]

> The first resistor is on the AC side of the diode and the voltage drop is different

Correct.

In PSD you can try to dummy-up the AC transformer internal resistance. (But it may not like such an "unrealistic" value.)

> Can anyone enlighten me on this a bit more?

No.

It is a "stupid" plan, in the engineering sense. You are wasting-away 360V to get 40V... that's like MASHing the go-pedal then using the brake to stay under the speed-limit.

Therefore it is rarely studied, and I have not seen any published rule-of-thumb design guides.

> Can anyone enlighten me on this a bit more?

Get a 10-pack of 47K. Series them up around 250K. Wire-up with your diode, cap, 5K-15K, cap, and 50K+50K trim network. See what you get. SHUT-DOWN before changes. Add or omit 47K resistors in the string until you get in the ballpark. Then count by 47K and go get one 2W resistor of that value.

If you are outside the usual range of gitar-amp AC voltages, the series resistor power rating may be different. It is a very ugly waveform, and ugly math to calculate.

[PRR]

This {below} is how to handle it in Duncan PSD.

The load is the sum of trimmer and stop-resistor. Generally you want minimum voltage to not-quite FRY the tube (so a trim accident won't cause instant meltdown) and a maximum voltage which will put the tube "OFF" (a good place to start). This is roughly a 2:1 range so set the stop-resistor about equal to trim-pot.

The "15K" has small effect on voltage. This circuit is more constant-current than constant voltage. As long as your 100K bias-trim network stays the same, 1K to 20K in front has little effect on bias voltage. (This is different than the small-resistance rectifier used with 50VAC bias taps.)

The "15K" is picked more for filtering. So you need cap values first. As a general ballpark you can use 1uFd per mA of load. We'll assume 50V-100V across the 100K string, around 1mA. 1uFd would work... except this is a half-wave circuit so we need 2X to 4X the cap to get clean. Say 4uFd. However the cheapest suitable caps are 100V-150V Electrolytics, and they are not sold much smaller (and no cheaper) than 10uFd. So use 10uFd.

However the simple calc on 270K+10uFd says 2.7 seconds for voltage to rise to 63% of final value. In fact the 270K is not pulling the cap half the time, so the rise is even slower.

It may not be necessary to have two caps. If the amp were perfectly balanced, grid-bias ripple would cancel. And there is a disadvantage: a quick on-off-on event may leave cathodes hot and bring plate voltage high before bias voltage slowly ramps up. In fact I suspect you should try one stage with first 10uFd and then 20uFd and see if the amp buzzes less.

If you do need to go as far as a 2-stage filter, then 1K resistance gives some cut of 60Hz buzz and big (2-stage) cut in the higher buzz harmonics. 10K and 10uFd gives quite a lot of filtering, more than you should need. 15K may have just been handy.

The PT's source resistance may be dummied-up, which is the same effect as an actual resistor.

The R-C-R-C filtering is very slow to rise, you need to set "delay" to at least 10 seconds. (If the wobble is slanty, it is still rising toward final value.)

Values shown will give 46.2V. This is the "OFF" bias. This is perhaps too low for most audio tubes fed 333VAC =~~ 450V DC. The "270K" needs to be smaller. 

[surfsup]

It is a "stupid" plan, in the engineering sense. You are wasting-away 360V to get 40V... that's like MASHing the go-pedal then using the brake to stay under the speed-limit.

This is tapping the same secondary feeding the power rail of the amp so its not a separate transformer or anything.

Get a 10-pack of 47K. Series them up around 250K. Wire-up with your diode, cap, 5K-15K, cap, and 50K+50K trim network. See what you get. SHUT-DOWN before changes. Add or omit 47K resistors in the string until you get in the ballpark. Then count by 47K and go get one 2W resistor of that value.

I want to avoid lengthy "guesswork" like this.

The "15K" has small effect on voltage

Yea I know this is just to get separation and another filtering stage

This circuit is more constant-current than constant voltage.

Yes, and this is what I'm really trying to find out, how much current is running through. So using the Trim+stopresistor value (50k + 47k), and I wanted -30Vdc:

30/97000 = 0.33mA current

So I can assume 0.33ma current is running through this circuit, but doing the math on the "other side" of the circuit (220k/15k) the math is wrong. That's what I'm struggling to understand how to figure out the voltage dropacross the first 220k and 15k resistors.

[jjasilli]

Probably not worth the effort for a formula.  TUT1 has a formula but I found it unhelpful, and the resistors are not all in the same places as in KOC's.  Probably best to plagiarize a known circuit.

I just dealt with this issue in an EL-84 Traynor Bassmate project, almost completed. Stock, it uses fixed bias tapped-off one of the HT windings.  I'm now using a bucking tranny to drop AC wall supply voltage; then a power resistor in the B+ rail to drop plate voltage from over 400VDC down to 300VDC. This messes with the voltage delivered by the stock fixed bias supply.  So, UNLOADED, I gerry-rigged away to put -18VDC or so the EL-84 grids, knowing this would drop with tubes in (loaded), as the AC supply voltage gets dropped down.  I kept the 15K filtering resistor for the reasons stated by PRR.  Added 2X 10K bias adjust pots -- one for ea power tube.  To get my component values in the bias supply circuit, the first dropping resistor (on the AC side of the bias diode) was replaced with a substitution box, at first set to about stock value.  The resistor to ground under the bias adjust pots pots was temporarily replaced by another pot in series with a small fixed resistor to ground.  I played with the substitution box and temporary pot to get a safely high bias voltage on the unloaded tube pins (while monitoring the voltage on the bias caps so as not to blow them up!).  Then popped-in the tubes, and fine-tuned the resistance values.  Then hard-wired in fixed resistors of the correct value to support the dual 10K adjust pots.  

[surfsup]

Yea i'm going to go with the schemo and tweak later. Maybe one day i will crack the voodoo behing this circuit.

[HotBluePlates]

Probably not worth the effort for a formula.

It is worth the effort. The problem is that there is no such thing as a simple formula to find the answer.

The bias circuit is roughly the same problem as you find with figuring the d.c. output of the main rectifier in the amp's power supply.

O.H. Schade (the guy that wrote up how 6L6's work, and likely invented them) did a bunch of research and compiled graphs to assist in answering this same question with regard to the B+ supply. The charts and the method are given in RDH4, and there's maybe 6-7 charts that you have to work through, one at a time. His "easy method" ain't as easy as you'd like.

And guess what? Of the many variables that you have to juggle, the resistance presented by the PT's primary winding (plus any series resistance) is one of the significant ones. And that resistance is tiny compared to the input resistor in a bias circuit (meaning it doesn't have as much impact on the B+ supply as it does in the bias supply).

... but doing the math on the "other side" of the circuit (220k/15k) the math is wrong. That's what I'm struggling to understand how to figure out the voltage dropacross the first 220k and 15k resistors.

If you want to do the math on the other side of the circuit, you probably need some trigonometry, some vector algebra and some calculus.

In a d.c. circuit, voltage and current are in phase and are constant at all times (we'll neglect step/transient analysis); additionally, you mostly think only of linear resistances. In an a.c. circuit, voltage and current are not in phase. Plus, you have a non-linear circuit element: a rectifier diode. The a.c. cannot be thought of in terms of RMS (as you typically know it: a.c. voltage * 0.7071), because the diode turns the a.c. into pulsating voltage and current, which charges the caps on the "d.c. side" of the rectifier and pulls pulses of current through the rectifier which are not solely related to the incoming a.c. voltage.

Strictly, RMS works, but the difference is that the waveshape is different, and so the normal conversion factors don't hold. The calculus part comes in when you need to calculate the effective amount of d.c. represented by the current pulses, so that you can conveniently calculate.

In any event, the rectifier's effect comes into play, the resulting waveshape, the size of the capacitors, the degree to which voltage and current are out of phase due to the balance of reactive and resistive components, and the amount of resistance on both sides of the rectifier (considered separately and with different effects). All of these considerations are why there is no one simple formula, why O.H. Schade's method had so many steps and graphs, and why PRR's suggestion of cut-n-try works faster than calculation.

It's also why he suggested, if you want calculated numbers before getting the soldering iron hot, that you tinker with the power supply simulator. Let someone else do the hard math for you.

[PRR]

> This is tapping the same secondary feeding the power rail of the amp so its not a separate transformer or anything.

I was speaking generally. Yes, it is used a lot in guitar amps.

> I want to avoid lengthy "guesswork" like this.

Me too. Look at my other posts; I'll calculate anything, even a cathode resistor, even a gambrel roof. But this is one case which defies practical hand analysis (even with a calculator).

I'll relent a bit. ASSUMING the desired DC is much-much less than available AC (otherwise it interacts like mad), it isn't too bad.

Find a Known-Case. I'll use my PSD example above. 333AC, 46V across 100K, or 50.6V across 100K+10K, which is 0.46mA.

_Pretend_ the "333V AC" were actually 333V DC.

The series resistor must drop 333V-50.6V= 282.4V, and at 0.46mA. 282.4V/0.46mA= 614K. But the (million point-by-point calculations simulation) says 270K needed. The "assume DC" number is 2.27 times higher than what we really need.

Why "2.27"? Well... about half because the diode only passes half the time, and less than half because it won't start to conduct until the AC wave rises over the 50.6V. The useful charging waveform is NOT a sine, or half-sine, but a slice of a sine. So the usual 1.414 or 0.9 factors don't give the right answer.

[surfsup]

Ok. Thanks guys for your willingness to share.

Why "2.27"? Well... about half because the diode only passes half the time, and less than half because it won't start to conduct until the AC wave rises over the 50.6V. The useful charging waveform is NOT a sine, or half-sine, but a slice of a sine. So the usual 1.414 or 0.9 factors don't give the right answer.

This is an eye-opener for me as i have a mathematics major and have done quite a bit of calc and vector analysis. I have little electronics knowledge though. I don't know what "tut4" is but i'll find out and try and get it. Interesting stuff. Once i feel i know what i need to calculate i should be able to set up a spreadsheet to do this automatically "before" the iron gets hot. Thanks again guys.

[sluckey]

With a couple clip leads and several resistors you can have an answer before your soldering iron heats up.

[jjasilli]

"tut4":  TUT = "The Ultimate Tone", a series of reference books written by Kevin O'Connor who owns a company called London Power in Canada. Worth googling.  The books cover various aspects of amps, components and related electronics and are labeled TUT1, TUT2, etc.  The author is quoted a lot and is often referred to as KOC.

Thanks to this thread I now see why I too was having trouble calculating a bias voltage in advance, and resorted to breadboarding. 

[surfsup]

I will probably pass on the TUT4 book. At $110, and just wanting to know more about one small circuit, I don't think it is the time to pony up that much $.

The books look nice though. Maybe for a xmas present from the wife. I'll post back later if I find out anything more on this.

[DummyLoad]

PSU designer get's us close. we fiddle with a few values and get results in less time than it takes to brew a pot of coffee.

--DL

[surfsup]

DummyLoad, in taking a schematic on a well documented/cloned amp which works and looking at the bias voltage setting recommended, the PSUD software does not even show the correct range to achieve the voltage for biasing the power tubes. I started this thread solely because the modeling software out there doesn't seem to like calculating the bias voltage. So yes it calculates the numbers quickly, but the numbers are wrong. (at least for the example I am working).

[PRR]

> it calculates the numbers quickly, but the numbers are wrong. (at least for the example I am working).

WHAT "well documented/cloned amp which works" example?

Ah, there's not that many variations. Marshall JCM 800 LEAD below, on plan and in PSUD.

The PT AC Volts is not noted, but we know the big DC, low-loss diodes, we can work out the AC Volts. While the PT is CT our bias only uses half. Half Wave. Now we'd normally go right to a cap, but as pointed out PSUD does have an R-C input option. The plan I found uses 56K fixed plus 22K trim, I took 67V as the trim-center (but you should check the 56K and 78K extremes, PLUS a +/-20% tolerance on about everything.

Expect 30V, get 29.87V.... close enuff.

[PRR]

Oh, PSUD "assumes" we want positive output. Obviously everything is the same if everything is upsidedown.

[DummyLoad]

the example i posted was/is based on a princeton reverb / vibrolux bias supply. it's pretty PSUD comes very close to fender values.

--DL

[surfsup]

WHAT "well documented/cloned amp which works" example?


Trainwreck. I must still be doing something wrong. TINA software is a little complex. I'll try PSU-D again.

[PRR]

> the modeling software out there doesn't seem to like calculating the bias voltage. So yes it calculates the numbers quickly, but the numbers are wrong.

The software is probably giving the right answer for the question you asked. The sim is hardly ever flat-WRONG.

The sim is dumb, and can't guess what you really want to know, and isn't the least bit interested in helping you ask the right question correctly.

SPICE will compute 8-digit answers 10 times faster than I can poke a sliderule; but I spend 10 times longer setting-up and re-re-re-checking the problem than I do finding an answer.

I don't do Tina; here's an older SPICE interface.

AC voltages are customarily "peak", not RMS or average, because SPICE supports non-sine wave sources. So "258VAC" is 365V Peak to SPICE.

SPICE offers several kinds of runs. New-fangled interfaces like Tina sometimes blurr the distinctions. You would think ".AC" would work; it won't. You must run a ".TRAN" command to handle non-steady AC puzzles.

I could have put the diode in the other way to get negative output, but it's too darn muggy to get fussy.

[PRR]

Here's 3 runs with trim-pot set up, center, down.