bias supply questions

[beinz]

ive have been reading recently on sites like valvewizard to better understand the different sections of guitar / bass amps.
i just finished reading about rectifier and power supply design, but there is no dedicated "read" on bias supply. ill start with a marsha!! / soldano style bias supply as an example because its what im most familiar with to this point.

when designing your power suppy, i understand that you want to begin with determining your p.t. specs first, then you know how to set up the circuit.
2204 used a c.t. high voltage secondary, and so used a full wave rectifier, and tapped the bias off the h.v. as there was no dedicated bias secondary,
the 2203 p.t. had a dedicated secondary for the bias, and used a bridge rectifier with a non-center tapped h.v. secondary.

this all i understand, which leads me to my questions about the specific resistors / capacitors in the bias supply.

i am trying to wrap my head around determining the changes in resistor / cap values when certain other elements change, for example.

ive been looking at schematics for soldano avenger 50, and slo 100.
both are tapped off the h.v. (360vac?) but the avenger has a
100k dropping res - diode - 10k smoothing res - 15k range res in series with a 25k pot - and 2 x 22u caps,
and the slo has a 220k dropping res - diode - 15k smoothing res - 56k range res in series with a 25k pot - and 2 x 10u caps.

both amps use 5881/6l6 tubes, so my question is why the different value of res / caps?
is it because one is 50w, one is 100w?
shouldnt they both need to reach the same neg voltage at the end of the bias supply, unless the h.v. is not the same on both, i am confused!

my other question is if the caps in combination with the resistors are what achieve the final neg voltage? the avenger bias has aprox half the total resistance, but double the capacitance of the slo, or am i just thinking about this all wrong?

ultimately im trying to understand a logic that applies from using one voltage source, that may apply when using another voltage source, ie a 360v h.v. tap vs a dedicated 100v secondary.

thx

[defaced]

The bias resistor network forms a voltage divider. Its not just the total resistance but also the ratio which determines the bias voltage. The resistors and caps form a RC filter. To keep the same frequency with half the resistance one must double the capacitance. The bias network resistance from the bias point to grpund (the bottom of the voltage divider as i usually say) also figures into the grid to ground resistance which is a specified value on the tube datasheet. Its also usually exceeded, often by a factor of two.

[beinz]

thanks for your help,
i spent some time reading about voltage dividers, and it is starting to make some sense, but as per the calculations i read about, im not sure they're coming out right.
the voltage out occurs between the bias range/pot combo and the bias smoothing resistor correct?
using the slo as an example, the bias range/pot combo are 56k/25k to ground, and the smoothing resistor is 15k.
using the equation V-out = V-in x Vb/(Vb+Va) Vb would be a min of 56k, max of 81k, but is Va only 15k, or does the 220k in series make Va 235k?

this is all probably pretty simple once i understand it...

if i omit the 220k resistor as part of this equation, then V-out = 360v x 81k max / 96k = 303.75v? or
V-out = 360v x 56k min / 71k = 283.94v?
now, is this even the right equation to find what im looking for?
if it is, is 283.94 - 303.75 the voltage out, or voltage dropped?

then i read this link

Series Circuits - MetroAmp Wiki

and now im not sure which way to calculate it... plus i think there are a few typos on this page...,
anyway, it would be awesome if you could push me in the right direction,

help is appreciated

[defaced]

The 220k needs to be included. Without it you're only working with half the circuit. You need it to figure out the current flowing through the resistors. With that you can directly calculate the voltage drop across any one resistor.

[beinz]

i decided to use this mark huss 2204 schem as an example to show where i am getting confused as it includes reference voltages at various different points on the schem.

this schematic is set up for 6550 tubes with replacement resistor values for el34's, but i will use his 6550 values, he states a -57v at the bias supply out.
after reading furthar on voltage dividers, it is the ratio (as you mentioned) that we are concerned with of a particular resistor to the total resistance.

that being said, if this 2204 is using 320v source, and we want -57v (in reference to ground im assuming) then 57/320= 0.18, we want the out voltage 18% of the source voltage, we need our resistor to gnd to be aprox 18% or the total resistance... correct?

since we have a 22k variable resistor along with the 47k, we have a minimum 47k and max 69k to gnd.
ill show values for each:
120+15+47=182
so, if i have understood the voltage dividing lessons, i divide 47 by 182 = 0.26 or 26% of total resistance. 26% of 320v = 83v
120+15+69=204
if i divide 69 by 204 = .34 or 34% of total resistance. 34% of 320v = 108v

also if i understand correctly, the diode positioned how it is will change the voltage to neg.
so depending on where the pot is dialed, i should end up somewhere between -83v and -108v,
but the schem states -57v????
im not sure if im calculating properly, or if im overlooking other details, just trying to understand and make sense so i can apply this idea to other schematics...

thanks again.

[MWJB]

You're on the right track, you're looking for specific -vdc (or range) out of the divider...rather than focussing on resistor & cap values, but you're missing out a stage in the AC to dc conversions...

Let's work backwards from the -57v to put it in perspective & break down the circuit. The adjustable voltage divider after the bias diode is what we want to look at first (this set up is common to most fixed bias amps, the fact that the bias is derived from the B+ secondary is irrelevant for now)...

At the mid bias trim pot setting we have (47K+11K)=58K as the load. 15K as the dropper, to work out the divider ratio:

load/(load+dropper)
58/(58+15) = 0.79

-57vdc/0.79 gives us -72vdc off the bias diode. Prior to rectification by the diode & filter cap we would have ~51VAC (72vdc/1.41) available at the bias supply (however it is derived). There is virtually no current in the bias supply so we can pretty well disregard the 150K grid load resistors...there will be a slight voltage drop accross them, but it is usual to read the bias voltage at their junction.

Regarding 6L6 amps, it's usual to see a C- voltage of 10%-12% B+ at the 220K grid resistor junction on Fenders for instance, the lower 10% for lower voltage amps (>400vdc), higher % for higher voltage amps (<520vdc plate voltage). Ultimately -ve voltage is a ball park affair, we set the bias to a milliamp based setting, starting out by adding a few negative volts to our target (colder bias) with tubes out, so we don't burn anything out whilst installing the tubes & fine tuning.

Back to your 6550 amp schem - RB2 value at 120K is just a value to get us in the region of 50VAC (-70vdc or so rectified) ...this value, within reason, is not critical & could be 100K to 150K, (or 45-60VAC) as we can vary the voltage divider values after the diode to compensate for higher/lower VAC.

Look at the 5F6A tweed bassman schem from 1958...it shows a 56K load & a 15K dropper. Now look at a (pre LTD model) 59RI bassman, that has the same 56K load but a 22K dropper. They are biased different? No. They are biased the same (C- is 11% of the B+ voltage), the difference in resistor value is to account for the fact the RI power transformer develops an additional 5VAC (55VAC) on the bias winding compared to the 1958 amp (50VAC). So at 470vdc on the plates for the RI we'd want to see -52vdc at the junction of the 220K grid loads, from 74vdc off the diode (0.7 divider ratio). Whilst for the 1958 amp we'd want to see -52vdc at the junction of the 220Ks, from -67vdc off the bias diode (0.78 divider ratio). It's the voltage (or range) we are looking for, the values in the divider are just a means to an end.

[beinz]

thanks for the clarification. it has made some things clear to me, but still a bit fuzzy on some other details...
i figure the diode might be part of my confusion.
anyways, looking at the 5F6A for example, the bias makes clear sense to me, there are only 2 resistors, and the secondary starts at 50vac which avoids the need of a resistor before the diode.
the 50vac x 1.41 = -70.5v after the diode. the 56k load and 15k smoother gives a .78 ratio as you stated, so -70.5 x .78 = -55v off the divider.

where im still unclear is when the third resistor is involved.

"At the mid bias trim pot setting we have (47K+11K)=58K as the load. 15K as the dropper, to work out the divider ratio:

load/(load+dropper)
58/(58+15) = 0.79"

in this explanation you use this divider ratio to get to -72v off the diode. would this be from the diode to gnd?
i was under the impression that the ratio was derived from 58k's percentage of the total ratio (120+15+58) not just the two values after the diode.
is it the diode that changes things?

if there was no diode (in theory) would the voltages be as i described previously?

[MWJB]

"in this explanation you use this divider ratio to get to -72v off the diode. would this be from the diode to gnd?
i was under the impression that the ratio was derived from 58k's percentage of the total ratio (120+15+58) not just the two values after the diode.
is it the diode that changes things?

Yes, that -72vdc would be read from the bias diode to ground. The ratio (for neary all fixed bias amps) is derived from the ratio between the 58K load & 15K dropper (though -vdc after the diode may be satisfactory to simply allow a load resistor alone, this could also be the case where you wanted a particularly cold bias, or wanted to use hot tubes & needed max -vdc to get a required plate current, e.g. some harp players like very cold bias to reduce feedback & gain in the output section, for the aforementioned 5F6A that might mean wanting -70vdc or more from the bias supply). The diode, doesn't know where the 50VAC comes from, whether it be a dedicated bias winding, or the B+ secondary/rectifier winding. The 120K (may be 100K or up to 180K depending on the amp) will not feature in every amp you see (again, compare a AA1164 Fender Princeton to a AB763/65 Deluxe - both amps run the 6V6s at reasonably similar B+, but the Princeton derives its bias from the B+ winding, the Deluxe from a dedicated bias tap. Either circuit could be used on either amp (assuming both amps had a bias tap). I did have a plan to build a parallel SE amp, each tube with its own independent bias supply, one from the bias winding, the other tapped from the B+ winding.

The diode converts the AC to dc (by a factor of 1.41), by installing it "reverse polarity" we get negative dc (note bias filter caps are installed backwards with respect to cathode bypass & power supply caps). If the diode was not there and we were just talking AC, then yes, the ratio would hinge on the 58K's percentage of the whole string - a 0.3 ratio & ~96VAC out...too much for what we are trying to achieve, before rectification (96VAC*1.41=135vdc), here.

Ultimately the 120K and the [15K & 58K] combo have different purposes. The 15K & 58K resistors down stream of the diode are setting the negative dc voltage to the power tube grid resistors. The 120K upstream of the diode is a makeshift 50VAC power transformer winding.

[beinz]

thankyou, clear now!

i was just getting jumbled on what was happening prior to the diode!
the diode removes anything prior from the division network.

thx for the help!