Bias Supply Droppers

[FourT6and2]

Quick question about the bias supply in a typical Marshall circuit. I understand what the usual 56K or 47K resistor does before the bias —sets the range of the bias control. But what about the resistor that comes right before the diode? I've seen various values here: 27K, 220K, 180K, and so on. What determines what value you should use? This is for a 50-watt amp, SS rectification, EL34s. PT secondaries are 350-0-350 and B+ will most likely be somewhere between 450v-470v. This is also for a dual-bias arrangement:

[V2]

As the thread title indicates, those are voltage-dropping resistors. The value of these Rs depend primarily on the voltage of the secondary. Think about it the same way you think about the HT supply. The first cap is a 'reservoir' cap that is followed by another PI filter (dropping R and cap). You'll end up with a negative DC voltage at the reservoir cap and another, lower DC voltage at the second cap (call this B-). B- will be affected by the preceding dropping resistors. If you tap off of the HT secondaries, you'll need larger dropping resistors to get an appropriate B-. If you have a dedicated 50V secondary for bias, then you'll need much smaller dropping resistors.

[V2]

And also think about what the bias circuit must do. On one end of the spectrum, it needs to basically 'turn off' the tube by supplying its grid with a "large" negative voltage (large in the absolute-value sense). At the other end of the spectrum, it must deliver at least some negative voltage to the grid so that the tubes don't melt down. The dropping resistors help to set the maximum negative voltage required to make the tube 'turn off', which is dependent on tube type. The third resistor in the Marshall circuit (56k in your example) is a shunt resistor (in series with bias pot to ground) that prevents the bias voltage from becoming too small when you dial back the bias pot.

[FourT6and2]

Thanks for the explanation!

So how does one determine what value is needed for that dropping resistor (or resistors, in the case of dual bias)? Is it possible to know what value to use before the amp is turned on for the first time? Is there a target B- I should aim for?

Really, I'm building a Ceriatone Chupacabra. But I'm using my own power transformer, which has higher voltage secondaries than the 325v one from Ceriatone. The Chupacabra is spec'd for a 325v-0-325v secondary, and the one already built that I have, has 337v coming into the bias dropping resistor, directly from the PT (123v mains). It uses the arrangement I posted above (220k dropper, 56k shunt).

But the Yeti, which is the same amp, but with 350v seocndaries, uses 180R dropping resistors and 47K shunt resistor. But if the voltages are higher from the PT, wouldn't the dropper need to be higher, not lower?

[sluckey]

But the Yeti, which is the same amp, but with 350v seocndaries, uses 180R dropping resistors and 47K shunt resistor. But if the voltages are higher from the PT, wouldn't the dropper need to be higher, not lower?

You have to look at the entire bias circuit. One uses a 47K to the bias pot, the other uses a 56K to the bias pot.

[V2]

Thanks for the explanation!

So how does one determine what value is needed for that dropping resistor (or resistors, in the case of dual bias)? Is it possible to know what value to use before the amp is turned on for the first time?

Merlin Blencowe has a forumula on his bias supply page (check out the derivation of the formula; I'm just skipping to the end):

Rseries = Rshunt * (.32 x mu -1), where mu for EL34 = 10.
Rseries = (25000+56000) * (2.2) = 178200.

R2 (15000) should probably be included in Rseries. So:
R1 = Rseries - R2 = 178200 - 15000 = 163200.

So try 150k or 180k for starters.

[FourT6and2]

Thanks for the explanation!

So how does one determine what value is needed for that dropping resistor (or resistors, in the case of dual bias)? Is it possible to know what value to use before the amp is turned on for the first time?

Merlin Blencowe has a forumula on his bias supply page (check out the derivation of the formula; I'm just skipping to the end):

Rseries = Rshunt * (.32 x mu -1), where mu for EL34 = 10.
Rseries = (25000+56000) * (2.2) = 178200.

R2 (15000) should probably be included in Rseries. So:
R1 = Rseries - R2 = 178200 - 15000 = 163200.

So try 150k or 180k for starters.

I think I understand. And I'm reading through Merlin's page. But let's say I pop in some 180K resistors and fire the amp up. What am I checking to make sure that value is appropriate? Is it merely if the tubes bias up to my target number?

EDIT:
Ok, I think I answered my own question. Merlin says the end result is to measure the screen-to-cathode voltage and divide by the mu of the tube in triode mode. For EL34s, that's 10. So essentially, I need to build the amp, try a resistor value, and measure the sreens to then determine if that resistor value is correct? So if the amp has 460v on the screens, I need to make sure I'm reading -46v?

But also, original Marshalls like the one I'm building is based on, used 27K droppers in the bias supply with 47K shunt resistors and 25K bias pot. But if I do the math using the above formula, it's:

(25000+47000)*2.2 = 158400 - 15,000 = 143400 = 143.4K

If a 143K resistor is needed, why do those amps use a 27K? I must be missing something.

[V2]

yep. Make sure you have -46Vdc (or less; -50 would be fine too) on pin 5 of each octal when your bias pot is rotated for maximum resistance. Season to taste.

On another note: whenever possible (e.g., designing a new amp), try to keep the resistor values small (with a separate bias tap). The output impedance of this circuit should really be lower than what the iconic Marshall circuit has.

[FourT6and2]

yep. Make sure you have -46Vdc (or less; -50 would be fine too) on pin 5 of each octal when your bias pot is rotated for maximum resistance. Season to taste.

On another note: whenever possible (e.g., designing a new amp), try to keep the resistor values small (with a separate bias tap). The output impedance of this circuit should really be lower than what the iconic Marshall circuit has.

Ok, I'll try 180K droppers with 56K shunt and see what happens.

[tictac]

Also, you can do a dual bias adjust arrangement with a single bias supply... you don't need two separate supplies...

TT

[FourT6and2]

Also, you can do a dual bias adjust arrangement with a single bias supply... you don't need two separate supplies...

TT

The board I'm using is already set up for dual bias

[tubeswell]

Think of the bias supply circuit as any other type of R/C power supply decoupling rail - only its oriented to provide a negative DC voltage instead of a positive one.

You have

1) the VAC source (The PT winding bias tap)

2) the rectification (The reverse biased diode providing 1/2-wave negative DC pulses)

3) the reservoir cap (the cap after the rectifier diode (that is oriented Positive-pole to ground)

3) the 1st dropping resistor (between the first reservoir cap and the next (smoothing) cap (also positive pole to ground)

4) the adjustable bias voltage resistance (the pot in series with the tail resistor)

The highest (negative) DC voltage will be present at the bias supply's (reservoir) cap's negative pole. The next highest negative voltage will be at the junction of the other end of the supply dropping resistor and the smoothing cap's negative pole. etc. You can alter the -VDC output by changing the values of resistance in this (negative) voltage divider. Reducing the resistance of the 1st (dropping) resistor will progressively increase the negative voltage up to the maximum possible negative voltage. You can get the dropping resistor down to about 1k, and still functioning smoothly if you use a 100uF (or greater) reservoir capacitance. The current demand on a properly functioning fixed-bias supply is negligible, so the values of the resistors in the overall bias supply dropping chain don't need to be extreme (i.e. it will run happily with resistances in the order of 10s of kOhm). So if you want a bias that goes as-negative-as-possible, try a 100uF reservoir with a 1k dropper going to another 100uF cap and a 50k pot with a 50k tail.

If that still isn't cold enough for you, convert the bias supply to a half-wave (negative) voltage doubler. But your bias supply options needn't stop there. Other types of bias supply are possible (including AC-coupled negative rectification from one of the High Tension winding ends). See the section of Merlin's website on bias supplies.

[FourT6and2]

Think of the bias supply circuit as any other type of R/C power supply decoupling rail - only its oriented to provide a negative DC voltage instead of a positive one.

You have

1) the VAC source (The PT winding bias tap)

2) the rectification (The reverse biased diode providing 1/2-wave negative DC pulses)

3) the reservoir cap (the cap after the rectifier diode (that is oriented Positive-pole to ground)

3) the 1st dropping resistor (between the first reservoir cap and the next (smoothing) cap (also positive pole to ground)

4) the adjustable bias voltage resistance (the pot in series with the tail resistor)

The highest (negative) DC voltage will be present at the bias supply's (reservoir) cap's negative pole. The next highest negative voltage will be at the junction of the other end of the supply dropping resistor and the smoothing cap's negative pole. etc. You can alter the -VDC output by changing the values of resistance in this (negative) voltage divider. Reducing the resistance of the 1st (dropping) resistor will progressively increase the negative voltage up to the maximum possible negative voltage. You can get the dropping resistor down to about 1k, and still functioning smoothly if you use a 100uF (or greater) reservoir capacitance. The current demand on a properly functioning fixed-bias supply is negligible, so the values of the resistors in the overall bias supply dropping chain don't need to be extreme (i.e. it will run happily with resistances in the order of 10s of kOhm). So if you want a bias that goes as-negative-as-possible, try a 100uF reservoir with a 1k dropper going to another 100uF cap and a 50k pot with a 50k tail.

If that still isn't cold enough for you, convert the bias supply to a half-wave (negative) voltage doubler. But your bias supply options needn't stop there. Other types of bias supply are possible (including AC-coupled negative rectification from one of the High Tension winding ends). See the section of Merlin's website on bias supplies.

Except, yeah... I'm talking about a bias supply fed from the Power Transformer's High-Voltage AC Secondary. Not a separate bias winding. The "dropping resistor" I'm referring to is BEFORE the diode—between it and the PT, which has no bias tap. The below schematic shows the circuit and values from a 50-watt, EL34 amp with 325v PT secondaries. My question is how to figure out a new value for the resistor circled, if I change the PT to a 350v-0-350v that, instead, maybe puts out an actual 365v AC, for example. Doesn't that dropping resistor need to change now?



I've read Merlin's page but I don't see where he describes this resistor or how to figure out its value. If I use his math, the result doesn't make sense with how the circuits I'm looking to build have been designed. The values don't match. And his math never takes the PT AC Secondary into account. It just ignores the incoming voltage. How is that possible?

Rseries = Rshunt(0.32 × µ - 1)

Nowhere in that formula does it mention the voltage drop required to function from HT AC Secondary. If the PT AC Secondary is 200v or 380v... doesn't it matter? How do I know what value to use based on the PT voltage? Because if I do the math, changing the PT voltage has no effect at all on it's solution:

Rseries = (56,000 + 25,000)*(0.32 x 10 - 1)
Rseries = 81,000 x 2.2
Rseries = 178,200ohms = about 180K

Where in any of that math is the PT AC High-Voltage Secondary? I'm certain I'm not understanding something. Because to me, it doesn't make sense.

[sluckey]

I suspect Merlin purposely avoided any calculations to determine the value of that resistor. Too involved and the result is often not perfect. Most circuit engineering is done through plagiarizing from reference books. The value of that resistor (often called the bias range resistor) is very quickly and easily determined experimentally. You already know that the higher AC voltage source will require a higher value resistor. It doesn't take much experience to know the resistor will typically be less than 1K when tha AC source is from a bias tap and it will typically be 100K or higher when the AC source is from the HT winding. You can either substitute different resistor values or even use a pot to dial in the proper value to give you the desired bias voltage range. The proper bias voltage depends on the output tube type. Typically for an EL34 a bias range of -30v to -50v will give you the exact bias voltage you need. Your R16 gets you in the ballpark and R20 will help set the amount of swing and the upper/lower limit. You don't even need the output tubes plugged in to get the bias circuit tweaked for proper operation.

[V2]

I don't know how Merlin derived his mu-based formula. I guess that is the missing piece of the puzzle. The aim, according to Merlin, is to pick Rseries and Rshunt values that will give you a maximum negative bias equal to HT / mu, where HT (i.e., B+) depends on the Vac of the HT secondaries. Rseries and Rshunt with obviously form a voltage divider that will lower the (negative) bias voltage; according to Merlin, his method will yield a max negative bias voltage equal to HT / mu at the bias pot. So you could just try it out or contact Merlin for the derivation of his formula.

Notice that Merlin's example computation contains an error. He is computing the Rseries for an EL34, which has a mu of 10. His end result appears to have used mu=10, but at one stage of the computation, he used mu = 20 (for EL84). That confused me for a while....