Fixed bias without standby switch

[c.stoffel]

Hello everybody! I built this circuit inside an old amp and it's sounding good but there's something I'm just a little bit worried about. I'm using fixed bias and there's no standby switch, nor a dedicated bias tap on the power transformer. I'm also using solid state rectification for the HV, so the B+ comes up really quick, almost instantly, but the bias voltage takes something like 5 or 6 seconds to get where it should be (thanks I believe to that 220k resistor after the diode on the bias supply...).


Some points:
1. I want to keep the front plate intact so I don't want to add a standby switch
2. Maybe cathode bias could solve my concerns but fixed bias worked much better in this amp because it has bias wiggler tremolo (applied to point A on the schematic), and I liked the tone much more once I converted it from cathode bias so...
3. Would a dedicated bias tap on the power transformer make the bias voltage go up as quick as the HV, due to a lower RC time constant? Maybe I could buy another power transformer with a bias tap, but it's an expense I'd prefer to avoid if possible.


I know there's a lot of debate on standby switches, and lots of things have been demystified, but I couldn't find anybody talking about this specific situation, so the question is: should I be concerned or just leave it alone and play the thing? 


Thanks!
Christian

[Paul1453]

I am wondering why the bias circuit takes that long to get to operating voltage.

It is also fed by a SS diode so it should be getting it's voltage almost as quickly as your HV circuit.

Maybe your e-caps in the bias circuit aren't up to snuff, and leaking?   

Your output tubes should be taking about the same amount of time for their heaters to free electrons for full conduction.

If your - bias is where it should be when the tubes warm up, it should be fine.

Only concern I might have is when the tubes are already hot and you turn it back on and the - bias takes that long to come up.

Then you could have excessive current flow in the output tubes until they get their proper bias.   

[Paul1453]

Then again, maybe because that 220K resistor comes before the diode on the Plexi's fixed bias circuit.   

[sluckey]

You can get a cheap/small 12v filament transformer from Radio Shack. Connect the 12v secondary to your existing 6.3v filament string. This will give you about 60VAC on the primary that can be used for the AC source for the bias supply, allowing you to use a much smaller (faster TC) resistor.

Or, you can run the amp cathode biased. Tremolo will wiggle the zero volts on the grids of the power tubes just as good as it will wiggle a negative bias voltage. I have a couple amps that do this.

Or, you can just play it like you stole it! 

[eleventeen]

That 220K is current limiting the charge path for the bias supply too much. Cut it by 80-90%, change to 22K to 47K.  20 uf is kind of light filtering for a bias supply, you may need more ufds. Bias uses almost zero current.


 


enough already.

[Paul1453]

Wouldn't reducing that resistor increase the - V ?

I thought HBP explained to me that that resistor is there to knock down the AC before the HW rectifying diode.
And that adjusting the value of that will change your - bias range.

I still need to figure out this part of my quad 6V6 Plexi circuit.

Putting that R before the diode might reduce or eliminate the delay?

You don't like the blinky emo thingys, 11?

I usually just put those in to add some color to the message.  LOL

[John]

If the bias voltage takes longer to "rise", meaning to get to -50 or whatever, doesn't that just mean the tubes aren't conducting current till that happens anyway?

[sluckey]

That 220K is current limiting the charge path for the bias supply too much. Cut it by 80-90%, change to 22K to 47K.  20 uf is kind of light filtering for a bias supply, you may need more ufds. Bias uses almost zero current.

There's plenty of Marshalls out there that use that exact bias circuit. The fact that the bias circuit is such a light load means that you can get by with small caps.

[Paul1453]

If the bias voltage takes longer to "rise", meaning to get to -50 or whatever, doesn't that just mean the tubes aren't conducting current till that happens anyway?

No.

Without the - V on the grid and the cathode connected to Gnd,

the tube will have it's accelerator pedal pushed to the floor.

Cathode bias puts the cathode at a + V as opposed to the grid.

[sluckey]

If the bias voltage takes longer to "rise", meaning to get to -50 or whatever, doesn't that just mean the tubes aren't conducting current till that happens anyway?

No. Just the opposite. The closer the bias voltage is to zero, the harder the tubes burn. 'Course, the tubes still gotta wait for the filament to heat the cathode before any current will flow.

It's an ECL82. "No reason to get excited." The thief, he kindly spoke.

[eleventeen]

The dwg below shows how a Cap charges when voltage is applied through a resistor.
The cap is considered "fully charged" after 5 * RC in seconds but is darn close after 4 RC, with R in ohms, C in farads. Or R in megohms and C in microfarads.


You have 220K * 20 uf, effectively. That is .22 Meg * 20 ufd in "RC units" thus each RC time unit (aka "seconds") is 4.4 seconds and you need at least 4 of them to get close to fully charging your bias caps. This is why your bias is taking so much time to charge no matter how many Marshalls use the circuit.

[Paul1453]

So if we move the resistor to before the diode,

then we have the cap connected with 0 R and the delay goes away, right?

[sluckey]

The dwg below shows how a Cap charges when voltage is applied through a resistor.
The cap is considered "fully charged" after 5 * RC in seconds but is darn close after 4 RC, with R in ohms, C in farads. Or R in megohms and C in microfarads.

You have 220K * 20 uf, effectively. That is .22 Meg * 20 ufd in "RC units" thus each RC time unit (aka "seconds") is 4.4 seconds and you need at least 4 of them to get close to fully charging your bias caps. This is why your bias is taking so much time to charge no matter how many Marshalls use the circuit.

I totally agree. I think the OP does too.

[eleventeen]

"So if we move the resistor to before the diode,then we have the cap connected with 0 R and the delay goes away, right?"

Not really, the diode does not much affect the RC time constant. It is the SIZE (value) of the resistor that the cap charging current has to flow through. Whether before or after the diode, every charging electron has to flow through that 220K (in this case) current-limiting, charge-limiting resistor. The delay would "go away" if we eliminated the R completely.  Not completely, of course, caps require some time to charge, but for all real world purposes, losing the Resistor would take the delay down to well under a second.


Since the ultimate aim of the bias rectifier is to produce "in the neighborhood" of 50 volts (forget the polarity) and a half wave rectifier off half a 300 volt winding should produce .7 * 300 = 210 volts which we then have to whack down to 50, we would like to be able to use 100 volt caps instead of 250 or 300 volt filter caps. So, by inserting the resistor either in front of the diode OR immediately after the diode, hardly matters, we can chop the voltage that ends up on the caps AND perform a bit of current limiting just to be nice to our parts. So somewhere between "zero delay" and "too much delay" is a 1-2-3 second time constant. I tend to use about 15K in that spot, but I *also* like to be able to develop enough (negative) bias volts so that I can completely or almost completely shut off the output. IMO it's handy for troubleshooting the preamp section if you have a hum problem which can be time consuming, you don't have to cook your output tubes while you chase it down. I like to be able to get as much as (negative) 65-75 volts on the grids and it also allows for component drift. The first time one applies power to a new build is what/when? Test the heaters, see if they light up? Check unloaded B+ or, better, throw a resistor calculated to pull say 40-50 mils across the main power supply just to be sure it's working and so you don't blow the main filter caps with too-high volts? Check bias? You bet. I like the idea of knowing that I have way too high negative volts (yes, sounds odd) to shut the output section hard off, just in case something fails 20 seconds after I turn it all on for the first time. The 22K I suggested would yield .4 seconds * 4 or 5 which is 2-3 seconds which should be fine.


Now we get to the voltage divider network, everything after the last cap in the chain. All you want to do here is to 1: provide enough adjustment range and 2: make it so that you CAN'T dial the bias so low that you redplate the output tubes. I don't know the proper bias volts for the given tubes, but for 6L6 where we want about 45-52 volts, I like to be prevented from going under say 30-35 volts. That's governed by the R UNDER the adjust pot. 

[MoparWade]

"No reason to get excited." The thief, he kindly spoke.

Probably my favorite lyric ever.

[Paul1453]

If we move the resistor to the other side of the diode,

then the 1st filter cap is connected directly to it's charging source, no resistance.

As you have explained, the second filter cap is still connected thru a resistor to it's charging source, so some delay remains.

But moving that resistor should greatly reduce the time it takes for the bias voltage to ramp up.  No?

[2deaf]

If we move the resistor to the other side of the diode,

then the 1st filter cap is connected directly to it's charging source, no resistance.

In an RC series circuit, it makes no difference what order the components are in as far as charging and discharging goes.  The resistor could be between the + side of the bias capacitor and ground and the time constant would still be the same.  Moving the resistor to the other side of the diode doesn't change the fact that it is in series with the capacitor.

[Paul1453]

I don't think so.

If the resistor is on the other side of the diode it is on the AC side.

The pulsing DC coming from the diode is connected directly to the e-cap.

No resistance remains on the DC side of the diode.   

[2deaf]

Very well.

[eleventeen]

Go back to the drawing, the schematic. The power source is a battery, a pure DC source that is on 100% of the time. The diode is a check valve which prevents any anything from happening when the diode is reverse biased. The rectifier circuit is thus a battery if and only if the diode is forward biased. So you have your battery but you have it only 50% of the time. Actually, 49.9975% of the time, and even less as charging goes on because the positive going part of the very first cycle of AC delivered to the diode anode can only forward bias the diode once it is over .7 volts positive relative to the cathode (bar end) and the cathode is climbing as the cap(s) charge. If the cap is charged to say 40 DC volts, then the diode anode has to reach 40.7 volts to conduct and continue passing current to charge the cap. So on the upswing of cycle #1 you lose .7/300 while the voltage is climbing and the diode again loses forward bias when that voltage falls below being .7 volt positive relative to the cathode. As the cap charges, there is charging activity less and less of the time because the diode is fwd biased less and less of each cycle and we assume that all the cycles are the same duration, thus take the same amount of time. Blah blah blah.


The circuit in effect is cut, chopped, interrupted half or more of the time. "Duty cycle". The circuit would not act like a battery at all, in any way, if the diode was not there, and the cap would act much like a short. Where the charge-limiting resistor is located in the chain doesn't matter one bit, just like where the switch is located doesn't matter. It could be two resistors, one on either side of the cap. Or three resistors. Their summed total resistance would be the operative number.


There is no circuit when the diode is non conducting. Charging the cap bears the requirement that the power source is DC. It's only DC half the time....or less.


The second cap, yes, is located at a greater resistance away from the point we think of as the DC source, and thus it does charge at very slightly higher delay from the first cap, but it is almost undetectible because when the first cap is not charging during the off part of the diode's conduction, the first cap is charging the second cap.

[Paul1453]

I get that the HW diode is only on 50% of the time.

So in the diagram, we are effectively turning the switch on and off 60 times a second.

With the resistor on the other side of the diode, there still is no resistance between the switch and the cap.

There is a very easy way to prove my theory wrong.   

Move the resistor to the other side of the diode.

You guys are saying that it makes no difference and it will take the same amount of time.

If it takes exactly the same amount of time for the bias voltage to ramp up, I am obviously wrong.

But if the bias voltage now comes up much quicker, I must be right.   

[2deaf]

You can take the 220K resistor out of the charging path and use 350V capacitors and then use a big voltage divider after that to get a real fast charge time.  I've never done that and I wouldn't do that because the time involved is not significant.  The bias from voltage doublers and full-wave bridges with no center tap using that capacitor to get bias voltage take an incredible amount of time to charge apparently with no ill effect.  Kinda like that quote from Mr. Zimmerman.       

[2deaf]

There is a very easy way to prove my theory wrong.

Move the resistor to the other side of the diode.

You guys are saying that it makes no difference and it will take the same amount of time.

If it takes exactly the same amount of time for the bias voltage to ramp up, I am obviously wrong.

But if the bias voltage now comes up much quicker, I must be right.

Try it.

[Paul1453]

This is not my amp we are talking about.

I just cited the difference of where the resistor is placed in this amp,
and where the resistor is on the Plexi circuit I recently built.

In effect I have already tried it, in that I didn't notice this delay in my Plexi's circuit.
I could have just overlooked it, so I flipped over my Plexi and connected my meter's leads.
Hit the power switch and immediately got - V that took less than 2 seconds to reach it's set point.   

That is less than half the time he said it is taking his amp to ramp up to his bias voltage.
Something must be different, and the only thing I see is where that resistor is in the circuit.   

[sluckey]

My Plexi 6V6 uses that exact circuit, except I used 25µF caps rather than 10µF caps. It takes an average of 14.8 seconds for the bias caps to fully charge. I used a Simpson 260 and Galaxy S5 stopwatch. The bias voltage just barely beats my cold GZ34 rectifier! If the GZ34 is still warm, B+ comes up faster than the bias. I think I'll change my caps to 10µF.

[Paul1453]

My Plexi 6V6 uses that exact circuit, except I used 25µF caps rather than 10µF caps. It takes an average of 14.8 seconds for the bias caps to fully charge. I used a Simpson 260 and Galaxy S5 stopwatch. The bias voltage just barely beats my cold GZ34 rectifier! If the GZ34 is still warm, B+ comes up faster than the bias. I think I'll change my caps to 10µF.

I didn't use a stopwatch.  I am using the 10uF caps, and a lower B+ voltage (not that the B+ should matter).  Your bias circuit took way longer than either of ours to charge up, but not as long as Eleventeen's posted formula suggests yours should take.

I don't know what is going on here, with 3 vastly different results, and none of them are exactly the result the formula suggests.
Usually, the formulas in the books can be verified by real world circuit results.   

[mresistor]

Couldn't he also use an off-stdby-on switch for a power switch? I know he says he doesn't want a standby switch, but to me it would be a very simple solution.

[sluckey]

He could do that. I don't think his situation is a problem.

[2deaf]

Exactly when a capacitor is fully charged is too subjective.  You need to time it to something like the 80% point or even 63.2%.  If setting up an experiment to determine the effect of resistor placement, use larger values for the resistor and/or capacitor so that the times are longer and easier to measure.

[c.stoffel]

Thanks everybody, lots of interesting responses.

Well, I tried the 220k resistor before the diode but the charging time of the bias supply didn't change at all. So we can conclude the RC time constant is the same with both arrangements.

I think I'm going to try cathode bias again because the preamp topology changed a lot since I experimented with it and didn't like the tone of the amp that much. That would be the simplest solution. But wouldn't a hot cathode biased push-pull amp be less sensible to bias wiggler tremolo as the cathode current counteracts the effect a bit, making for a less pronounced tremolo effect? Well, I guess I'll try it and see.

A small transformer backwards from the 6.3V sounds like a good idea, if I don't like the results of cathode bias. Food for thought!

The idea of using 350V capacitors on the bias supply is nice too, cause the 220k resistor could be removed and the time constant would drop tremendously, right? I never realized that using lower voltage capacitors on the bias supply was a matter of economics, but it makes sense!

The off-stdby-on switch is a cool idea too, except that I want to keep the front plate intact (just for visuals) and the original power switch is, well, different... take a look at the attachment.

Talking about bias supplies, has anybody seen how it is done on the Vox AC50/4? That's a weird circuit, I'm not sure how it works!

I'm gonna try cathode bias and see if it sounds good!

Thanks!
Christian

[sluckey]

Your bias circuit took way longer than either of ours to charge up, but not as long as Eleventeen's posted formula suggests yours should take.

Well I stopped the timer when the voltage was within 1 volt of fully charged. It takes a long time to juice up the last little bit. It really takes 58 seconds to reach absolute full charge, such that the meter reading ceases to increase. That's pretty close to the 5RC rule. .22 x 25 x 2 x 5 = 55 seconds. Your short charge time is the one I question.

[sluckey]

But wouldn't a hot cathode biased push-pull amp be less sensible to bias wiggler tremolo as the cathode current counteracts the effect a bit, making for a less pronounced tremolo effect?

The cathode current is stabilized by the cathode bypass resistor. Your bias wiggler should work very well with the ECL82s. Here are a couple of my circuits that work very well. The Maggie is ECL86s (similar to your tubes) and the Ampeg is 6V6s (a bit harder to wiggle than the ECL86s but still a very strong tremolo).

     http://sluckeyamps.com/lil_maggie/Magnatone_M2.pdf
     http://sluckeyamps.com/RCA/Ampeg_J12B.pdf

[jjasilli]

You can take the 220K resistor out of the charging path and use 350V capacitors and then use a big voltage divider after that to get a real fast charge time.  I've never done that and I wouldn't do that because the time involved is not significant.  The bias from voltage doublers and full-wave bridges with no center tap using that capacitor to get bias voltage take an incredible amount of time to charge apparently with no ill effect.  Kinda like that quote from Mr. Zimmerman.     

The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path.  Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used.  This saves not only money but lots of space.

[Paul1453]

You can take the 220K resistor out of the charging path and use 350V capacitors and then use a big voltage divider after that to get a real fast charge time.  I've never done that and I wouldn't do that because the time involved is not significant.  The bias from voltage doublers and full-wave bridges with no center tap using that capacitor to get bias voltage take an incredible amount of time to charge apparently with no ill effect.  Kinda like that quote from Mr. Zimmerman.     

The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path.  Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used.  This saves not only money but lots of space.

This is a concise and clear explanation of what I thought I understood.  Knocking it down early and using smaller caps.

I don't mind being wrong about something like this.

It is what techs like me do.  We put out our theory based on what we think we know.
Then we devise a test that clearly either proves or disproves our theory.
If proven wrong, it is back to the drawing board to formulate a different theory and another test for that theory.

What does bother me is not understanding why I was wrong on this.
So far I have not heard a clear and compelling reason as to why my logic was faulty.
Some techs don't bother to try to figure out why their theory was wrong.
This stuff kind of eats at me even after I have fixed the problem.
Why was my theory, which I still think logically makes sense, proven wrong by testing?
Where is the gap in my understanding that this situation is falling through?
How can I fill in that gap?

What makes me happy is that I seem to have the fastest charging bias circuit.
I don't know why, but I do know that it is not taking my circuit near as long as yours to stabilize at it's set point.

[2deaf]

The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path.  Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used.  This saves not only money but lots of space.

The charging path would be from a winding tap, through a diode, across a capacitor, then to another winding tap (or the other way if you like conventional current flow).  The voltage divider is then placed in parallel with the charging path.  This arrangement will charge up so fast that you will have trouble measuring it.  Putting a 220K resistor in the charge path will allow you to use smaller, cheaper capacitors, but the trade-off is going to be time.  I don't think the time involved here is any problem.  How much could a tube possibly red-plate in a couple of seconds?

[jjasilli]

I don't mind being wrong about something like this. . .  What does bother me is not understanding why I was wrong on this.
I'm thinking it doesn't matter where the resistor(s) goes:  E.g. (other circuits with rectification):  1) in a B+ circuit a dropping resistor cold go between CT & ground, instead of in the B+ rail; 2) in an LED circuit, a current limiting  resistor could go before or after the LED; in either position it limits current & drops voltage.

Maybe another reason bias voltage delay is not a problem is that there's a similar delay in the duty cycle of the B+ filter caps???

The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path.  Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used.  This saves not only money but lots of space.

The charging path would be from a winding tap, through a diode, across a capacitor, then to another winding tap (or the other way if you like conventional current flow).  The voltage divider is then placed in parallel with the charging path.  This arrangement will charge up so fast that you will have trouble measuring it.  Putting a 220K resistor in the charge path will allow you to use smaller, cheaper capacitors, but the trade-off is going to be time.  I don't think the time involved here is any problem.  How much could a tube possibly red-plate in a couple of seconds?

I'm not clear on this.  A diagram would be helpful. 
Perhaps the Vox circuit -- posted by Paul, with the stacked caps before the bias diode -- is a variation on this theme.

[Paul1453]

He could do that. I don't think his situation is a problem.

I agree, I don't think this is a problem.

With all our circuits, we seem to have some - V on our grids almost immediately.
With the Cathode tied to Gnd, and at least some - V on the grid,
that should help keep some of the newly awakening electrons from rushing to the Plate.
If our bias is rising in relation to the B+ filter caps also charging and the rising Plate voltage,
then we really don't have a condition where our tubes accelerator is mashed to the floor.

I try to protect my vintage tubes as best as I can, so I do understand his concern when he noticed this.
But without any evidence of the tubes red plating, even for only a second or two,
then I don't think this is a real problem.

[c.stoffel]

Well today I tried cathode bias in this amp again and I really don't like it. I wanted to like it but I can't! I've experimented with it before and now I remember why I changed to fixed bias... The clean tone is nice but the overdriven tone in cathode bias is just... I don't know, there's a fizz on top of the notes and it's really ugly. It gets too bright in a non-musical way, totally uninspiring to play. And I've experimented with:
- different values for cathode resistor (tried around 70% plate dissipation and about 90% too to see if the tone would improve)
- different values for cathode bypass capacitor (10uF, 22uF, 100uF, 220uF, 1000uF)
- even tried using separate cathode resistors/bypass cap for each power tube.

The only situation where it improved a bit was using a common cathode resistor and NO bypass cap. Anybody has a clue on why? Maybe that's just the nature of the ECL82/6BM8? Even when I tried copying some successful power amp topologies like the 56T (6BM8 version) and the Little Wing I've never been able to get those good tones. The tubes I'm using are Winged C 6F3P, are these any different in practice from the other ECL82's?

Using fixed bias is a completely different story. The overdriven tones sound really good, and no sign of that fizziness...
I found out that Laney uses a separate winding on the power transformer to get the bias voltage for some of their amplifiers. It's the same situation as mine: solid state rectification, and no standby switch. The CUB10 and CUB12 use the same power transformer, and searching I found that it has a 225V winding and a 20V tap for bias. Since the 10 uses 6V6s and the 12 uses EL84s they have slightly bias arrangements... maybe a separate bias transformer would be my best bet?

I have no problems messing with this circuit cause it was all rebuilt already, it's just the aesthetics of the amp that I want to keep original. My concern is that I want this amp to be reliable, that's why I'm worried about the bias taking so long to go up.

[2deaf]

I'm not clear on this.  A diagram would be helpful. 

[2deaf]

If the resistor is on the other side of the diode it is on the AC side.

The pulsing DC coming from the diode is connected directly to the e-cap.

Think about the current going through the resistor and the diode.  It is the same.  The resistor and the diode both have the same pulsing DC no matter which side the resistor is on.   There is no AC on either side of the diode.  It may look like there is if you hook your 'scope to it, but the probe makes a circuit with AC that isn't there otherwise.

[Paul1453]

The Cub 10 has a FW rectifier feeding the bias circuit.

If you are still concerned about this bias voltage delay,
maybe making yours FW would reduce the caps charge time?

What do you guys think, time reduced or not?

[jjasilli]

I'm not clear on this.  A diagram would be helpful. 

But according to eleventeen, the charging time should remain the same.  Que passa?

[2deaf]

What do you guys think, time reduced or not?

I doubt it.  There is no getting around the RC constant.

If you make a full-wave bias circuit by putting a 220K resistor and a diode on the other end of the winding, the DC voltage will be significantly more negative than with a half-wave circuit.  The full-wave version will have significantly less ripple than half-wave. 

How about if we use 47K resistors between the windings and the diodes and then use a lower resistance from bias to ground to get the desired range?  The circuit will draw more current, but still negligible.  The 47K's will allow the capacitors to charge up much faster.  The lower resistance to ground will increase ripple, but the lower ripple with full-wave will counter that. 

[2deaf]

But according to eleventeen, the charging time should remain the same.  Que passa?

Orale vato, where did he say that?

[Paul1453]

What do you guys think, time reduced or not?

I doubt it.  There is no getting around the RC constant.

If you make a full-wave bias circuit by putting a 220K resistor and a diode on the other end of the winding, the DC voltage will be significantly more negative than with a half-wave circuit.  The full-wave version will have significantly less ripple than half-wave. 

Why would we get significantly more negative voltage?

The way I look at it, we would just not be flipping the on off switch 60 times a second.  The second diode is going to give us the same voltage pulsing DC output as the first one gave us.  It will only fill in those blank spots from the 1st diode by giving us another pulse when the 1st one is shut off.

The way HBP had explained this circuit to me before, reducing that resistor is what would give us a significantly more negative voltage.
That seems to make sense to me.   

[Willabe]

Why would we get significantly more negative voltage?

The way I look at it, we would just not be flipping the on off switch 60 times a second.  The second diode is going to give us the same voltage pulsing DC output as the first one gave us.  It will only fill in those blank spots from the 1st diode by giving us another pulse when the 1st one is shut off.

Yes and that's why. The cap charging refresh rate is now double. As it fills in those "blank spots" the caps do not discharge as much.

So it keeps the cap(s) charged up higher.

[Paul1453]

Yes, but they can't charge up more than the voltage level we are supplying them.   

We don't get double the voltage from a FW rectifier than we do from a HW.

The way I see it, the caps will just charge up twice as fast. 

[Willabe]

The way I see it, the caps will just charge up twice as fast.

Right, but your not seeing the peak charge from the rectified dcv, your seeing the average dcv. The caps hold the charge up to an average which is what we see when we take a B+ reading.

The longer the time between the cap getting charged the more the cap has to release it's charge to hold up the B+ average dcv.

So the cap discharges. As this happens, over and over again, the result is a lower dcv average. It can't keep up the dcv, so the B+ dcv average drops.

But double the charge feed to the cap and there's less discharge of the cap between charges, so the dcv average stays higher with less ripple.

[eleventeen]

"We don't get double the voltage from a FW rectifier than we do from a HW.


It depends whether you are talking about a FW *bridge* or a "two diode" FW rectifier like the type used on the main B+ supply like a Fender Twin, where the two HV legs are fed to two diode anodes (or two series strings of 3 diodes each) and the cathodes (bar ends) are tied together. BOTH types are called "FW". The type using the 4 diodes in the (as usually but not always drawn) diamond is called a "FW bridge".


But the "2 diode" rectifier configuration requires a CT to work. Now I know you are going to say "but the tranny DOES have a CT"....well, yes it does but for the bias supply, the way we are stealing a single HV lead through a single diode is as a half wave rectifier and we are making that circuit between the CT which is grounded, using only half the HV winding, and completely independent of how we are using BOTH halves of the HV winding elsewhere in the amp. So, looking only at that half of the HV winding, across that half-winding, we DO NOT have a CT. Thus we/you can not thinking we have a "2 diode" FW rectifier and thus, if you say "FW" you MUST be referring to a bridge rectifier, and sure enough, a bridge WILL put out double the volts of a half wave. But the final conundrum is that you can not use a FW bridge for this purpose because the "ground" of the bridge rectifier would fight with the "ground" of the other rectifier. Half of the output of a bridge that you tried to use for this purpose would be shorted out. You can't use EITHER type of FW rectifier for the bias without resolving the ground conflict when you are stealing one HV leg to feed the rectifier. You HAVE TO use the half wave.




"The way I see it, the caps will just charge up twice as fast. "

The way the electrons see it, they won't.

[Paul1453]

So we can't just hook up another - diode with the center tap connected to ground, I get that.

If we disconnect the HV CT, and then connect the - diode, then we would have a FW bridge with the + going to B+ and the - going to the bias.
Equal and opposite now as referenced to Gnd.

We get double the voltage from the FW bridge by using the - as the GND reference.

The only problem I see with using a FW bridge by disconnecting the CT, is the path the current will now be force to take to GND.
I'm not sure if that would cause all our B+ current to flow back through bias circuit now or not.