Fixed bias without standby switch

[eleventeen]

It would disable the main rectifier. 6L6's will make no output with -50 on their grids and zero on their plates. That is a perfect certainty. But we could have a lengthy discussion about it, I guess. 

[Willabe]

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

Yes we can. We put it on the other hi voltage wind secondary leg.

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.

No FWB has 4 diodes.

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

No that's not why.

[Paul1453]

6L6s 

This was the circuit he posted about, ECL82 output with PS shown.

Maybe we could discuss this circuit?

[Willabe]

6L6s 

This was the circuit he posted about, ECL82 output with PS shown.

Doesn't matter what power tube, disconnect the CT no electrons can be brought into the PT secondary wind for current to flow out. 

Just like disconnecting a tubes K to ground, no output electrons because there's no input electrons. 

[HotBluePlates]

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

I think I'll blissfully stay out of this one. But regarding "RC" the "R" is anything/everything in the path between the "C's" two leads.

So winding -> diode -> cap minus -> cap plus -> R -> ground -> winding will be the same effect on time-to-charge as the stock arrangement (which is what I think 2deaf was saying much earlier).

This was the circuit he posted about, ECL82 output with PS shown.

Maybe we could discuss this circuit?

Your drawing, in Reply #2 is functionally-same as c.stoffel's drawing in the original post. If it charges too slowly for his liking, it is because of the 220kΩ resistor (the "R" in "RC"). Sluckey suggested getting an independent winding for bias via an added transformer, of lower voltage, so the "R" could be smaller and speed up the charging, while still maintaining the same voltage output for the overall circuit.

Remember, the definition of "ampere" is a rate of electron flow (1 coulomb past a point per second equals 1 ampere). Resistance slows current, so for same voltage you get less A's. If current is slowed by big-R, then cap charging (which is dumping a bunch of electrons on one of the cap's plates) also slows.

[Paul1453]

6L6s 

This was the circuit he posted about, ECL82 output with PS shown.

Doesn't matter what power tube, disconnect the CT no electrons can be brought into the PT secondary wind for current to flow out.

Really Willabe?

I may not be an electrical engineer or a tube amp designer, but I'm not an idiot.

I was a damn good repair tech, back in the day.
I'm still quite rusty, but things are starting to come back to me.

You've come into this discussion and posted a few things my Army training tells me are just plain wrong.
I haven't attacked you about it, but come on, give me a break.
I'm just trying to have a civilized discussion about these concepts to stimulate my memory.

[Paul1453]

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

I think I'll blissfully stay out of this one. But regarding "RC" the "R" is anything/everything in the path between the "C's" two leads.

So winding -> diode -> cap minus -> cap plus -> R -> ground -> winding will be the same effect on time-to-charge as the stock arrangement (which is what I think 2deaf was saying much earlier).

I had been hoping you would have weighed in here by now.

You seem to have a way of distilling complex subjects down to something I can understand.

[Paul1453]

I also think Sluckey's idea about adding another transformer is probably the easiest and best solution. 

[HotBluePlates]

I modified my earlier post to add some extra information.

If you have a specific question, I'll try to help. I think everyone else was giving good, correct answers, but perhaps not in a way that turned the lightbulb on for you.

[Paul1453]

I recall in the 6V6 Plexi thread you had indicated that adjusting that resistor would change the voltage level supplied to the bias circuit.

I'm still trying to figure this circuit out completely,
so I can make a reasonable guess as to what resistor I need to put in to my 120 VAC tap to supply my bias board with 25-30 VAC.

[HotBluePlates]

... I'm still trying to figure this circuit out completely, so I can make a reasonable guess as to what resistor I need to put in to my 120 VAC tap to supply my bias board with 25-30 VAC.

I don't know. I'm sure there's a formula out there somewhere, but I don't know what it is. And that's because while you'd think "Use Ohm's Law," the diode makes the process non-linear and current pulses non-continuous, which makes the math more involved. So what I typically do is guess way higher than I think it should take, and trim the resistance down on test until I get the voltage I want.

I do notice your board specifies an input of 65v. I'd probably be looking for a small transformer which provides that 65v or less to feed the board.

[2deaf]

Why would we get significantly more negative voltage?

If the HW and FW versions were both charging a purely capacitive load (no resistor to ground), they would both charge up to the same DC voltage which would be slightly less than the peak AC voltage driving the circuit.  If you introduce a resistive load in addition to the capacitive load (a resistor to ground), the FW version will have a higher DC voltage than the HW version.  If you increase the resistive load (decrease the value of the resistor to ground), the ratio of HW DC voltage to FW DC voltage decreases.

[eleventeen]

There are 3 distinct issues with the ckt you just posted. One is that the input 220K is largish and thus it slows down charging the bias filter caps to their intended steady-state value because the RC time constant of the 220K > 20 uf (effective) is over 4 seconds and you need about 4 of those periods to get the caps near their final desired voltage. To solve this, lower the 220K to some 5-digit value, like 15K, 22K, 33K, 47K.


The second issue is that the way the "adjust" pot is connected to the output of the bias supply, adjusting that pot changes the apparent load on the entire supply and possibly changes the bias supply output. This (meaning, the undesirable effect of this) is exacerbated by the over-large 220K feed resistor. We don't want that. We want the bias supply to see a fixed load and thus (as much as possible for an unreg supply) to provide a fixed voltage. That load should be a stack of three resistors. In idealized form, the middle R is a pot and we take the bias from the wiper. As we adjust the wiper, the supply sees the constant load of the 3 stacked resistors and thus its output does not change. For bias, we are taking only the most miniscule current, we just sticking a charged electrode into the tube. We just "pick off" a voltage of our choice over the range of that pot, but unlike the dwg you show, we do not change the overall load the supply sees. The lowest resistor is the "preventer" that keeps you from being able to adjust the bias so low that the output tubes overcurrent and redplate. That's the third issue. The adjust pot (as wired) changes the load on the whole supply and yes, it will affect the bias delivered to the 470K/470K junction but in a goofy way.




Now you will see bias supplies with the wiper of the adjust pot connected to one end of the adjust pot, making a "rheostat". Fine. Two things about that. One is, that should the wiper fail which is a thing that happens once in a great while, the tubes would lose their neg bias and go right away to overcurrent. So if we connect that pot as a rheo, then the worst that can happen with such a failure is that the bias voltage snaps to the voltage available at one end of that "pot". That means, we would prefer to connect the wiper to the more negative end of the pot so that if it fails, it fails to under versus over current.


And finally, to do the rheo, there needs to be the "preventer" resistor underneath it.


As an extra bonus, a possible 4th problem is that the twin 470K resistors feeding the output tube grids are probably too large. On most amps, those are 220Ks. This may not matter.

[2deaf]

Boy, I had this drawing all ready to post and then I read eleventeen's post.  I think I did some stuff that he said not to do.

If you are going to take the bias from the wiper, you should connect a large resistor from the wiper to the more negative side of the pot.  That way, if the wiper fails open, the tube bias reverts to the most negative voltage available from the supply.

I tied the wiper to the more negative side of the pot.  The change in the load is only 5K, so I'm not losing any sleep.  Should the wiper fail open, the bias reverts to the most negative voltage available from the supply.

This circuit will probably use ten times as much current as the one in the OP.  Still only a few ma's and no sleep deprivation.  If we went down to 15K, I might get a little concerned about the current consumption. 

The ripple should be very small due to the full-wave rectification. 

The actual values might need to be adjusted a little if a circuit like this was ever used, but they should be pretty close.

This circuit will charge up lickety-split.

[2deaf]

There is a 220K resistor between the bias supply and the 470k resistors in the OP drawing.  That needs to go away so that the audio signal coming from the PI will have a direct path to ground at the junction of the 470k's. 

[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.

I changed the caps to 10µF. Now the voltage ramps up to -30v in only 4.25 seconds. Still takes about 15 seconds to reach full charge of my operating bias which is -36v. I'm not nearly as excited now! 

[c.stoffel]

There are 3 distinct issues with the ckt you just posted. One is that the input 220K is largish and thus it slows down charging the bias filter caps to their intended steady-state value because the RC time constant of the 220K > 20 uf (effective) is over 4 seconds and you need about 4 of those periods to get the caps near their final desired voltage. To solve this, lower the 220K to some 5-digit value, like 15K, 22K, 33K, 47K.

The second issue is that the way the "adjust" pot is connected to the output of the bias supply, adjusting that pot changes the apparent load on the entire supply and possibly changes the bias supply output. This (meaning, the undesirable effect of this) is exacerbated by the over-large 220K feed resistor. We don't want that. We want the bias supply to see a fixed load and thus (as much as possible for an unreg supply) to provide a fixed voltage. That load should be a stack of three resistors. In idealized form, the middle R is a pot and we take the bias from the wiper. As we adjust the wiper, the supply sees the constant load of the 3 stacked resistors and thus its output does not change. For bias, we are taking only the most miniscule current, we just sticking a charged electrode into the tube. We just "pick off" a voltage of our choice over the range of that pot, but unlike the dwg you show, we do not change the overall load the supply sees. The lowest resistor is the "preventer" that keeps you from being able to adjust the bias so low that the output tubes overcurrent and redplate. That's the third issue. The adjust pot (as wired) changes the load on the whole supply and yes, it will affect the bias delivered to the 470K/470K junction but in a goofy way.

Now you will see bias supplies with the wiper of the adjust pot connected to one end of the adjust pot, making a "rheostat". Fine. Two things about that. One is, that should the wiper fail which is a thing that happens once in a great while, the tubes would lose their neg bias and go right away to overcurrent. So if we connect that pot as a rheo, then the worst that can happen with such a failure is that the bias voltage snaps to the voltage available at one end of that "pot". That means, we would prefer to connect the wiper to the more negative end of the pot so that if it fails, it fails to under versus over current.

And finally, to do the rheo, there needs to be the "preventer" resistor underneath it.

As an extra bonus, a possible 4th problem is that the twin 470K resistors feeding the output tube grids are probably too large. On most amps, those are 220Ks. This may not matter.

Cool, looks like this would make a better and safer bias supply. And I think you're correct about the 470K resistors feeding the output tube grids. The ECL82 datasheet specifies a max. value of 2M for cathode bias and 1M for fixed bias, and the 6F3P (which I'm using now) specifies 1M for cathode bias and 500K for fixed bias. I think I'm gonna change them to 220K and double the coupling caps.

There is a 220K resistor between the bias supply and the 470k resistors in the OP drawing.  That needs to go away so that the audio signal coming from the PI will have a direct path to ground at the junction of the 470k's. 

Well, this amp doesn't have a depth control for the tremolo, so that 220K resistor is there to act like a 'fixed' depth control, otherwise as I see it the tremolo couldn't be applied to that point, am I wrong? Take a look at the Fender 5G9 and the Gibson GA-5T (this one also doesn't have a depth control). By the way, the value of this resistor and the others on the bias supply should be considered as part of the resistance from the output tube grids to ground, right?

[HotBluePlates]

... Well, this amp doesn't have a depth control for the tremolo, so that 220K resistor is there to act like a 'fixed' depth control, otherwise as I see it the tremolo couldn't be applied to that point, am I wrong? ...

You could look at it that way if you're applying tremolo.

The bias circuit will charge just as fast with/without it. That resistor (and the bias feed resistors you're talking about dropping from 470kΩ to 220kΩ), as well as the coupling caps feeding the output tube grids, will determine how fast the output tubes recover from an overload input signal. Too slow a response leads to farty blocking distortion. Too small looks like a heavy load on the phase inverter outputs, making it hard for the phase inverter to deliver a clean output signal of the required size.

p2pamps has a tagline: "Everything affects everything." This is an example; don't try to design "the best ever," but instead use existing amps as a guide for what might be "good enough" in terms of all the tradeoffs, and see if the amp needs adjustments after it's built.

[2deaf]

Well, this amp doesn't have a depth control for the tremolo, so that 220K resistor is there to act like a 'fixed' depth control, otherwise as I see it the tremolo couldn't be applied to that point, am I wrong? Take a look at the Fender 5G9 and the Gibson GA-5T (this one also doesn't have a depth control). By the way, the value of this resistor and the others on the bias supply should be considered as part of the resistance from the output tube grids to ground, right?

I went back and read your original post with the note about Point "A".  Disregard my comments on the 220K resistor. 

[Paul1453]

I appreciate you all having patience with me and my hard head sometimes.

I need to read over most of your comments many times for it to sink in for me.

Your explanations help me fill in my big gaps I am missing after a long dormant period.   

RC charging, ahhh can you say series circuit, Paul?
Sometimes I go off in the weeds, and can't see the forest from the trees.
KISS, Paul, KISS.  Keep It Simple Stupid.

I'm really trying to get a strong grasp on all these tube amp PS and bias circuit fundamentals.
Because as was just said "Everything affects Everything" and this is the foundation for it all.
I used to think that e-caps just filtered DC and that's it.
Now, I'm coming to realize that the uF of filtering can influence and tune (sort of) the response of the tubes performance.
"Everything affects Everything", KISS, Paul, KISS.
There is probably damn good reason they used certain cap and resistor configurations.

I'm really getting a little anxious, as I'm about to finalize the modifications to the Plexi circuit to accommodate my quad 6V6s.
And start wiring things in place.  I don't want to wreck my vintage stuff with one of my stupid mistakes.
I want to use circuit designs that protect my components while helping them perform at their best.
And that is not as easy as it sounds because, "Everything affects Everything", KISS, Paul, KISS.   

[c.stoffel]

I went back and read your original post with the note about Point "A".  Disregard my comments on the 220K resistor. 

That's alright! I'm learning a lot from this discussion.

Another amp I found that uses fixed bias, SS rectification and no standby switch is the Fender Pro Junior. It takes the bias voltage from the HV winding but with a capacitively coupled bias power supply. Is this kind of circuit faster to charge than the one I've been using with a large resistor from HV? Or maybe Fender just don't care if the bias takes a little to go up?

[eleventeen]

Yes, it will charge faster, practically immediately, because there is no series current limiting resistor that ALSO, as it limits current, creates an RC time constant as we've discussed. That .047 cap is not really necessary.


If at any time you wish to seriously increase any remaining bafflement you may have on this topic...go look at a Bogen power supply, where the main B+ comes from a half-wave voltage doubler.


http://el34world.com/charts/Schematics/files/bogen/bogen_chb100.pdf


Once again, we are stealing the bias from one of the HV lines that creates the B+ for the amp. In THIS configuration the "blocking" cap that feeds the "backwards" bias rectifier diode ABSOLUTELY MUST be present. And it generally has to be a FAT capacitor, like a .22.


Why do we need that cap? Because at the junction of the stacked diodes that make up the doubler, we have all kinds of pulsating DC flying around, and that would get into the poor, innocent bias supply and ruin its day.

[c.stoffel]

Still talking about my amp, does anybody have an idea on why I get such ugly tones I mentioned earlier when cranking the volume using cathode bias (shared or single resistor), but it gets better when I remove the cathode bypass cap(s) and using fixed bias it changes completely with no sign of that nasty tones? And it happens using exactly the same circuit, just changing the bias method.

[Paul1453]

From my rudimentary understanding, the bypass cap increases gain in the tube.

You may be getting too much gain when cranked up that is distorting your signal in a bad way.

Fixed bias doesn't have any bypass cap, as far as I know, so that even when you go cathode biased but no bypass cap the tubes response is similar?   

[2deaf]

The Pro Jr. bias is the old Ampeg thing that Marshall also used in my 900 and Bogen used a variation of.  The capacitor and resistor before the diode forms an AC circuit with a real ugly signal.  The diode then rectifies this AC into negative DC.  These things typically take a lot of time to charge up.

Not only is C12 necessary, but its value is critical.  If you change it from .047uf to .022uf, the absolute value of the (AVOT) bias voltage will be dramatically reduced.  It you increase C12 to 0.22uf, the AVOT bias voltage will be increased.

The value of R28 is also critical.  Values significantly greater than or less than 56K will both reduce the AVOT bias voltage.

Increasing the value of R29 will increase the AVOT bias voltage and decreasing the value of R29 will decrease the AVOT bias voltage.

Unlike many things in guitar amps, these bias circuits have components that must be the value shown and in the positions shown or the circuit won't work.  It is an extreme case of everything affecting everything.

The Pro Jr. is a little odd in that it doesn't have a load resistor after the second filter capacitor so that C13 and C14 will charge up to the same voltage.  R30 forms a series RC circuit which will slow the charge rate.  For some reason I seem to feel that the whole thing is rather wimpy and that contributes to the long charge time.

[2deaf]

The voltage doubler in the Bogen CHB100 is full-wave at B+ with respect to ground.  It is half-wave at the junction of the two 40uf capacitors.  The bias circuit is the same idea as the Pro Jr. one.  This time a voltage divider is used in the AC portion, but otherwise it is the same.

If you try to take the bias from the other end of the winding, it won't work because there is a straight-wire path in parallel with the bias circuit that is much more appealing to electrons.  With a FWB without a center-tap, you can take the bias from either leg. 

[Paul1453]

As an Army repair tech we were trained to try to break circuits down into their smallest logical units for troubleshooting.

So for power supplies we would tend to say we have an AC module and a DC module.
AC was pretty much everything up to the rectifier, DC was everything after.
I've noticed this is causing me to miss the bigger picture sometimes.
We were never allowed to modify or improve any circuits design.
So that method was helpful in quickly isolating and identifying faulty components.

But in an "Everything affects Everything" environment I seem to have blinders on now.
I'm hoping I can widen my view now to see how the PS works as a whole unit.

Thank you for the detailed explanations that illuminate the bigger picture for me.
Even if I seem to still be in the dark.
I will reread your detailed explanations until I finally see the light.

Best regards,

[jjasilli]

Not only is C12 necessary, but its value is critical.  If you change it from .047uf to .022uf, the absolute value of the (AVOT) bias voltage will be dramatically reduced.  It you increase C12 to 0.22uf, the AVOT bias voltage will be increased.


Yes, I'm thinking that the cap input circuit  [cap > shunt resistor > diode] is serving double duty as both a voltage divider & a filter, though I don't have my head completely around this yet.  See:  http://forum.allaboutcircuits.com/threads/resistor-role-in-a-high-pass-filter.47449/      &     https://en.wikipedia.org/wiki/Voltage_divider

[sluckey]

Look at page 6 of this pdf...

     http://sluckeyamps.com/misc/Amp_Scrapbook.pdf

[HotBluePlates]

Not only is C12 necessary, but its value is critical.  If you change it from .047uf to .022uf, the absolute value of the (AVOT) bias voltage will be dramatically reduced.  It you increase C12 to 0.22uf, the AVOT bias voltage will be increased.

Yes, I'm thinking that the cap input circuit  [cap > shunt resistor > diode] is serving double duty as both a voltage divider ...

Concur with 2deaf & Jjasilli. The cap and the resistor-to-ground which follow it form a voltage divider for a.c. The cap's portion of that divider is its capacitive reactance (in ohms) at the frequency of the pulsating d.c. applied from the PT/bridge. To recall an earlier discussion, this voltage division can only happen if a.c. is applied to the cap/resistor, because the diode is after them.

Sluckey's info also shows how the baseline is shifted to allow an apparent negative voltage to be rectified by the diode.

[jjasilli]

Look at page 6 of this pdf...

     http://sluckeyamps.com/misc/Amp_Scrapbook.pdf

Very cool!  A picture is worth 1000 words 

[shooter]

Look at page 6 of this pdf

Sluckey, you need to post when *updates* are available
(doc's updated, thx)

[Paul1453]

Look at page 6 of this pdf

Sluckey, you need to post when *updates* are available
(doc's updated, thx)

Yes, thank you for that resource Sluckey!   

OK, just one more question for the slow learners like me.

We try to knock the voltage down to use lower voltage e-caps, saving money and space.

I already had cheap 10uF 400V ecaps from China I used in my bias circuit.

The formula did not indicate that this would make any difference.

Does using 400V ecaps where 100V are indicated have anything to do with charging time?   

[shooter]

have anything to do with charging time

volts determine when it explodes, farads determine how long it takes to charge

[jjasilli]

Caps are "analog devices". They have to be physically larger to handle more of something, be it Volts or Farads.  That is, they need more of the material they are made from; and hence must be bigger in physical size.


Caps are made of different materials a such as ceramic, silver mica, electrolytic, poly, etc.  This affects physical size from type to type.  E.g., a ceramic cap may be physically smaller than a poly cap of the same value. 

[Paul1453]

OK, so I didn't think the 400V would make a difference.

Thank you for confirming that.

I didn't accurately time my charging, and I probably just considered them fully charged when they got to that 80+% and significantly slowed their charge rate.

Knowing that this amount of - grid V would prevent the tubes from red plating.

I wasn't concerned that my vintage output tubes would be damaged.

In the future if I'm concerned about protecting fixed bias vintage output tubes,
I will look to have a separate tap that has the least amount of R in the charging of the bias caps as possible.

Thank you all for helping me understand these concepts better.   

[EKDENTON]

I don't get the original point or question. If billions of amps are already working great with the bias supply circuits that have already been designed in the past, it would seem much more logical not to re-invent the wheel and just use what has always been used. If i blow a tube a little sooner than its expected life span I just order a new one, its not a problem they are not that expensive. LOL

[jjasilli]

I think a quest for a deeper understanding of circuits is driving this thread.  It is disconcerting to come to the realization that bias voltage is not instantaneous.  It does not appear that anyone is trying to reinvent the bias circuit. 

[EKDENTON]

I am interested in understanding this. Given this info that the bias capacitors take some time to charge ...would it make any difference in an amp  tht has a standby switch? The bias caps would be charged and tubes heated, before you hit the standby.






And for those who prefer not to have a standby switch,  then it is better to have smaller capacitors with a lower voltage rating that charge faster?

[jjasilli]

Yes.  And from a purely practical perspective I agree with both your posts.  When all is said & done, the B+ supply is also delayed because of the duty cycle of its R-C networks, and filaments take time to heat up.  So bias delay is not fatal.  It's easy enough to plagiarize an existing bias circuit; but it's also good to have a deeper understanding. 

[Merlin]

The Cub 10 has a FW rectifier feeding the bias circuit.
What do you guys think, time reduced or not?

Yes, connect another diode from the other leg of the transformer winding and the rise time will increase, but the final voltage will remain the same. Quick, simple, done.

You can also connect a diode between the grid and cathode of each power tube to reduce how hard they conduct as the coupling caps charge, before the bias supply chokes everything off.

[2deaf]

The Cub 10 has a FW rectifier feeding the bias circuit.
What do you guys think, time reduced or not?

Yes, connect another diode from the other leg of the transformer winding and the rise time will increase, but the final voltage will remain the same.

The only time that a HW and FW bias supply will result in the final voltage remaining the same is when there is a capacitive load, but no resistive load.  The bias supply being discussed here has a resistive load, so the FW version will result in a more negative voltage than the HW version.  This is not just my opinion.  A quick, simple experiment will verify my statements.

The time it takes to charge the capacitive load to 63.2% of the maximum voltage is going to be real similar if not the same whether you use FW or HW.  Don't take my word for (nobody was going to, anyway), set up an experiment.  Like I said before, use larger values to increase the time so that it is easier to measure.

[eleventeen]

In what will undoubtedly not the be the final word on the topic of whether a cap is required between the HV feed and the input to the HW rectifier supplying bias, here is a page from the 1963 RCA receiving tube manual showing a 7027-fired push-pull amplifier with the cathodes of the 7027s grounded and a perfectly straight-ahead bias supply fed via NO CAP to the cathode of the bias-rectifying diode. I have not experimented with whether that diode can be fed DIRECTLY from the HV feed, it is entirely possible that might not work the same. But the drawing clearly shows that only a voltage-dropping resistor is needed. I myself always place a resistor in this position for the reasons I have stated, it allows lower working voltage caps to be used in the bias supply and provides current limiting into the bias supply.

[sluckey]

a perfectly straight-ahead bias supply fed via NO CAP to the cathode of the bias-rectifying diode.

The AC voltage feeding that bias rectifier has a zero volt baseline. IOW, the sinewave swings positive and negative. There is no need for a coupling cap to shift the baseline in this case.

[HotBluePlates]

To say what Sluckey said, in a different way: The circuit just posted has a full-wave rectifier (which a grounded center-tap on the high-voltage winding). The previous circuit which lead to discussion of the cap feeding the bias circuit uses a full-wave bridge rectifier, with no center-tap for the high-voltage winding.