'59 Bassman Bias Supply Math......

[Jack_Hester]

I know that it's a simple circuit that's found in any number of amp designs.  I started thinking about it while troubleshooting the bias supply in my Oliver PA-100X.  So, I started working the math to see how a 50vAC supply at the input becomes -48vDC on the output. 

The voltage divider, made up of the 15K and 56K resistors, is easy enough to work.  But what stumps me is how the 50vAC becomes the voltage that arrives at the junction of the diode and the first 8uf capacitor.  Working backwards from the -48vDC, I get right at -61vDC. 

But, I can't seem to figure out how that works back to 50vAC.  My thinking is that one diode should give me a half-wave 1/2 of the peak, or .707 x 50vAC = -35.35vDC.  I just can't seem to figure out how this voltage almost doubles, by the time it reaches the voltage divider.  I'm hoping someone can shed some light on this for me. 

Now, let me share what I did do.  I built this on my breadboard, and recorded the readings.  Here's what followed:

The components used for the test on the Bassman Bias Supply:
 
 2 - 8uf-150v capacitors
 1 - 15K resistor - Carbon Comp - Measured @ 16.71K (I call this R1)
 1 - 56K resistor - Carbon Comp - Measured @ 60.11K (I call this R2)
 1 - 1N4007 diode
 
 Circuit was wired according to the Fender '59 Bassman schematic, and AC voltage was applied (by auto-transformer) until 50.03v (Vin) was reached.
 
 DC voltage was measured at the junction of the diode and the first 8uf capacitor, and found to be -65.4v (I call this V2).
 
 DC voltage was measured at the junction of the 15K (16.71K), 56K (60.11K), and the second 8uf capacitor, and found to be -51.13v (Vout). 
 
 I calculated in reverse, through the 15K/56K voltage divider and got -65.34vDC.  That is, using the actual resistor values. 
 
 I am more than satisfied with these results.  Confirms that the AC supply to this circuit is 50vAC.

My question that I hope someone here can answer is, how to do the math behind the 1.306 multiplier that steps the voltage up to that entering the voltage divider. 

Thanks for any input.  Have a good one.

Jack

[sluckey]

My thinking is that one diode should give me a half-wave 1/2 of the peak, or .707 x 50vAC = -35.35vDC.  I just can't seem to figure out how this voltage almost doubles, by the time it reaches the voltage divider.

You used the wrong multiplication factor converting RMS to peak. Peak = 1.414 x RMS, so in your case 1.414 x 50 = 70.7. This is always true regardless of whether you use a half wave rectifier or a full wave rectifier.

Your half wave rectifier passes a 70.7v negative voltage half cycle to charge the filter cap. Then it blocks the next half cycle. So you end up with a 70v pulse, skip a pulse, 70v pulse, etc. (A full wave rectifier passes every half cycle, so there are no skipped pulses. Still only 70.7v peak). Bottom line, all the pulses applied to the filter cap will be 70.7v regardless of which type rectifier you use. And if there is no discharge path for the cap, ie, no load, the cap will charge to 70.7v and hold it, again, regardless of which type rectifier you use.

Now, whenever you put a load on the filter cap, the cap will discharge some also, so the 70.7v peak may never actually be reached. But even if it does, the cap will be discharged some before the next pulse can come along and pump it back up. A full wave recto passes a pulse every half cycle, so the cap will be pumped up twice as often as a half wave recto. IOW, with a half wave recto, the cap will discharge (drain) to a lower voltage before the next pulse can pump it back up. All this stuff means the AVERAGE dc level of all these pulses will be higher for the full wave recto than it will be for a half wave recto.

[Jack_Hester]

You used the wrong multiplication factor converting RMS to peak. Peak = 1.414 x RMS, so in your case 1.414 x 50 = 70.7. This is always true regardless of whether you use a half wave rectifier or a full wave rectifier.

Your half wave rectifier passes a 70.7v negative voltage half cycle to charge the filter cap. Then it blocks the next half cycle. So you end up with a 70v pulse, skip a pulse, 70v pulse, etc. (A full wave rectifier passes every half cycle, so there are no skipped pulses. Still only 70.7v peak). Bottom line, all the pulses applied to the filter cap will be 70.7v regardless of which type rectifier you use. And if there is no discharge path for the cap, ie, no load, the cap will charge to 70.7v and hold it, again, regardless of which type rectifier you use.

Now, whenever you put a load on the filter cap, the cap will discharge some also, so the 70.7v peak may never actually be reached. But even if it does, the cap will be discharged some before the next pulse can come along and pump it back up. A full wave recto passes a pulse every half cycle, so the cap will be pumped up twice as often as a half wave recto. IOW, with a half wave recto, the cap will discharge (drain) to a lower voltage before the next pulse can pump it back up. All this stuff means the AVERAGE dc level of all these pulses will be higher for the full wave recto than it will be for a half wave recto.

Makes good sense to me.  Actually, I used the 1.414 multiplier.  But because I came up with the 70.7vDC answer, and not understanding the function that takes place with the capacitor, I just divided my actual reading of 65.4v by my actual input of 50.03v to get the multiplier I used. 

Your reply got me to thinking about averaging and form factor, from another fellow who commented on my PA-100X bias, elsewhere. 

From him, and the following article, I know how the form factor is derived.

http://www.electronics-tutorials.ws/accircuits/ac-waveform.html

That's when I decided to take the peak value (crest value, in the article) and divide it by the form factor.  This get me so close to the value of V2 that I'm looking for.  As well as the Vout value. 

And, my actual unloaded Vout may drop ever so slightly, at the two sets of two resistors in series, before the power tube grids.  Maybe by a volt or so. 

Am I headed in the right direction with the form factor?  Sure looks right.  Thanks for your help.

Jack

[PRR]

> Am I headed in the right direction with the form factor? 

With this circuit? The waveform (or factor) isn't very important. First cap WILL charge to the PEAK of the input wave. If wave is Sine, "50V" is 70.7V Peak. That's what the cap charges to. Deduct 0.6V for diode drop, 70V.

The cap will sag between peaks. If the rectifier is half-wave it will sag more. But 71K load on 8uFd cap at 60Hz is pretty light, low sag. I would not expect sag to 61V. I wonder if your cap could be leaky?

This circuit is a Peak Catcher. Unless very lossy, math from the peak should be correct. There are other circuits with large resistance into the rectifier. Now Form Factor is critical. Which is when I throw-up my hands, use cut-n-try design.

[Jack_Hester]

> Am I headed in the right direction with the form factor? 

With this circuit? The waveform (or factor) isn't very important. First cap WILL charge to the PEAK of the input wave. If wave is Sine, "50V" is 70.7V Peak. That's what the cap charges to. Deduct 0.6V for diode drop, 70V.

The cap will sag between peaks. If the rectifier is half-wave it will sag more. But 71K load on 8uFd cap at 60Hz is pretty light, low sag. I would not expect sag to 61V. I wonder if your cap could be leaky?

This circuit is a Peak Catcher. Unless very lossy, math from the peak should be correct. There are other circuits with large resistance into the rectifier. Now Form Factor is critical. Which is when I throw-up my hands, use cut-n-try design.

Thanks ever so much for your input. 

The whole point of this is not to find precision in the math.  Just close to see a good effect on component changes.  Cap(s) could be leaky, but they are brand new, as are the resistors.  Took only a few minutes to build this on the breadboard.  My auto-transformer is very coarse, when trying to adjust to 50vAC.  Much smoother on the upper end.  I played with it for a bit, and finally got it stabilized at 50.03v. 

Under the actual conditions, and using the actual measured values of the carbon comp resistors (of unknown origin), (and not knowing how to apply the effect of the cap values on the circuit), I calculated in reverse from my measured output, and came up very close to the measured value of V2.  That's when I started playing with the math, to find as accurately as possible the value of Vin.  Once I accomplished that, I easily straightened the math out, going in the forward direction, from Vin.

Maybe this afternoon, after Church, I'll test the math more, by varying Vin through a wide range (not to exceed the cap voltage ratings of 150v).  I'll report my findings. 

Thanks again, PRR and Sluckey, for your thoughts on this.  Have a good one.

Jack

[Jack_Hester]

Took some readings and made a chart.  Also, put the calculations in a small program.  It's all been a good mental exercise for me, plus it makes it handy tool to see value changes at a glance, if needed. 

Anyway, take a look if you have an interest in such.

Jack

[Jack_Hester]

And, I'm working on the same, for my Oliver bias supply circuit.  As soon as I can round up the components for that circuit, I'll make the same test.  Haven't put my hands on all the needed components, yet.  But, it too will go on the breadboard.  Program already written to run the calcs.  I'll build the voltage chart for it, as well. 

A friend of mine is collecting parts to build a Soldano SLO-clone, and I noticed a note on that bias supply stating that that a 220K resistor must be inserted before the diode, if an HV tap is used.  Other than that, it's still the Bassman configuration with a means to vary the output. 

Jack