Voltage Divider Math in a Bias Supply.........

[Jack Hester]

far out, that's a cool unit.

You might explore voltage drops across the many connections. All the interconects might result in a volt or two loss through the circuit. That will throw off any precise rations if that is what you are seeking.

Not really looking for precise readings. Just confirmation of what to expect. And in the case of my breadboard test, I took the time to measure the resistors, as they were carbon comp. I know from past experience that off-the-shelf components such as carbon comps are not precise.

Anyway, using the measured values of these resistors, and the measured voltages, I am able to calculate what I should expect. I came up 1.3v low as compared to the actual. This is good.

The whole point of this exercise has been this: I want to see what change would occur when I replaced say the 56K with a 24K fixed resistor, and a 25K pot. As I am a hobbyist programmer (Delph 7), I wrote a small program to demonstrate this. The voltages and resistor values are such that they can be changed at any time. And, I can calculate from the input to the output, or just the opposite. Once I ironed out the multiplier that I needed for the first rectified voltage, and where the multiplier came from, I am now able to see the results of these component value changes, along with voltage changes.

For my multiplier, I used the Form Factor notes that David H. gave me, and applied it as follows:

V2 = Negative rectified voltage at the junction of the diode and the first capacitor
= [ -Vin x Sqrt(2) ] / [ (1 / Sqrt(2)) / (2 / Pi) ]
= [ 50vAC x 1.414 ] / [ .707 / .6366 ]
= [ -70.7v / 1.11 ]
= -63.662vDC

I'm still not sure that I'm doing this right. But, the numbers are looking good.

Jack

[Jack Hester]

I've attached a voltage chart and schematic for my breadboard tests, yesterday afternoon (Mon). Also, a very small companion program that I used to run the numbers for comparison with my actual readings. The schematic is the typical '59 Bassman, but found in many amp designs. For my calculations, I show the 15K as R1, and the 56K as R2. The negative voltage reading taken at the junction of the diode and the first capacitor, I label as V2.

For the test, I used the actual measured value of the resistors, which are different from the indicated values. I've experienced this with carbon comp resistors, before. These values are noted on the chart.

The reason for the input voltages not being exactly on the round number is that my auto-transformer is a bit coarse in adjustment.

I am pleased with the results, as it gives a reasonably accurate indication of voltage changes for resistor changes. Once I learn how the capacitors interact with the resistance values, I'll revise to include that. Just a good mental exercise for me.

Jack

[Enzo]

This is interesting, but allow me a suggestion. In math there is something called significant digits. Your calculated answers can have no more significant digits than the data. So in the first entry, your extra digits on the calculation are a false precision. Your data has four significant digits, so so must your calculation. In other words your calculated answer was also 12.89. Dead on. And that 15 digit output is more precise than your data by a long ways. Stop at 10.08. That 0.02v could easily have been probe tip resistance.


You need to have a filter cap on the circuit right after the rectifier for the readings to be useful in practice.

The filter cap allows the rectified AC to charge up to peak value




https://www.physics.uoguelph.ca/tuto...ig/SIG_dig.htm

[Jack Hester]

This is interesting, but allow me a suggestion. In math there is something called significant digits. Your calculated answers can have no more significant digits than the data. So in the first entry, your extra digits on the calculation are a false precision. Your data has four significant digits, so so must your calculation. In other words your calculated answer was also 12.89. Dead on. And that 15 digit output is more precise than your data by a long ways. Stop at 10.08. That 0.02v could easily have been probe tip resistance.


You need to have a filter cap on the circuit right after the rectifier for the readings to be useful in practice.

The filter cap allows the rectified AC to charge up to peak value




https://www.physics.uoguelph.ca/tuto...ig/SIG_dig.htm

Good stuff. Actually, for this kind of calculation, 1 or 2 decimal places for the resulting answers is probably more accurate than necessary, as the voltages drifted a bit up and down. Could have been my auto-transformer, or the thermistor warming up. Or, the carbon comp resistors warming up, as well. A completely rounded number is the best. I'll continue to use the lengthy values of Pi, Sqrt(2), etc., but the calculated voltages may only be 1 or 2 places. It's just a matter of putting it in the program code just how far out I want it to display. The program actually uses something over 10 decimal place accuracy, regardless of how many I tell it to display.

On the next test, which will be a breadboard version of my Oliver Bias Supply circuit, I'll tweak the program down to a couple places, and observe the results.

Jack

[ewd]

I know that this is an old thread, but seems appropriate for my question. I'm building a 5F6-A Bassman schematic amp and will be installing an adjustable bias. As I like to understand what I'm actually doing, I have a dumbass question (I've occasionally been called a dumbass, by ex-wives and current friends). The Mojo752 PT that is sold as a replacement PT for the 5F6-A has a 50VAC bias tap that will be rectified into negative DC voltage by a 1N4007 diode. I was doing some research to see how 50VAC can be rectified into what I calculate is the approx -70VDC needed to do go into the Resistor/Bias pot voltage divider circuit to get the desired negative DC bias voltage range to the power tubes. What I found was a statement that the diode, in addition to supplying the negative DC, also multiplies that rectified DC voltage by an approx 1.4 factor. I know that a diode provides a voltage drop in the forward direction, but how does it multiply voltage when reversed to give negative voltage? That 1.4x multiplier factor, along with the voltage divider resistor values and bias pot, appears to provide the negative bias voltage range I've seen on other schematics using the original Fender PT, when doing the math. I just don't understand how the diode multiplies the negative DC voltage.

[Enzo]

The single diode rectifies the AC into DC. The resulting DC is 1.414 times the RMS voltage of the AC.


There are voltage multipliers (look up voltage doubler) but they require two or more diodes. SUch a circuit would not be needed with your 50vAC tap.

[g1]

And I'll add that it is not really multiplication, but 2 different measurements of AC voltage. The tap is measured in RMS volts, but the diode allows the peak voltage through. AC peak is 1.414 times the RMS.

[ewd]

Thanks Enzo & g1, I believe I now understand the answers to my question.