Adjust supply voltage based on divider resistance

In summary: The voltage signal is the AC signal, the current signal is the DC signal. The negative supply voltage is the Vgk voltage of the PMOS, increasing it decreases the current, decreasing it increases the current. The problem I'm having is that as the AC signal amplitude increases the operating point changes, I'm trying to stabilize it by adjusting the negative supply voltage based on the resistance of R12 (which is a potentiometer controlled by a digital signal). What is U3 doing? It's an integrator opamp used for DC offset control. What does changing the negative supply do for you? Changing the negative supply voltage changes the operating point of the stage
  • #1
coinmaster
41
0
Hello, I'm working on a project and I need some assistance. I need to adjust the negative supply voltage to an integrator opamp that I use for DC offset control based upon the resistance in a voltage divider elsewhere in the design.
Draft24-2.png

R12 is the trimmer resistor to adjust the CVS circuit and V5 is the power supply that I need to adjust based upon the resistance of R12. So I need some sort of mechanism that will measure the resistance of R12 and then increase the power supply voltage as the resistance of R12 decreases. I need amount of adjustment to be adjustable as well.
Any ideas?
 
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  • #2
Perhaps use a dual gang pot for R12. Eg as used for the volume control on some stereo amplifiers.
 
  • #3
Unfortunately R12 must be an opto-isolator because the circuit will be digitally controlled.
 
  • #4
A question well stated.. (something has to be pretty true to achieve platitude status)...

okay , i see R12 with his wiper position controlled by "Mr servo" .

Is R12 a potentiometer or a rheostat ?
We don't know much about "load" ?
Where is that "null output" you're trying to achieve measured ?

To the question as i read it -
If a rheostat, R11 and R12 form a half-bridge
differencing amplifiers might report voltage across each , they might be ratio'ed

or an R11B and R12B string connected from V6 to load would give you a whole bridge whose unbalance you could difference-ampify ,
 
  • #5
jim hardy said:
okay , i see R12 with his wiper position controlled by "Mr servo" .
Actually in the audio circles a servo refers to anything that changes something based on another change in something. The "servo" is the device I'm trying to make.
jim hardy said:
Is R12 a potentiometer or a rheostat ?
It's an LDR so technically it's more like a potentiometer.
jim hardy said:
Where is that "null output" you're trying to achieve measured ?
Not shown in this schematic. A change in operating point in the stage (done by the CVS) requires adjustment to the DC offset "servo" in order to maintain the null.
 
  • #6
You might try adding a current mirror to isolate the current through the resistor. Then with the voltage and the current... Not an elegant solution.

Yet since R12 is controlled, perhaps you can estimate it from the control current? This may lead to a more elegant design.
 
  • #7
Jeff Rosenbury said:
Yet since R12 is controlled, perhaps you can estimate it from the control current? This may lead to a more elegant design.
Yeah that may end up being the solution, I'll have to use a digital control and offset. The problem then is how do I adjust the power supply. Maybe a depletion mode mosfet or something, I don't think opto-isolators work with negative voltage..
 
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  • #8
coinmaster said:
It's an LDR so technically it's more like a potentiometer.
LDR? Light Dependent Resistor?

I'm old fashioned
to me a two terminal variable resistor is a "rheostat"
a three terminal variable resistor is a "potentiometer"

though the terms really apply more to how they're connected than how they're built

00343x01.png


upload_2016-2-12_16-59-52.png


potentiometer vs rheostat connections - mnemonic for me is
potentio- , implying potential difference or voltage + meter = measure , as with a Wheatstone bridge
and rheo = flowing, + stat = control, as with a series resistor to control current flow

All that aside - thanks for the clarification.

draft24-2-png.95710.png
Not knowing anything about load,

Looks like current through R11 and R12 hovers around a few tens of microamps to bias PMOS on ?( a few volts across R11)
so decreasing R12 would turn on PMOS harder, turning on Q3, dumping more current to Load, increasing Vload to restore constant current through R11&R12...

Looks to me like Vload then is a rough measure of R12
well maybe an inverse measure

if this is inside a closed loop might you get away with something so crude ?

Just a thought.

old jim
 
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  • #9
This is the description of the circuit
CS-Load-Servo-Wavebourn.gif

At DC: its a Voltage source

At some subsonic frequency:
Its a gyrator of aproximately
equal and constant current.

Above 20Hz: Its Anti-Triode
or Aleph or whatever...
So in essence It's a CVS triode load, at least that's what I take from it.
 
  • #10
now current exiting R2 can split to Q1 unlike when Q1 was a FET...

is R3 still servo controlled ?

moving target here

time to try an experiment i'd say. Are you building this or simulating it ?
 
  • #11
Referencing the original picture:
I'd like to help, but I'll be darned if I can figure out what you are really trying to do (never figured out the previous thread either). What is U3 doing? It's an integrator, running open loop at DC, and saturates at the negative or positive supply depending on the input. What does changing the negative supply do for you? Doesn't change anything if you are not saturated.

You want the adjustment to the negative supply to depend on the value of R12 independent of the components around it? You are controlling it digitally based on what? Can you use the digital input to affect the supply?

You are leaving something out that I'm not able to assume. I'm lost (and that's usually pretty hard to to).
 
  • #12
jim hardy said:
now current exiting R2 can split to Q1 unlike when Q1 was a FET...
It was highly recommended on this thread to make Q1 a fet http://www.diyaudio.com/forums/tubes-valves/140699-anti-triode-sepp-how-do-best.html
jim hardy said:
time to try an experiment i'd say. Are you building this or simulating it ?
Both, haven't gotten to the build phase yet. In the simulation, the operating point shifts with the AC signal amplitude unlike a normal gyrator load, I have to look into that. Worst case scenario I revert to a normal gyrator load, at the end of the day I just need to be able to adjust the voltage and maintain the current.
meBigGuy said:
I'd like to help, but I'll be darned if I can figure out what you are really trying to do (never figured out the previous thread either).
I'm trying to have digitally adjustable load line operating points. I'm using a CCS on the cathode and a CVS on the anode.

meBigGuy said:
What is U3 doing? It's an integrator, running open loop at DC, and saturates at the negative or positive supply depending on the input. What does changing the negative supply do for you? Doesn't change anything if you are not saturated.
Adjusting the negative supply adjusts the voltage output for the DC offset. Its purpose is to maintain DC null at the output of the stage.
Draft24-1.png

For example, if I push a 6080 tube to 250v anode voltage I'll need -150v grid voltage for a proper bias.
As I decrease the value of R12 I increase the anode voltage which then requires me to increase the supply voltage to U3 in order to have an increased bias voltage.
 
  • #13
6080 ? Low mu triode?

Are you trying to build a transformerless tube amp ?

I remain firmly convinced the inductance of the output transformer is necessary for that "warm tube" effect
and inverting triodes has to be hard on the grid.

i like low hum low distortion , 7199/6973

I hope you'll let us know how it sounds
old jim
 
  • #14
jim hardy said:
I remain firmly convinced the inductance of the output transformer is necessary for that "warm tube" effect
and inverting triodes has to be hard on the grid.
I'm not really interested in the warm tube effect, I'm more interested in realism, this topology offers that in spades at least in my experience.
 
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1. How does adjusting the supply voltage based on divider resistance work?

Adjusting the supply voltage based on divider resistance involves using a voltage divider circuit, which consists of two resistors in series. By changing the value of one of the resistors, the voltage output can be adjusted.

2. Why would I need to adjust the supply voltage?

There are several possible reasons for needing to adjust the supply voltage. One common reason is to match the voltage requirements of a specific component or circuit. Another reason may be to control power consumption or to compensate for voltage drops in a long circuit.

3. What is the relationship between divider resistance and supply voltage?

The voltage output of a divider circuit is directly proportional to the ratio of the two resistors. This means that as the divider resistance increases, the supply voltage decreases, and vice versa.

4. How do I calculate the necessary divider resistance to achieve a specific supply voltage?

The formula for calculating the output voltage of a voltage divider circuit is Vout = Vin * (R2 / (R1 + R2)). To achieve a specific supply voltage, simply rearrange the formula to solve for R2. Keep in mind that the resistors used must have a higher power rating than the maximum power dissipation of the circuit.

5. Are there any limitations or risks to adjusting the supply voltage based on divider resistance?

Yes, there are some limitations and risks to consider when adjusting the supply voltage. If the resistance is too high, the output voltage may be too low and not meet the requirements of the circuit. Additionally, using resistors with values that are too low can cause excessive power dissipation and potentially damage the circuit. It is important to carefully calculate and select appropriate resistor values to avoid these issues.

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