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This is for a part of a system that provides feedback to a voltage reference of 0.6 V. The output voltage has a range of 125 V to 500 V.

Using a voltage divider, 414k and 2k will provide the 0.6 V when the output is 125 V. Similarly a 1664k and the 2k will provide the 0.6 V when the output is 500 V.

In order to control the feedback, I plan on using a resistor network adjustable between the 414k and 1664k for the top resistor. I have split that resistor into a fixed 390k resistor and a 24k - 1274k variable resistor network.

What approach should I take to solve this problem?

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  • \$\begingroup\$ Consider voltage breakdown rating and Pd Pd series as well as keeping stray noise away from feedback R . This leads to trade pods for min/max values. Then choose a shunt pot with a fixed R to compute the correct gain and offset using a formula or spreadsheet . There are also programs to help choose standard values from ratios. Keep working on it. \$\endgroup\$ Commented Mar 27, 2018 at 5:07

3 Answers 3

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First off it appears that your calculations are incorrect. I also think you are using the wrong approach. You will want to come up with an approach that puts your variable resistor in the low voltage side of the divider so that the pot does not get exposed to the high voltage.

So assume you have the typical voltage divider configuration:

enter image description here

For your two endpoints:

Vin = 125V Ra = 220K Vout = 0.6V Rb = 1061.09K

Vin = 500V Ra = 220K Vout = 0.6V Rb = 264

It will be a whole lot easier to use a readily available 1K pot with a series resistor of about 60 ohms. The pot can easily adjust down to ~200 ohms to get the control range you need.

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  • \$\begingroup\$ I was just looking at the MCP41 series and was bummed when I realized the voltage on the resistor pins are limited to about 5 V. Your suggestion would likely work with this digital rheostat. I'll come back and post what I decided on once I get it together. \$\endgroup\$
    – A.S.
    Commented Mar 27, 2018 at 3:17
  • \$\begingroup\$ You were right about my calculations, the lower resistor should be 2k and has been corrected. How did you decide on 220k for the base resistor in your suggestion? \$\endgroup\$
    – A.S.
    Commented Mar 27, 2018 at 3:23
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    \$\begingroup\$ I selected the 220K resistor for a number of basic reasons. First off resistor values in the megaohm region are best avoided in most circuits if possible. The currents become very low and the input impedance of loads on the voltage divider will affect the voltage divider accuracy. The second reason was that the 220K resistor resulted in the use of a 1K pot which to me is a good standard value in most pot technologies. Lastly the component needs to be a reasonable device that you can purchase with a good tolerance range whilst still being able to handle (continued) \$\endgroup\$ Commented Mar 27, 2018 at 12:31
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    \$\begingroup\$ (continued from above) ~500V across the device. Plus 220K is a bog standard value. In applications like this you also want to weigh the above considerations against the power lost in the components to limit the effects of self heating. The 220K value @500V results in a little less than 3mW in the resistor and that should be of little effect in a resistor that has a physical size to handle 500V. You may also want to consider the SPOF (single point of failure) concept in your circuit design too. If the resistor to 500V should fail to a short or a smaller value in ohms it could (continued) \$\endgroup\$ Commented Mar 27, 2018 at 12:39
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    \$\begingroup\$ (continued from above) result in damage to your lower voltage circuits in the lower leg of the divider. One design technique is to put two resistors in series instead of just one to the 500V. The other thing you can do us to place a zener or tranzorb device across the lower leg. \$\endgroup\$ Commented Mar 27, 2018 at 12:42
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You really haven't provided enough information for a definitive answer. One simple approach is a ordinary variable resistor (potentiometer) in the 1 to 2 megohm range. By the way, your title and your actual requirements don't match: 10 kilohms to 1 megohm vs 414 kilohms to 1.664 megohms vs 24 kilohms to 1.274 megohms. You haven't specified the resolution you need or if the resistance is to be manually adjustable or computer controlled. For low resolution, a single turn potentiometer can be used. For higher resolution, use a 10 turn pot.

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  • \$\begingroup\$ The title is an approximation, 10k~24k, 1M~1.274M. The 414k and 1.664M are split into a 390k and the aforementioned 10k-1M. A resolution of 1k would be enough. \$\endgroup\$
    – A.S.
    Commented Mar 27, 2018 at 3:19
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Following the advice from Michael Karas, I looked into a low side controlled divider, using a MCP41XX series digital potentiometer.

The voltage divider is composed of a fixed resistor on the high side, and the digipot and a fixed resistor on the low side (with the fixed resistor between the digipot and ground).

Here are factors I need to consider when selecting the resistor values. -The Vout of the divider needed to be less than 5 V, per the digipot limit. Still, I wanted to get a high enough voltage swing to improve resolution. -The low fixed resistor is what prevents the output voltage from shooting sky high. -The bias current should be in the 500 uA range.

My final design was the following. A 620 kohm high resistor, a 5 kohm digipot, and a 750 ohm low resistor.

With these values, the max voltage is fixed at 500 V per the 750 ohm resistor. The min value is 65 V when the feedback is across the 5750 ohm low combination. The bias current is between 104 uA and 800 uA when 65 V and 500 V, respectively. The maximum voltage on the digipot high side is 4.59 V, close enough to my desired range.

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  • \$\begingroup\$ If this is the correct answer then you can accept it yourself to mark it as such. Don't forget to upvote any answers that were helpful. \$\endgroup\$
    – Transistor
    Commented Mar 27, 2018 at 11:50
  • \$\begingroup\$ @Transistor I will soon, system won't let me accept my own answer until tomorrow (48 hour wait period) \$\endgroup\$
    – A.S.
    Commented Mar 27, 2018 at 18:48

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