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I said voltage divider on the title, but could be anything.

I have a DC source of 19.7V.

I need an output that varies from 0V to 18V.

I have:

  • Resistors (1/8W);

  • 1k potentiometer;

  • 5k trimpot;

  • 2N3904;

  • LM358;

  • Capacitors.

My only requirement, besides part count, is that both potentiometers be used: the 1k being the main, and the 5k acting for fine tuning.

And the adjust... not too sensitive. The more linear, the merrier, and taking advantage of the whole excursion of the potentiometer.

In the example below I could not obey the requirement of sensitivity and excursion.

The transistor I'm trying to use has a cutoff around 680mV and saturation around 820mV. Very small window. Hence, my trouble.

Example

More:

  • The output current can be small. 5mA will work;

  • This whole system will control the GND pin of a LM7805, and later on, of a 7905;

  • LM317 unavailable;

  • The 1k potentiometer can be turned by hand and is what I have available. The 5k trimpot needs a screwdriver.

History:

At first I thought a voltage divider could do it, but in my attempts, the voltage hardly approaches 15V.

Still open for this type of design, though (just bear the requirements if you'd like to help me this way)

After giving up, I tried using OpAmp. It worked. But later I will need a negative version of this structure I'm trying to build, but 19.7V-(-19.7V) my opamps won't support.

Maybe I could reduce the supply voltage of this OpAmp, but then... part count.

Third attempt, more promising but difficult, was with a transistor.

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    \$\begingroup\$ is this a school assignment? \$\endgroup\$ – jsotola Jun 28 at 23:31
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    \$\begingroup\$ Designers don't start with a few parts and say what it's for, they start with specs like the voltage, current, load min, max tolerance, stability, max heat rise. Why is 5k fine-tune and 1K coarse when it should be the other way around. Because its smaller? Is 50mA to load enough? 500mA? then you need a big heatsink. Is the pot rated for 1/2W? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 at 2:37
  • \$\begingroup\$ Hello. Edited the OP to explain those details. It's a personal project (regulated supply), and "what I have" is what answers most of the questions. \$\endgroup\$ – Lucas BS Jun 29 at 4:42
  • \$\begingroup\$ @sunnyskyguy EE75, "designers don't start with a few parts" - actually, design engineers should start with all requirements simultaneously, which are cost/availability, reliability, electrical, environmental, mechanical, service, etc. Electrical's just one and, to me, usually the easy bit to sort out. Engineering's them all. Lots of different situations in different companies, or in being a student/hobbyist which I imagine it's here. \$\endgroup\$ – TonyM Jun 29 at 7:42
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    \$\begingroup\$ @TonyM We understand each other, you have my profile, I was saying exactly the same meaning, Spec= requirements 1st, choose parts design and parts next. BTW did you understand how I made assumptions on specs and created a solution that works. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 at 15:57
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schematic

simulate this circuit – Schematic created using CircuitLab DONT USE ABOVE poor simulator part of EE.SE enter image description here Don't burn it out by over loading the pot with 1/4W type or excess current.

Use this simulator instead

Every part serves a purpose.

If you understand this design and tried it out, say OK (+1)

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  • \$\begingroup\$ The TO92 NPN can handle 50mW @ 200’C/W rise with 10V drop and 5mA to meet your requirement \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 at 16:35
  • \$\begingroup\$ Do you understand my solution uses Emitter Follower? So the 1k Pot with current gain of 100 now looks like a 10 Ohm pot \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 at 16:39
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    \$\begingroup\$ Very good, upvoted \$\endgroup\$ – TonyM Jun 29 at 21:02
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    \$\begingroup\$ @TonyM I wish more professionals would recognize a good example as you did. But it seems wanting on EE.EE W5VO take note. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 29 at 21:24
  • \$\begingroup\$ It works perfectly on the simulator. I'll see if I can mod yours to get around the high current on the pot. Meanwhile: how did you come to this design and the values ? \$\endgroup\$ – Lucas BS Jun 30 at 21:06
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I'll share my own solution, which loses on two aspects:

  • Voltage will never reach my desired minimum of 0V;

  • The part count is a little higher. Luckly, it's all resistors.

The only advantage here is the low current through the components.

I encourage those interested to enhance this model.

My next goal is a negative counterpart: 0V to -18V from a -19.7V supply

Simulation

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Is this an assignment or similar.
If so, we will still help but in a more 'tutorial' manner.

Here is a fairly complete outline.
Purposefully with no circuit and somewhat skeleton detail assuming an assignment. Ask more if wanted.

The opamp is your great friend.
An UNLOADED pot will give you the whole voltage range.
Add a trimpot at the top and you can reduce the range.
If the pot is linear (and not log or semilog) you get linear Vout with pot position.

Use the opamp to drive the transistor as en emitter follower.
Compare pot voltage and vout (scaled into common mode range of opamp) and drive transistor (drive scaled to allow for opamp not being a rail to rail one).

Divide Vout by a factor to bring it into opamp common mode range and compare it to the pot voltage. Pot Vout or pot total Voltage range can be scaled down to lower voltage to match opamp Vout related input.

Result should be reasonably superb.
Instability is possible.
Judiciously placed small cap MAY be needed somewhere if it oscillates.

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  • \$\begingroup\$ It works. And works smoothly \$\endgroup\$ – Lucas BS Jun 30 at 21:15
  • \$\begingroup\$ Damn site that allows only 5 minutes to edit ! Anyway: In my first attempt I did this exact design. It works perfectly, and will definetely resolve the problem of others in a similar situation. Problem is (and I should have written more about it in the OP), that I want to make a symmetric counterpart. The only OpAmps I have are DUAL. So, in order to be ecomic, I could use a single chip for both sides. But my rail to rail voltage is 19-(-19)V. Too high for an OpAmp \$\endgroup\$ – Lucas BS Jun 30 at 21:22
  • \$\begingroup\$ If you scaled down the pot ref voltage and the transistor drive voltage then you can use a voltage usefully less than V_amp_max/2. The LM324 has max supply V of 32V or +/- 16V. So if you work on say +/- 12V you can scale down measured voltages to inside the common mode range and scale up drive voltages to suit. Ask if that doesn't make enough sense. \$\endgroup\$ – Russell McMahon Jul 1 at 13:18
  • \$\begingroup\$ Think I understood. Something like leveling and boosting the opamp output so, with the help of the transistor, the system spans the desired voltage. The challenge this time would be to make, using your example, +/-12V. Would require 4 resistors. Apart from being a little more complex than the other answers on this thread, it's a beatiful solution. \$\endgroup\$ – Lucas BS Jul 2 at 14:16

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