I am designing a Digital Logic Board and need to add a DC signal output from 0 to 5V (with potentiometer) with a constant 10mA. I have checked designs with LM317 and LM334 for adjustable current upto 10mA, but no results. Can someone please provide an easy to understand schematic as i m not very good at analog designs?

Update due to comments:

I'm trying to control a DC output voltage from 0 to 5V, with a simple potentiometer 5kohms, using voltage divider. Then i am using a 520 Ohm Resistor to limit the current to 10mA (R=U/I -> 5/0.01 = 500ohm -> available 520ohm). So when the the potentiometer is on 0, the output is 5V and current limited to 10mA, but when the ratio changes, the voltage will drop to reach 0mA, and the current with also drop proportionally. My question is, are there a solution where I can get constant 10mA regardless of the outputed voltage?

Here is a litte schematic to help understand my request: enter image description here

Edit 2:

The Digital Logic Board I m trying to make is a simplified version of this one http://hps-systemtechnik.com/wp-content/uploads/content/english/3910.pdf, and the DC output i m trying to make is this one:

enter image description here

  • 4
    \$\begingroup\$ a DC signal output from 0 to 5V (with potentiometer) with a constant 10mA That doesn't make much sense. You either have 0 to 5 V or you have a constant 10 mA. You will have to explain this more clearly by adding a schematic (of the circuit you want to connect to) and describing more clearly what you need. \$\endgroup\$ – Bimpelrekkie Oct 14 '18 at 12:02
  • 2
    \$\begingroup\$ You can control the output voltage or the current. You can set a limit on both (e.g., up to 3.8 V at up to 10 mA) but you can't specify both. The load will determine the relationship between them. \$\endgroup\$ – Transistor Oct 14 '18 at 12:03
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    \$\begingroup\$ If ohm's law (U=RI aka V = ΩI) confuses you, then change the variables to something you understand better, such as mass = density × volume. What you are technically saying is that the mass may be changing from 0 kg to 5 kg and the volume will always be 10 cm³. This means that the density obviously will have to change in order for both of those to be true. In other words, the resistance in ohm's law will have to be variable. You can't control the resistance, therefor you can't control 0 - 5 V and 10 mA. It's one or the other you can control, not both. \$\endgroup\$ – Harry Svensson Oct 14 '18 at 12:15
  • \$\begingroup\$ We're trying to point out that your thinking is faulty. If you can explain why you need this we can probably help better. What is your load? A resistor? An LED? Something else? Please specify. By the way, there is a built in CircuitLab schematic editor on the edit toolbar. \$\endgroup\$ – Transistor Oct 14 '18 at 14:59
  • \$\begingroup\$ @Transistor I m new to stackexchange, and not yet familiar with all the tools, sorry for any inconvenience. It can be any type of load, as it is an output from the board. I edited my post and added a link to one board with the 0-5V output, to better understand. \$\endgroup\$ – awakrim Oct 14 '18 at 15:05

What you are looking for is a 0 - 5 V variable voltage power supply with a current capability of 10 mA. It is not a constant-current power supply but rather a maximum output current that will be supplied at any setting of the voltage.


simulate this circuit – Schematic created using CircuitLab

Figure 1. A simple potentiometer circuit.

The simple potentiometer circuit has a few problems.

  1. With the wiper set at anywhere other than min or max, the output voltage will vary with the value of RL. That means you've failed your constant voltage requirement.
  2. There is no current limit.

Adding a current limiter presents its own problems. As the current limiter is going to drop at least a volt or two your supply will have to be that much higher than 5 V.

enter image description here

Figure 2. Basic Low Voltage Regulator (VOUT = 2 to 7 Volts). (Figure 4 of datasheet).

Have a look at the old LM723 voltage regulator. If you can figure it out you will be improving your analog skills. In particular, I suggest you start with the schematic in Figure 4 and have a look at the formulas on Table 2 - particularly the current limit formula.

  • \$\begingroup\$ So if I understood correctly, to be able to have output from 2-7V, I will have to put a potentiometer in R1: R1=10k, R2=10k, R3=(R1*R2)/(R1+R2)=5k. And Current limit must be 10mA, Rsc=Vsense/Ilimit = 1V/0.01= 100ohms? \$\endgroup\$ – awakrim Oct 14 '18 at 16:26

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