DC voltage summing type circuit

Question

I hope someone can help me. I'm not an electrical engineer by occupation. I have a need for a circuit that can do the following:

Provide a steady 8 V DC output at rest. Add to that output, the voltage seen on a potentiometer wiper (0 to 7.8 V). In other words:

Potentiometer Vout	Circuit Vout
0 V	8 V
1 V	9 V
2 V	10 V
3 V	11 V
etc.
7.8 V	15.8 V

From the research I've done so far, it seemed as though a couple of inverting op-amps would be the answer (the first an inverting summing op-amp so that the 8 V baseline supply input and the potentiometer input wouldn't interfere with each other, the second with a unity gain to simply invert the output of the first).

Is there an easier way to achieve it or is this the best approach?

I've tried breadboarding it with an LM324 and I'm not seeing the results I'm expecting, due possibly to my running the op-amp from a single 18 V DC supply instead of a +16/-16 V supply.

---

Thanks to everyone for taking the time to answer and explain. What a great community!

Apologies for the delay in responding. The nature of my work means I am away from mobile and internet access until I get home :(

I will try the suggested circuits, using the LM358 and a 24 V supply.

To follow up on some of the comments, here is the circuit I had tried breadboarding before. All resistors used were 1% metal foil. The variable 0-7.8 V input is the resultant voltage from a 10k linear potentiometer attached to a speed controller circuit.

I'll be sure to post my progress using the suggested circuits Thanks again for your time and expertise to help me with this!

Answer 1

At the op-amp's non-inverting input is the average of the two inputs and this voltage is multiplied by 2 by the non-inverting amplifier. The buffer is required to prevent R2 from loading the pot.

To use a +16 V supply you would require a rail-to-rail output op-amp which the LM358 isn't.

Answer 2

All you need is to sum two positive voltages. A summing non-inverting amplifier will do it.

The idea is:

• Take an average of input voltage with 8V: \$V_1=(V_{in} + 8{\,\rm V})/2\$.
• Multiply the average by x2: \$V_{out}=V_1\cdot2 = V_{in} + 8{\,\rm V}\$.

The R3-R4-R5 divider generates 8V from 18V. The R1-R2 branch adds the input voltage and causes the divider to output the average of 8V and the input voltage. OA1 with R6-R7 form a gain-of-2 non-inverting amplifier.

^{simulate this circuit – Schematic created using CircuitLab}

The adjustment potentiometers are not optional. With 1% resistor tolerance, you'll need them - and also because of the offset voltage of the op-amp.

To adjust the circuit:

1. Set ZERO and GAIN trimmers to center position.
2. Power up.
3. Let the circuit warm up for 10 minutes.
4. Set input to 0V.
5. Verify that the voltmeter reads 8.0V+/-0.1V.
6. Set the voltmeter to relative measurement and take reference (null) reading. The voltmeter should read 0.000V.
7. Set input to 5V.
8. Adjust GAIN for output of 5.000V+/-1mV.
9. Set input to 0V.
10. Reset the voltmeter to absolute mode.
11. Adjust ZERO for output of 8.000V+/-1mV.

The adjustments are independent of each other.

Assuming if the resistors and potentiometers had ideal values, the input-to-output transfer curve would look as follows:

Answer 3

I presume you mean "metal film" not "metal foil" resistors. Metal foil resistors are generally extremely expensive extreme precision resistors, and one would not want to use them anywhere near an LM324.

This should work if you're not too fussy about precision. You can buffer the input with an LM324 voltage follower if you want, however the 8V output precision will be affected a bit because no single supply op-amp can pull right down to the negative rail and the LM324 has trouble handling more than a few tens of uA sinking to the negative rail. A negative supply would eliminate the problem, some other op-amps are better in this regard without a negative supply (but ones that are both good, and comfortable with an 18V+ supply are less common and cheap).

The downsides of using the pot directly as shown (no buffer) are that the linearity is affected a bit (because the pot has some source resistance, which varies with pot position)- the magnitude depends on the pot element value and if the pot wiper or connection opens up, then the output voltage will rail to the positive side. If you have a buffer you can add a resistor so that if the input is disconnected the output will go to 8V. Depending on the application this may be desirable. The linearity will not be affected much with the shown circuit if your pot is something like 10K since the source resistance only varies from 2.5K at the middle to about zero at the ends, and it's in series with 144K.

^{simulate this circuit – Schematic created using CircuitLab}