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I am planning to using a buck / boost converter to run a DSLR camera from my USB powerbank AND other power sources. For that module I want to build with the buck / boost converter, I need a simple LED voltage indicator circuit, since I need to adjust the potentiometer from the converter whenever I change the source of power.

Example 1

input: 5V from regular USB powerbank
output: 8.4V to the camera coupler

Example 2

input: 9V from DIY battery pack
output: 12V to other application

What I want the circuit to do is, when I turn the potentiometer, I want a LED to light up when the output voltage is between 8.4V and 8.6V ( ready for camera ). And when it's in an other range, I want an other LED to light up.

I've heard something about a combination of resistors and z-diodes to do the trick ?

Additionally it would be perfect, if the indicator circuit is not depending on whether the output circuit is closed or not ( camera is pluged in or not ).

Simple sketch:

enter image description here

Pleas tell me your ideas and solutions.

EDIT

I've found this link that shows this circuit enter image description here

How does that work and how do I calculate the values for the parts ? I didn't understand the text on the linked page.

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  • \$\begingroup\$ Your output voltage is the same in both cases. Why do you need to adjust anything? \$\endgroup\$ – Nick Johnson Jun 22 '15 at 14:53
  • \$\begingroup\$ I am pretty sure that the output voltage will change depending on the input voltage, if i don't change the potentiometer. Doesn't it ? \$\endgroup\$ – Ace Jun 22 '15 at 15:06
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    \$\begingroup\$ The whole point of a voltage regulator is that it produces the same output voltage regardless of load current and input voltage, within its limitations. You don't say what you're using, but it's unlikely that it needs adjusting. And, if it does, you should get a better regulator. \$\endgroup\$ – Nick Johnson Jun 22 '15 at 15:07
  • \$\begingroup\$ okay thank you, I will check that as soon as I can. But it would be nice anyway to be able to check if the voltage is in the range I want, since I don't want to overpower my camera... also maybe I want to use the regulator for other applications, where I need other output voltages, since I'm building it modular. Therefore it would be nice to know a circuit that lets me light up a LED for each range I choose. \$\endgroup\$ – Ace Jun 22 '15 at 15:12
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    \$\begingroup\$ What's wrong with the obvious solution of buying a cheap LED panel voltmeter off EBay and hooking that up? \$\endgroup\$ – Nick Johnson Jun 22 '15 at 15:13
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As ox6d64, motioned, you can use a window comparator. The trick is, what to compare to, since you have varying supply voltages.

Here I am using a LT1461 voltage reference that can handle input voltages up to 20V and puts out 3.3V with 0.04% accuracy. So it is like a very precise voltage regulator but lower current (50 mA). That is still enough to drive a some LEDs besides serving as a reference.

enter image description here

It then uses two comparators. The input voltage is fed into two voltage dividers; the top one yields 3.29 volts when fed an input of 8.6V:

$$\frac{6.04kΩ}{9.76kΩ + 6.04kΩ} \times 8.6V = 3.29V$$

and the bottom one yields 3.31V when fed an input of 8.4V:

$$\frac{6.49kΩ}{10kΩ + 6.49kΩ} \times 8.4V = 3.31V$$

I couldn't get exactly 3.3V with either divider since the exact resistor values weren't available. The standard 1% resistor values are here.

So the bottom comparator will output a high when the voltage is above 8.4V, and the top comparator will output a high when the voltage is below 8.6V. These two conditions are ANDed together using the NAND gate. I use the latter so when it is asserted, it outputs a ground and turns on the LED.

You can modify the circuit to work with whatever other voltages you need to measure.

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  • \$\begingroup\$ And putting the (open-collector) comparators the right way around will allow you to wire-OR them, saving the logic gate. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 23 '15 at 16:08
  • \$\begingroup\$ Also note that the comparitor outputs need pull-up resistors or the circuit (as shown) won't work. \$\endgroup\$ – Dwayne Reid Jun 23 '15 at 16:32
  • \$\begingroup\$ @IgnacioVazquez-Abrams Noted (and good suggestion), however I left the circuit as since it is easier to understand with an explicit AND gate. \$\endgroup\$ – tcrosley Jun 23 '15 at 17:08

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