LED indicator for DC voltage

converterdiodesledpowervoltage

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.

Best Answer

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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