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I want to design a circuit that has 3 LEDs connected to it.

When the voltage of the power source connected to the circuit is...

3.6 volts I want only the first LED to light.

7.2 volts I want the second LED to light. The first LED can light also.

14.4 volts I want the third LED to light. The first and second LED can light also.

Is this kind of thing possible? and how do I make a start?

Thanks,

Karl

EDIT: sorry guys, the dashes were confusing. I'm trying to measure positive voltage. What I'm making is an enclosure system that allows you to make battery packs of various voltages, 3.6, 7.2 and 14.4v. I want to make a measurement tool that confirms there is power and indicates what voltage the pack currently has.

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    \$\begingroup\$ You can google "bar graph LED schematic". There are many possible approaches. \$\endgroup\$ – Wesley Lee Apr 5 '18 at 16:53
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    \$\begingroup\$ Ar those voltages actually negative or are you just setting them off with dashes in front? \$\endgroup\$ – Spehro Pefhany Apr 5 '18 at 17:16
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Here is a straightforward Zener-based circuit which accounts for the forward voltage drop over the LEDs themselves:

schematic

simulate this circuit – Schematic created using CircuitLab

D1 is a 4.7V device, and D2 is an 11V device. These should give enough margin to work with different colours of LED, not just red (which tends to have the lowest forward voltage).

Note that since the full voltage is still applied to the first LED's limiting resistor, that will be the brightest one even when a higher LED is in fact lit. For that reason I do recommend that you use different coloured LEDs, taking advantage of the fact that red appears less bright than other colours.

An alternative circuit using transistors as LED drivers avoids this problem, and yields LEDs with identical brightness:

schematic

simulate this circuit

In this case, D1 is a 5.1V type and D2 is 10V. The precise values are less critical than with the first circuit; the forward voltage that has to be added to the Zener voltage to obtain the threshold is that of the transistor (typically 0.7V), not the LED.

Notice also that all three LEDs now draw from the same limiting resistor. This makes the brightness of each LED more nearly the same regardless of which battery is connected, since although the current through the resistor is increased with higher voltage, that current is shared by more LEDs. However, because the LEDs are now effectively connected in parallel, they should all be the same colour (and hence forward voltage).

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  • \$\begingroup\$ Why do we need to add Zener voltage + transistor Fwd voltage for the threshold value instead of the LED? \$\endgroup\$ – CNA Apr 9 '18 at 3:48
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    \$\begingroup\$ Just look at the second circuit - the Zener is in series with the transistor's base-emitter path, but not with the LED. So the Zener will conduct when the battery voltage rises above the Zener threshold plus the base threshold, and that's when the transistor will turn on, driving the LED. \$\endgroup\$ – Chromatix Apr 9 '18 at 16:51
  • \$\begingroup\$ Hello, thank you very much for your elegant solution. I'm new to this and only beginning to understand. This application I'm speaking of is a battery voltage tester. Does the transistor set-up increase the current draw of the circuit? I'm asking because this circuit will need to draw as little current as possible in order to do its job. \$\endgroup\$ – Karl Peters May 23 '18 at 11:43
  • \$\begingroup\$ @KarlPeters The second, transistorised circuit should consume less power than the first, Zener-only circuit with the higher voltage batteries, and the same amount with the lowest voltage battery. The vast majority of the power goes to lighting the LEDs, and you can tune that by varying the resistor value. \$\endgroup\$ – Chromatix May 24 '18 at 8:44
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The cheapest and easiest solution I can think of at the moment is using zener diodes. Get the zener diodes for the voltages you want to indicate 3V, 7V, and 14V approximately.

Putting an led in series with the zener, and a current limiting resistor should work. The first light will turn on at 3V, at 7V the second light will turn on and the 3V one will remain on, and so on.

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  • \$\begingroup\$ This seems like a good answer, but don't forget to account for the LED's forward drop voltage (typically 1.8 to 3.3 V depending on colour) when choosing your Zeners. You probably don't need a Zener on the 3V indicator; if you're using red LEDs, a 5V and a 12V Zener will correctly select the upper two. \$\endgroup\$ – Chromatix Apr 6 '18 at 18:22
  • \$\begingroup\$ If the first led lights at 3.6v, wont the current become excessive on that led at 14.4v? \$\endgroup\$ – Drew Apr 6 '18 at 18:46
  • \$\begingroup\$ It'll certainly be brighter, but you should choose your limiting resistor for the highest voltage anticipated. I've posted an answer of my own which includes a slightly more sophisticated circuit. \$\endgroup\$ – Chromatix Apr 6 '18 at 19:14
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The device you need is called "dot bar graph driver", most commonly implemented as LM3914 integrated circuit. Here are instructions how to use it. You just need to adjust voltage dividers to suit your particular levels.

enter image description here

The LM3914 can safely operate up to 25V, so just connect the net labeled (5V) to positive terminal of your source, and ground net to negative terminal of the source you need to monitor. The circuit, however, will consume maybe 10-20 mA, so the source should be capable of supplying this current without affecting its voltage.

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  • \$\begingroup\$ All of which don't operate with negative voltages. This isn't an answer, just a copy-and-paste. Downvoting, I'm afraid. \$\endgroup\$ – TonyM Apr 5 '18 at 17:10
  • \$\begingroup\$ (sorry for my rubbish writing, I meant:) To be fair, I don't have to be - you wrote an answer to the question. I just compared the question and your answer then commented. But if you doubted the polarities, how comes you wrote an answer? Using an LM3914 is a way of doing it but please then show a circuit to the OP that does the whole job, rather than imply it. \$\endgroup\$ – TonyM Apr 5 '18 at 18:56
  • \$\begingroup\$ Take care when flagging answers, use the downvote first, use low quality or not answer only if it fits the bill. See this: meta.stackexchange.com/questions/86984/… \$\endgroup\$ – Voltage Spike Apr 5 '18 at 19:01
  • \$\begingroup\$ @TonyM cut people some slack, remember SE's policy is 'be nice'. People can edit they're answers. \$\endgroup\$ – Voltage Spike Apr 5 '18 at 19:44
  • \$\begingroup\$ @laptop2d, I completely endorse the rules and am one of many very polite on the site, ready to discuss and to reach friendly agreement. You can follow that chain here, including frankly silly and snide replies with no goodwill. Those wear out patience. Answers can indeed be corrected, which I've been encouraging but not seen. We're representing professional engineering here, sad to see petulance and impatience. This had all run its course some time back, though. \$\endgroup\$ – TonyM Apr 5 '18 at 20:59
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Use three comparators with inputs at -3.6V, -7.2V and -14.4V.

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Well, this is the simplest as possible. Since diodes are current elements, it matters to them, that enough current is flowing through, which for a regular led is something between ~1mA and no more than 20mA.

(I'm often using LEDs, that you can see, if they're turned on at already 1mA)

3.6V - D1, resisor 960 R, the current will be about 3.75mA 7.2V - D2, resistor 1920 R, the current will be about 3.75mA 14.4V - D3, resisor 3840 R, the current will be about 3.75mA.

This schematics will work just fine and diodes and none of the diodes at voltage 14.4V will not be harmed.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ This won't work as required. Any of the three specified voltages applied will light all three LEDs, with the second and third ones simply being dimmer. \$\endgroup\$ – Chromatix Apr 6 '18 at 18:15

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