# Can someone make LED voltage level indicator, no zeners!

Discussion in 'Homework Help' started by annahughes, Apr 16, 2016.

1. ### annahughes Thread Starter New Member

Apr 16, 2016
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0
Hi! I was wondering if someone could show me how to make a voltage level indicator with LED diodes, resistors, and voltage sources. Please do not include zener diodes. On my lab sheet they feature a voltage divider circuit but they have resistors on the branches with the diodes, I was wondering if they were necessary. I have attached the link to the lab and the info sheet.
http://www.bu.edu/eng/courses/ec410/documents/ec410-lab2-spring08.pdf

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,395
497
The resistors in the branches are there to control the current through the diode. If you are using regular/cheap red led then the maximum current should be 20 mA, most people like to run them at 15 mA. If you got 20 mA or less going through the branch, then you can skip the resistor.

The formula for calculating resistor value of the diode current control resistor is:
$
R_{1,2,3}=\frac{V_{IN}-V_{1,2,3}}{I_D}
$

- Vin is the voltage entering the circuit, you, as designer, should know what it is.
- V1,2,3 is the voltage source in each branch, since you are the one designing the circuit, you should know what it is.
- Id is the current through diode that you, the designer, want. You don't want too much current because you will burn the diode with it. The regular red led has a max of 20 mA, green is a little higher if I remember right, I have no idea what blue led use. You should be able to google the info.
- So, you know all the variables, plug the numbers in and solve for R(1,2,3) of each branch.

Last edited: Apr 16, 2016
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3. ### annahughes Thread Starter New Member

Apr 16, 2016
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0
Thanks this explains everything!

What about the first resistor that was in series with the input voltage? Does that affect the calculation?

How does Rx place in the R1,2,3 calculation?

Last edited by a moderator: Apr 17, 2016
4. ### dannyf Well-Known Member

Sep 13, 2015
1,825
364
The calculation laid out earlier is not the right approach to follow.

First of all, you have to look at your LEDs and get two parameters: at what point (of Ifwd / Vfwd) that you consider an LED is "lit up"?

Most people would consider an LED to be reasonably on at 1ma and up and Vfwd varies. Let's say that you are using white leds, and you think 1ma on @ Vfwd = 3v.

So you should pick the serial resistors so that at the lowest Vin, the first LED reaches 1ma@3v. Let's say that we want that to be 5v. So the voltage drop over R1 is 2v (=5v-3v) and its resistance is 2K(=2v/1ma).

For the 2nd LED, you want it to kick in at 9v. That means the voltage drop over R2 is 6v (=9v-3v), and its resistance is 6K.

....

Once you have picked them all, you can pick R0 (the overall serial resistor) so that it can deliver sufficient current all all levels. It is not a bad assumption to make it a fraction of R1//R2//R3. Like half of it.

The critical thing here is that the design is crude in that the brightness does vary, as the leds don't have a sharp on / off point.

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,395
497
Yes, I noticed that problem when I did a quick simulation of the circuit. In my simulation I did not use Rx at all. That way my Vin was in parallel with the three diode branches, therefore the same voltage was across each branch. Using that I calculated the R1,2,3 resistors for each branch. For V1,2,3 I used AA/AAA batteries so I have V1=1.5 volts (one AA/AAA battery), V2=3 volts (two AA/AAA batteries in series), V3=4.5 volts (three AA/AAA batteries in series).

Regarding Rx.
The three led branches form a Current Divider. If you use the red led in your bar graph, the max. current through the led is 20 mA, normally people run 15 mA so I will use 15 mA. So our current divider divides current into three 15 mA streams. The total current in the current divider will be 15+15+15=45 mA. Rx is located in front of the current divider so all of the 45 mA will pass through it. Now just pick Rx, say 1 Ohm (or 2 Ohm or 100 Ohm, etc.). Apply Ohm's Law: 1 Ohm*45 mA=45 mV. 45 mV will be the voltage drop across Rx. You take your Vin, subtract 45 mV, you just removed 45 mA from your Vin, this is your New Vin, plug it into formula I gave you earlier and calculate R1,2,3 for each branch. Basically the questions is: How much voltage do you want to remove from Vin before the voltage enters the bar graph portion of your circuit? If your Vin is 100 volts, you will want to remove a lot, and you do it at the Rx portion of your circuit.

Last edited: Apr 16, 2016