# LM339 set-points

Discussion in 'General Electronics Chat' started by Dritech, Dec 1, 2013.

1. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
756
5
Hi,

How can I calculate the values for the resistors which will be used in a voltage division configuration to set 6 different voltage set-points?

So basically I know Vcc, the set-points voltages and need to determine the resistors' values.

Sep 20, 2005
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3. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
What voltages do you need ?
You can calculate a quick divider string by thinking of it as 1000 ohms per volt (for example).

Last edited: Dec 1, 2013
4. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
756
5
Thanks for the replies.

But that can only be done when having 1 voltage output right? In this case I need 6 outputs

The voltages needed are 50mV, 200mV, 250mV, 300mV, 450mV, 500mV.

5. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
What is your supply voltage and available current?
And is the supply voltage well regulated ?

Last edited: Dec 1, 2013
6. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
756
5
The supply voltage will be 12V which will be regulated to 10V using a zener diode. Is the available current of the supply? If so, I will be using a bench top power supply so the current can go up to 5A.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
You can use this voltage divider.
And for R7 you can use 180K + 10K or 180K + 51K (precision trim potentiometer). For 3K you can use 3x1K resistor or two 1.5K resistors.

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8. ### #12 Expert

Nov 30, 2010
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I did it with the 1 ma theory.

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9. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
756
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Thanks for the reply Can you please tell me how the values were determined? Is the value of R1 assumed?

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes you can assume R1 value and then use Ohm's law to find the current
I = 50mV/1K = 50uA And next use Ohms low and find the remaining resistors.
R2 = (200mV - 50mV)/50uA = 150mV/50uA = 3K and so on.
Or you can assume current as #12 did. But he made a small error.
He assume 1mA.
So from then R1 = 50mV/1mA = 50R;
R2 = 150; R3 = 50; R4 = 50; R5 = 150; R6 = 50; R7 = (10V - 0.5V)/1mA = 9.5K = 9.1K + potentiometer

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11. ### tubeguy Well-Known Member

Nov 3, 2012
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That's where thinking of the divider as 1000 ohms per volt is very useful.
Ohms law states that 1volt/1000ohms = 1ma. 1ma is a convenient low current which works well in circuits like this.
The 1ma current flows through the whole resistor string and each resistor has a voltage across it based on it's resistance.
So, 1000 ohms = 1 volt(1000mv) drop, 50 ohms = 50mv drop etc,,,

One thing that can easily trip you up is that you need to calculate the drop individually across each resistor in the stack.

Last edited: Dec 1, 2013
12. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
756
5
Thanks for the replies Jony130 and tubeguy.