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Voltage divider with 10kΩ potentiometer

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SPDCHK

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I can calculate the volt drop over a given resistor (in a series circuit) using Ohms and Kirchoff's laws. e.g. figure below...
**broken link removed**

but for the life of me I cannot seem to reverse engineer the formula to calculate values for two unknown resistors with known volt drop values.
**broken link removed**

How do I calculate the values for resistors R1 and R2, to give me an output voltage on Potentiometer VR1 of between 2V and 4V with the value of VR1 = 10kΩ, supply voltage = 5V (Output form PIC micro controller).

I want to use this output voltage to drive a IRFP450 N-Channel MOSFET. According to the datasheet my VGS is 2V (min) and 4V (max).

Thank you
 
You're confusing yourself with the pot! - pretend it's just another resistor, with the slider at the top the bottom resistor is increased by 10K, and at the bottom the top resistor is increased by 10K. So work out those two seperate circuits! - this makes the circuit the same as your top one.
 
That does unfortunately not solve the problem. I still have the two unknowns R1 and R2. Even if I assume a value for the pot. I still cant calculate Rtotal

Eg Rtotal = Pot(x)Ω + R1 (?) + R2 (?)

The values for R1 and R2 will be static once used within the circuit. Using Ohms law to calculate the circuit current for one assumed value for the Pot will obviously not be the same for the pot at full stroke.

I found this **broken link removed** where the author calculates the values for R1 and R2. I would like to know how he did it.
 
You're thinking about the problem the wrong way (thinking you need Rtotal). Nigel told you exactly what you need to do. Let me reword your thinking for you:

You have two unknowns, therefore you need two equations. You have the two circuits Nigel described:
1. One with the pot wiper all the way at the top with an output voltage of 4V
2. and one with the pot wiper all the way at the bottom with an output voltage of 2V.

Each circuit will give you one equation. You have two unknowns R1 and R2.
Two equations + Two unknowns = solution!
 
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Think of the voltage drops across each resistor. The left drawing (attached) shows this - 1V across R1, 2V across RV, and 2V across R2. The bottom end of the potentiometer is thus 2V higher than 0V (0V + 2V = 2V), and the top end of the pot is again 2V higher than the bottom end (2V + 2V = 4V)

There are a couple of ways to calculate R1 and R2; let's start with Ohm's law to calculate the current through the resistor chain. It's the same current passing through all three resistors (Kirchhoff's current law). Since we know the voltage across RV, and the resistance of RV:

[latex]I = {\frac {V_V}{R_V}} = {\frac{2V}{10k\Omega}} = 200\mu A[/latex]

We now know the current through R1, and the voltage across it, so we can calculate R1's resistance:

[latex]R_1 = {\frac {V_1}I} = {\frac{1V}{200\mu A}} = 5k\Omega[/latex]

Similarly for R2:

[latex]R_2 = {\frac {V_2}I} = {\frac{2V}{200\mu A}} = 10k\Omega[/latex]

You could calculate them another way - by understanding that the voltage dropped across each resistance is in proportion to the resistance. That is, the ratio of resistances is equal to the ratio of voltages dropped:

[latex]R_1 : R_V : R_2 = V_1 : V_V : V_2 = 1 : 2 : 2[/latex]

This means that RV and R2 have the same resistance, and R1 is half of that resistance. The values are easy to see intuitively.

A third way is to use the potential divider rule to derive equations that can be solved by rearrangement:

[latex]V_1 = \frac{5R_1}{R_1 + R_V + R_2} = 1V[/latex]

[latex]V_V = \frac{5R_V}{R_1 + R_V + R_2} = 2V[/latex]

[latex]V_2 = \frac{5R_2}{R_1 + R_V + R_2} = 2V[/latex]

I'll leave you to figure those out.
 

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Aha she cried!!

Thanks Cabwood, I finally see the light. I now know what I've been doing wrong the whole time. I kept on changing the value of RV from 0 to 10k and trying to calculate with the different resistance values, where all along, the RV value NEVER changes within the series circuit. The total current throught the whole loop will always be 200μA, no matter where the POT wiper is.

Thanks guys ;)
 
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