Voltage dividers

[windozeuser]

I don't get how to calculate the voltage division to divide the main voltage between n number of resistors. Can anybody please explain the principles and theory of voltage dividers.

 

[Nigel Goodwin]

I don't get how to calculate the voltage division to divide the main voltage between n number of resistors. Can anybody please explain the principles and theory of voltage dividers.

Have you considered ohms law?.

 

[DigiTan]

First find the equivalent resistance through the entire resistor network, then use: I = V/R to get the current through the entire network. From there you can find the voltage drop on any resistor using V=I*R. If any resistors make a formation, you'll need a special formula.

If you only have two series resistors things are much easier. If you know the supply voltage, and you want to get this divided voltage from "Ra," just use this...

Vdivided = (Vsupply) (Ra)/(Ra + Rb)

...so you can see--for example--if Ra and Rb are equal, your voltage is divided by two.

 

[Megamox]

Current in divider = total voltage/total resistance (ie R1+R2+R3...)

Multiply this current by each resistor value to find the voltage drop across it and then just add the potential differences up from ground voltage to find the voltage at any point you're interested in.

Megamox

 

[windozeuser]

Yeah, I was using ohms law for the circuit current (which is constant in series).

I had to design a circuit that had a 12 volt source, and 25mA. There had to be three resistors, with a voltage drop of 8VDC between R1 and R2, and a voltage drop of 2VDC between R2 and R3.


This is the way I did it, Is there an easier way? It seems to work.

The resistors run nice and cool. The total circuit power consumption is
Pt = 12V * 0.025 = 300mW

Considering I'm using half watt (500mW) resistors.

 

[Megamox]

Looks fine, you could easily drop those down to quarter watt ratings without a problem.

 

[Styx]

Looks fine, you could easily drop those down to quarter watt ratings without a problem.

and he said the power he calculated was 300mW :roll:

how you figure he can use 250mW resistors?

 

[Megamox]

Well he's using 3 different resistors and the biggest drop across any of them is 6v, with an I of 25ma, so max power in that would be 150mw..

 

[mstechca]

If we multiply 12 volts by 0.025 amps (25mA), we get 0.3 watts (300mW).
This means that a 1/4 watt resistor won't work without problems.
Use 1/2 watt resistors.

 

[windozeuser]

Hey Thanks for all your replies .

Well he's using 3 different resistors and the biggest drop across any of them is 6v, with an I of 25ma, so max power in that would be 150mw..
Very Happy

Yeah I was starting to wonder why you said 150mW? Cause the individual power ratings of the resistors add up.

PT = PR1 + PR2 + PR3...

Or you can just Multiply the Source Voltage by the Circuit Current (which is the power of the entire circuit).

Although, When I check the power of each resistor:

PR1 = 4v * 0.025 = 100 mW
PR2 = 6v * 0.025 = 150 mW
PR3 = 2v * 0.025 = 50 mW

Ignoring TOTAL Circuit power consumption (which doesn't displace across any of the resistors. Each resistor has a power of less than 250 mW; so I can use 1/4 watt resistors safely.

 

[DigiTan]

Now you're fishing with dynamite! 8)

 

[mstechca]

Ignoring TOTAL Circuit power consumption (which doesn't displace across any of the resistors. Each resistor has a power of less than 250 mW; so I can use 1/4 watt resistors safely.

If I was building your circuit, I will use 1/2 watt resistors.

If you can get enough heat from your circuit, share some of it with the homeless. It is better than throwing a faulty circuit in a garbage can :wink:

 

[DigiTan]

Why 1/2 watt? They're space hogs.

 

[Megamox]

Single resistor = 300mw/500mw = 60% of its maximum power rating
3 Resistors, Max power = 150mw/250mw = 60% of max power rating
So using 3 1/4 rated resistors gives the same margin of power useage as a single 1/2 watt which you would find acceptable, looks the same to me

Megamox

 

[mstechca]

Single resistor = 300mw/500mw = 60% of its maximum power rating
3 Resistors, Max power = 150mw/250mw = 60% of max power rating
So using 3 1/4 rated resistors gives the same margin of power useage as a single 1/2 watt which you would find acceptable, looks the same to me

Megamox

so how does 150mW enter the picture when the wattage we are dealing with is 300mW?

The only way that 1/4 watt resistors can be used is to increase the resistance of at least 1 resistor.

Why 1/2 watt? They're space hogs.

Because in Canada, 1/2 watt is the closest to 1/4 watt resistors.

If in your area, you can find any resistor that can handle 300mW and the maximum resistor rating is less than 1/2 watt, I like to see a picture. and don't show me a 1/4 watt one.

 

[DigiTan]

so how does 150mW enter the picture when the wattage we are dealing with is 300mW?

The only way that 1/4 watt resistors can be used is to increase the resistance of at least 1 resistor.

If in your area, you can find any resistor that can handle 300mW and the maximum resistor rating is less than 1/2 watt, I like to see a picture. and don't show me a 1/4 watt one.

I can't provide any Kodak moments, but to put things in focus: 300mW is the total combined dissipation on R1+R2+R3. Right now, only the individual dissipations are in the limelight. In fact, you could crop things down to a 1/8th watt resistor for R3 and R1. Panning back to the diagram, windozeuser said his supply is 12V @ 25mA constantly, so our money shot is:

R1 = 4V/25mA = 160Ω
R2 = 6V/25mA = 240Ω
R3 = 2V/25mA = 80Ω

Zooming in on the individual dissipations, you'd get (25mA)²*160Ω = 100mW on R1; (25mA)²*240Ω = 150mW on R2; and (25mA)²*80Ω = 50mW at R3 for the photo finish. A snapshot of windozeuser's results reveals these power figures are double-prints. The 300mW figure doesn't steal the spotlight until all three dissipations are added up.

Just be sure to use my favorite resistor type: carbon film. (One last photography pun there. :lol: )

 

[_3iMaJ]

In case math works better for you

Here is a mathematical description of voltage division:

 

[audioguru]

Gee, I didn't know they made resistors from old Kodak film.
Do they over-expose it to make poly-capacitors?

 

[mstechca]

I hate it when greek leters enters math.

 

[DigiTan]

Gee, I didn't know they made resistors from old Kodak film.
Do they over-expose it to make poly-capacitors?

That's what I figured, but the thought makes me shutter. :lol: