Need help choosing the right resistors for a circuit

[Ian]

Hi
I need a resistor calculator for calculating the right resistors you need for a circuit.

or if you know how to do it here is the voltages and stuff. I am supplying voltage for two fans. Thats all. Also if you know where to get a electronics guide please tell me.

My input voltige is; 32 940m

output i need 24v 20ma ( i am going over a little what i need for my fans i wont matter)

input voltage 32 940ma

output voltage 12v 22ma

please i need help

[avrfreaks]

I believe this is what you want.

Use Ohms Law: V = I x R or Voltage = Current x Resistance

24V / 20mA = so 24V divide 20e-3 = 1.2K ohms

12V / 22mA = so 12V divide 22e-3 = 545.45 ohms

This also might help: http://www.tscm.com/elecform.pdf

[cybergibbons]

Maybe best to take a step back and ask what exactly you are doing.

Are you saying you  have a 32V supply, and wish to supply two fans, one which is rated 24V/20mA, the other 12V/22mA?

That sounds like a very, very small current for a fan though. Are you sure it is right?

The calculations in the second post aren't quite right.

If you have a 32V supply, and want the fan to see 24V when it is pulling 20mA, that means that you need to drop 12V over a resistor when 20mA flows through it. i.e. 12/0.02 = 600?. You then need to take into account power - P = VI, so 12 * 0.02 = 0.24W - so a quarter watt resistor would be OK.

This is all very similar to working out LED resistors.

I'll leave it as an exercise for the reader to work out the other.

Anyway, back to the original issue -using resistors isn't the best way of dropping voltage to supply a fan. Firstly, you waste all the power as heat in the resistor. Secondly, a fan that says "20mA" on it doesn't always draw 20 mA. It will use more when it starts up, more when dust starts to block it, and there will be a lot of variation anyway.

With the start-up, the fan will draw more current, causing more voltage to be dropped over the resistor, reducing the supply voltage, and in turn making it harder for the fan to start. With a DC brushless fan, it may not even start up at all.

Why not use a regulator?

[saturation]

Hi CG,

I concur completely.

I think our friend is very green, so lets finish the need, and hopefully decides to look into electronics more.

On the second resistor you need 900 ohms and 0.4W.  However, its best to derate the power rating on resistors, so you'll need 1 Watt to reduce the risk of resistors overheating and breakdown.  On the 0.24W resistor mentioned earlier, I would use 0.5 watt.  Since resistors are cheap in that watt range, its a conservative move when you dissipate power continuously.

If you have a 32V supply, and want the fan to see 24V when it is pulling 20mA, that means that you need to drop 12V over a resistor when 20mA flows through it. i.e. 12/0.02 = 600?. You then need to take into account power - P = VI, so 12 * 0.02 = 0.24W - so a quarter watt resistor would be OK.

This is all very similar to working out LED resistors.

I'll leave it as an exercise for the reader to work out the other.

[Ian]

Hi
Thanks so much.

[cybergibbons]

On the second resistor you need 900 ohms and 0.4W.  However, its best to derate the power rating on resistors, so you'll need 1 Watt to reduce the risk of resistors overheating and breakdown.  On the 0.24W resistor mentioned earlier, I would use 0.5 watt.  Since resistors are cheap in that watt range, its a conservative move when you dissipate power continuously.

Very true - a normal resistor will get warm at full rated power - and I think (though correct me if wrong), they are only rated for that power when mounted on a PCB (i.e. free air, and copper track to dissipate heat). And for the money, 0.25W->0.5W isn't going to break the bank.

[saturation]

Hi CG,

Yes that's right.  Officially you can 'spec' a derating, akin to this:



But practically speaking if you double the calculated required wattage you're good to go for most conditions.  If the resistor is too hot to touch, its not properly derated.  The derating matters most if the ambient temperature rises, such as a poorly ventilated or cooled case.

A real problem is not the resistor, but the voltage regulation provided by this approach.  If the heat rises, the resistance drops, causing more voltage to appear across the load, making things a little hairy for these fans.

Although a semiconductor approach can provide much better control, the resistor method is simple, direct, and if derated, more than adequate.

On the second resistor you need 900 ohms and 0.4W.  However, its best to derate the power rating on resistors, so you'll need 1 Watt to reduce the risk of resistors overheating and breakdown.  On the 0.24W resistor mentioned earlier, I would use 0.5 watt.  Since resistors are cheap in that watt range, its a conservative move when you dissipate power continuously.

Very true - a normal resistor will get warm at full rated power - and I think (though correct me if wrong), they are only rated for that power when mounted on a PCB (i.e. free air, and copper track to dissipate heat). And for the money, 0.25W->0.5W isn't going to break the bank.

[Strube09]

Ian.

Check out this little bit of software. I use it to quickly develop divider networks and many other circuits.

It also makes sure it selects from real value resistors.

The only bad thing is that if you to pay for it you have to wait like 30 seconds for the results. But it seems to work pretty good.

http://drn.digitalriver.com/free/Electronics/Resistor-Selector/60261.html

[jahonen]

For divider and parallel optimizations, I wrote a little free utility called resistor optimizer (works at least in Windows 7 and XP).

Regards,
Janne

[Zad]

Or use 12v and 24v 3-terminal regulators.

[Zero999]

For divider and parallel optimizations, I wrote a little free utility called resistor optimizer (works at least in Windows 7 and XP).

Regards,
Janne

Looks good but surely there's a faster method than trying 1000s of possible resistor combinations?

There's a formula which can be used to calculate E values, or better still use a look up table.

Once you've worked out the divider ratio, loop varying R1 from 1R to 10R, calculating the value of R2 rounded to the nearest preferred value, then you can scale, sort the results and trim to the user's preference. This will cut the number of combinations tried to the E-number specified. You could also have an option to use say E96 for R1 and E24 for R2 and possibly another option to use both E96 and E24 values.