What resistor to drop from 5V to 1.4V? - help

[zGhost]

I'm trying to drive my temp monitor off of the 5V line out of the PSU instead of off a battery. Currently the unit is using a 1.4V battery and doesn't seem to draw but about 100uA. And of course I can't remember the damn formula to calc this out so... help! Please...

 

[CTho9305]

unless your current drain is very constant, using a resistor wont supply a very steady voltage. you *could* use a few diodes (5 .7v diodes) at a current that low, but they probably would also not be good enough.

 

[pm]

The formula is V=IR. In this case, you want a voltage drop of 5V - 1.4V = 3.6V. You say that it draws 100uA - seems a little high to me - that's 0.1mA... seems high. But anyway if it is 0.1mA, then it's R = V/I = 3.6V / 1E-4 = 36kOhm resistor. But CTho is right. If the current draw varies this method will result in a fluctuating voltage. Also if your guess is off by say, a factor of 10 (it draws 10uA instead) then you will end up with 5V - .36V = 4.64V and this voltage may be high enough to blow up the chip.

If you want 1.4V, then I'd recommend taking the 3.3V off of the motherboard voltage regulator and using that as a "GND" signal and then use 5V as the source voltage. Then you'll get 1.7V which is surely close enough for tolerance. Alternatively, you could grab a CPU Vdd/Vcc pin and use that instead. Most CPU's run in the 1.5V range nowadays.

 

[blahblah99]

That's easy.. get a 1kOhm resistor in series with 2 1N4148 diodes to get you the 1.4V drop.

The circuit goes something like this:

+5V--------resistor----->|-->|-------GND

Your temp monitor goes between the resistor and the first diode. You can pick up all these stuff at any electronic parts store, or radio shack.

If 1kohm doesn't give you the 1.4V, then lower it to around 800, or 560.

 

[Sahakiel]

I think 1kOhm is too little.
If I'm calculating correctly, and the monitor only draws 100uA, then the effective resistence is 14kOhms. Sticking a 1kOm resister in front of that would result in about 4.67V drop across the monitor.

Try something like this:

+5V
|
[27 Ohm]
|
+----[10 Ohm]----GND
|
[Temp Monitor]
|
GND


That should give you slightly under 1.4V with perfect components. Very likely, the component tolerences would cancel things out to give close to 1.4. Just make sure the current is 100uA, cause if I remember correctly, Radio Shack's resistors have a 1/2 W tolerence.
Or, if you're lazy, hook up a rheostat to your voltmeter and set it to output 7/25 of the input voltage.


This is all assuming I haven't forgotten my EE. Feel free to correct me if I'm wrong.

Edit: One more thing. Feel free to up the resistence by an order of 10 or more. As long as the ratio is the same and you don't raise it too high, current will still flow enough to drive the monitor and the voltage across the monitor will still be 1.4V.

 

[zGhost]

thanks for all the input.... All the resistor values were giving me fluctuating voltage outputs.... So I'm guessing that my current draw varies. Guess I will have to go with a diode of sorts....

pm: great idea about taking the CPU voltage.... think I'll try that.

Oh and by the way right after I posted I remembered that R=[Vs-Vo]/I was what I was trying to remember
R - Resistence
Vs- Voltage Source
Vo- Voltage required/out
I - Current

 

[Moohooya]

You could either use a zener diode and a resistor, or use two resistors in series and use the center tap. Using a zener should give you a very stable voltage.

 

[Yossarian451]

Maybe it has been too long since I did any electronics, but couldn't he use a capacitor to stabilize it then use a resistor to change it?????

Been I while since I tried to teach myself elctronics.

 

[element]

Originally posted by: Yossarian451
Maybe it has been too long since I did any electronics, but couldn't he use a capacitor to stabilize it then use a resistor to change it?????

Been I while since I tried to teach myself elctronics.

Erm no. The capacitor will only clean choppy DC at best. That is what it is used for in DC power supplies. What he needs is a voltage regulator, but of course PM's post is right on in that he can just use the available voltage regs on the PS/mobo to accomplish this. A lot easier than coming up with your own voltage regulator circuit.

 

[Yossarian451]

Sorry I thought it was DC, was just think that is how I had always heard you can clean dc, 4 diodes and a capcitor then a resitor.

 

[JackMDS]

Silicon diodes when conducting current forward reduces the voltage by .7V regardless off the current. The .7v is the potential needed to open the Silicon gate.

It is a little cumbersome, but connecting 7 diodes (you can buy the cheapest they go for few cents) using serial connection to the temp gizmo, the 5V will be 1.5V stable.

5V -Diode - Diode - Diode - Diode - Diode - Diode - Diode - 1.5V

BTW: Germanium diodes reduce the voltage by .1V

 

[IRJack]

pm has a good suggestion with using the 5V and 3.3V as ground. If 1.7V isn't close enough for your tolerance, you could add a series diode (germanium type - thought the drop for that was *theoretically* 0.3V) then throw a couple of decoupling (parallel) capacitors on the output for stabilization. I'd use something like a 10uF and 0.1uF. That's kind of optional, but I'd recommend at least using the 0.1uF cap.

Circuit:

5V - - - - (series diode)- - - (parallel cap to ground) - - (parallel cap to ground) - - - - Output
3.3V = ground

Just don't actually tie your 3.3v to your PS ground.

Don't actually count on a consistent 0.7V drop across all silicon diodes neither - most that I use run in the 0.5 - 0.6 range.