Why can't I use a voltage divider?

I am working on building my own firework controller. the whole thing will be powered by 6v dc, with the exception of the test circuit this going to run on 1.5v (I am using an optocoupler to separate it from the 6v). I was just going to use 2 resistors to make a voltage divdered to get the volatge down to 1.5. I just read however that "voltage dividers should not be used to supply power to a load". My question is why cant I use this method? When do you actually use this method? and what should I use to properly reduce the voltage? 

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-max-4 months ago

Output resistance. (ESR) Thats why.

An ideal DC voltage suplly has 0 ohms output resistance, which means the voltage does not sag as you draw power. A betttry is going to have a resistance much less than 1 ohm if its fresh (50mOhms is a realistic ESR)

But with a voltage divider, the output resistance is equal the the resistance of your reistors as if they are wired in parelell, plus the ESR of the DC supply you use. This means that if you know how much current the load draws, you can very easily figure out the voltage drop the supply has. And with something like several THOUSAND ohms of ESR, I can tell you, it's gonna suck!

If you really need a quick and dirty 1.5V supply, try just connecting a resistor directly in series with your load, basing the resistor value on Vdrop between the supply and load, and the expected maximum current draw. Choose a resistor value smaller than that, and connect a 2 or 3 diodes in series "shorting" the 1.5v output side to ground. The diodes will act as a crude 0.6x3 voltage shunt. They will create a hard voltage limit. Alternatively you can use a IR or red LED, which also gas a Vf of 1.2 -- 1.8V or so. I do this all the time wgen I need additional voltage rails and dont give 2 sh*ts about efficency or heat.

-max- -max-4 months ago

Of course, also make sure the diodes can handle the surplus current when the load is drawing the minimum current. Any current the load is not drawing is taken by the diodes. 1N400x series are good.

seandogue1 year ago

Voltage dividers are *generally good only as references. Adding a load resistor effectively reduces the lower legs' resistance by paralleling the load with that resistor.

By doing so, the output voltage is also reduced.

In general, a load resistance applied to a simple voltage divider should be no less than 100 times the value of the lower divider resistance to minimize the effects.

Suppose for instance, that you have a voltage divider where

R1 = 10K

R2 = 10K

Therefore, with no load

Vo = Vi x (R2/(R1+R2) = Vi x 10K/(10K + 10K) = Vi/2

Now, apply a load resistor of 10K ohm

from the resistance in parallel equation

R2||Rload = (R2 * Rload)/(R2+ RLoad) = 10K*10K/(10K+10K) = 5K

the **effective output voltage is now

Vo = Vi x (5K/(10K+5K) = Vi/3

Quite a difference aye?

You expected 1/2 Vi, but you got 1/3 Vi. Oops.

Now try the same math with a Rload = 100K.
And then try it with a Rload= 1000K (1M) .

I think you'll see the point.

Most often, when we professionals use v-dividers, we follow them with a buffer which has very high input impedance and very low output impedance so it can drive current without impacting the desired voltage.. Opamps are often emplyeed for the purpoise.

iceng1 year ago

The opto-coupler is an LED so just use the LED3 but add a series resistance like 470 ohms or LED3 and the NPN will be damaged.

You don't want to use a voltage divider where the ignitor resistance is not guaranteed to be consistently the same from unit to unit because it will change the voltage..

Do I understand correctly, that there is a say a 30ma test if the ignitor is wired correctly and later fired by applying unrestricted 6 volts DC ?

Klick the Pic to see whole image.

Mpc1055 (author)  iceng1 year ago

I like this configuration, Thanks!!!

Show your circuit. There ARE occasions when a simple divider can be used. This might be one of them, usually when the current required is very low. A better method is to use a resistor and zener diode,

Mpc1055 (author)  steveastrouk1 year ago

My apologies with the schematic doesn't make sense, I am very new to electronics. Where it says "e-match" there will be a FalconII ignitor hooked up.http://www.firingsystems.us/Electical-Igniters-1-...

What i have read on the falcon igniter is that they work in 6-12volts and you shouldn't exceed 50ma when you test. This is one of the reasons I chose an optocoupler. My thinking is that if i pass 1.5v with low current there is now way it will heat up the nichrome filament but it will activate the led in the optocoupler.

Screen Shot 2016-02-03 at 7.57.59 PM.png

You'd need less than 10mA to make a red LED glow very brightly, the opto is redundant here, all you really need is a resistor > 480 Ohms in series with the igniter and the LED. Even a dead short across the igniter terminals would still allow only 12mA through.