Electrical – How to calculate current(I) when I don’t know voltage drop

ohms-law

So everyone knows Ohms law (\$I=\frac{V}{R}\$) .

Let's say I have a 9V battery and a LED. Furthermore the LED only needs 1.9v so the voltage drop of my resistor would be: \$9-1.9 = 7.1\mathrm{V}\$

I have a 1k resistor (1000 ohms) so now to calculate current we have to plug it all in…

$$I=\frac{7.1}{1000} = 0.0071\mathrm{A}\space\space (7.1\mathrm{mA})$$

But what if I change my mind and I don't want my 1k resistor to drop 7.1V but 7.2V instead, so now it would look like this:

$$I=\frac{7.2}{1000} = 0.0072\mathrm{A}\space\space (7.2\mathrm{mA})$$

And now this is the part where I get confused. Because I can't control the voltage drop of my resistor so it can either be 7.2V or 7.1V. But I don't know which one is the right one.

I could say I want my 1k resistor to drop only 1v so then again:

$$I=\frac{1}{1000} = 0.001\mathrm{A}\space\space (1\mathrm{mA})$$

So and if I don't know what my voltage drop is then I don't know current either. So can someone explain this to me?

Best Answer

Normally, you don't want a fixed voltage for your LED. You want a current. (\$20mA\$ not uncommonly.) So you are focused on the wrong goal.

But let's say you really do want to set up some fancy voltage regulator for your \$9V\$ battery. (Instead of doing the sensible thing about setting up the current, instead.) You can try this. Just use a \$100\Omega\$ resistor where the LED goes, at first, and adjust \$R_2\$ until you measure a voltage across the \$100\Omega\$ resistor that you like (as shown, where the (-) and (+) appear in the schematic and where you place your voltmeter for measuring.) Then you can replace it with the LED and see how that goes.

schematic

simulate this circuit – Schematic created using CircuitLab

You can play with the potentiometer, \$R_2\$, to then adjust the voltage at your LED. Should only waste an excess of \$2mA\$, regardless of the set voltage. The maximum output voltage will be \$6V\$ with a current compliance of up to \$100mA\$. (\$Q_4\$ may need to be a TO-220 packaged type, just in case.) The problem will be that you will have a very hard time adjusting the potentiometer without having a serious impact on the LED brightness. But it can be a learning experience.

Or you could just use a variable voltage regulator IC.

But you really should just shoot for setting a current using a simple resistor, using the formula \$R=\frac{9V - 1.9V}{20mA}= 355\Omega\$. Then go get a \$330\Omega\$ or a \$390\Omega\$ resistor. It will work just fine. Then measure the voltage across your LED, too. See where it is.