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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?

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  • \$\begingroup\$ Could you please rephrase to make your question clearer? Try making it more objective and then showing examples of your thought process. Not both at the same time. \$\endgroup\$ – Wesley Lee Sep 7 '16 at 1:28
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    \$\begingroup\$ You need to look up "load line" on the intergoogleweb, methinks. When you mix the LED VI characteristic with the load line, you will find your operating point in V and I. As in here: en.wikipedia.org/wiki/Load_line_%28electronics%29 \$\endgroup\$ – Sredni Vashtar Sep 7 '16 at 1:35
  • \$\begingroup\$ yes and im sorry... lol \$\endgroup\$ – CosnotraLF Sep 7 '16 at 1:36
  • \$\begingroup\$ @Sredni Vashtar: so since you know this... can you tell me what the voltage drop would be across the 1k resistor when i have a led that needs 1.9V and 20 mA of current \$\endgroup\$ – CosnotraLF Sep 7 '16 at 1:47
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    \$\begingroup\$ @UmarMahmood: My calculator says you'd need 355 Ohms for 20 mA. \$\endgroup\$ – Peter Bennett Sep 7 '16 at 2:16
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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.

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  • \$\begingroup\$ I want to give you +1 for printing out that LEDs need a specified current, but I don't think you should have indulged the rest of the question... \$\endgroup\$ – Mark Ch Sep 7 '16 at 21:25
  • \$\begingroup\$ @MarkCh: I probably shouldn't have engaged it, at all. It was crazy-minded at the outset. But sometimes a push like this gets them off the dime, so to speak. Anyway, that's how I played it. \$\endgroup\$ – jonk Sep 7 '16 at 21:28
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There is nothing you can do except maybe get a variable resistor to control your current

Also if you were to drop 7v with the 1k resistor then you would be expecting 7.0 mA

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  • \$\begingroup\$ oh ok but is it possible to drop all 9v with the 1k resistor so that the led would get 0v... Can you tell me how to do that.. \$\endgroup\$ – CosnotraLF Sep 7 '16 at 1:42
  • \$\begingroup\$ No, the LED will drop the voltage and so will the resistor. The LED will always drop 1.9v. You should change your voltage supply if possible. A resistor could be used to limit the amperage below the operating target of the LED. Could you please expand on what exactly you wish to do. \$\endgroup\$ – Umar Mahmood Sep 7 '16 at 1:48
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    \$\begingroup\$ Your LED will drop approximately 1.9 volts if there is any significant current flowing through it. The voltage will drop slightly as the current is reduced, eventually reaching zero as the current reaches zero. \$\endgroup\$ – Peter Bennett Sep 7 '16 at 2:27
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But what if I change my mind and I don't want my 1k resistor to drop 7.1V but 7.2V instead

If you're using a 9V supply and a 1.9V LED, then any resistor will always have 7.1V across it.

Perhaps you're not familiar with the "voltage divider" concept? The nine volts of the power supply will automatically distribute itself across the chain of components connected to it it. If we have a 1.9V LED and a resistor, the resistor will always have 7.1V across it, where 7.1V plus 1.9V adds up to be 9.0V, the same as the power supply voltage.

In other words, you don't get to change your mind and have 7.2V instead. The LED gets to decide, not you. (Well, I guess you could instead raise the power supply up to 9.1V instead of 9.0V. Or, instead use a 1.8V LED with your original 9V supply.)

Note that all of the above is describing a simplified case. It's using an ideal LED: an ideal LED which always shows 1.9V across its leads, even at zero current. A real LED won't have a fixed voltage, but instead will exhibit an exponential V-I curve like any real-world diode. A real LED might have 1.9V at ten or twenty mA ...but this voltage will be much lower at a microamp, and much higher at 500mA. If you want to really understand everything that's going on, you'll have to knuckle down and study the exponential diode equation of voltage versus current.

That, or go and play around with a circuit simulator which uses a full-blown mathematical model of an LED, and not just a simplified approximation where the LED Vd is fixed at 1.9V.

is it possible to drop all 9v with the 1k resistor so that the led would get 0v

Just replace the 1K with an infinite-ohms resistor! In other words, an open switch. But this won't work for a simplified ideal LED, where the voltage across it remains 1.9V forever, even when the battery is removed. :)

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  • \$\begingroup\$ I'm not sure this is all that helpful. It assumes an diode with a perfect Vd. If you change the resistore, the voltage drop across the diode changes. It might be small, but it changes. \$\endgroup\$ – Scott Seidman Sep 7 '16 at 20:59
  • \$\begingroup\$ OP appears to assume that LEDs have a perfect Vd, then wants to change the resistor voltage without changing the Vd. It's an oversimplification, and the oversimplification is the source of OP's confusion. To understand the odd behavior of real diodes, instead look at Shockley Equation I = Is * e^(V/vt), see what it says about diode voltage at various currents. \$\endgroup\$ – wbeaty Sep 8 '16 at 2:02
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You can't "change your mind". There is a linear relationship between voltage and current for a resistor, and there is a non-linear relationship between the two for a diode. Both relationships can be expressed as an equation of the form I=f(V). Thus, there are two equations for the system. The simultaneous solution of both equations tells you where the system will live.

From my answer, https://electronics.stackexchange.com/a/151653/11684, which points to http://i.stack.imgur.com/1cUKU.png enter image description here, you can see that this can be solved, or at least conceptualized, graphically as the intersection of the two lines. You change the resistor, that changes the slope of the line, and the new intersection point tells you your new equilibrium point. You need a certain voltage drop? Draw a vertical line at that voltage. Connect the point where the vertical line intersects with the diode curve, connect that point to Vcc, and now you know what resistor you need to acheive that from the slope of the resulting line. In other words, you pick your parameter, and the set of simultaneous equations fills in the rest.

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  • \$\begingroup\$ Oh... Thx...+1.... 👌🏻 \$\endgroup\$ – CosnotraLF Sep 7 '16 at 21:35
  • \$\begingroup\$ What's vdd and vd... I think Ik what vd is... Is voltage drop but what's vdd \$\endgroup\$ – CosnotraLF Sep 7 '16 at 21:36
  • \$\begingroup\$ @CosnotraLF Vdd is your power supply voltage, and Vd is the voltage drop across the diode. \$\endgroup\$ – Scott Seidman Sep 7 '16 at 21:46
  • \$\begingroup\$ ok i got it.. i know how it works know but... there is a little problem... how would that function look: Forward Current(I)=f(Forward Voltage(V)) or I=f(V) like is thre a way to calculate those numbers.. because this app called "EveryCircuit" does it and i wanna know how it calculates it.. \$\endgroup\$ – CosnotraLF Sep 8 '16 at 16:38
  • \$\begingroup\$ like would it be y(I)=2^x(V)... or no? \$\endgroup\$ – CosnotraLF Sep 8 '16 at 16:39
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Find out what current, I, your LED needs from its data sheet.

Do the calculation R = (9 - 1.9) / I

Use a resistor of value R in series with the LED to provide the correct current.

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