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i have a electrical circuit with a 9v battery,LED(0.018Amps),ressistor(500 Ohm) and potentiometer. And im trying to work out the correct ressistance(for the"pot")for the LED to light up at optimium level for the longest life span.Is this formula/calculation correct ?

                        V=IxR
                        V=Ix(R1+R2)        R2 is the potentiometer    
                        R2=(IxR1)/V
                        R2=(0.018x500)/9
                        R2= 1 Ohm 
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  • \$\begingroup\$ YOu also have to factor in the forward voltage of the LED. That is, subtract that from the 9V available. \$\endgroup\$ – Trevor_G Oct 14 '17 at 16:03
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    \$\begingroup\$ If your only goal is to maximize the LED lifespan, increase the resistance to maximum (or just disconnect the LED entirely). If you have some other requirement (like the LED must produce X lumens), then set it to the minimum current to achieve that requirement. \$\endgroup\$ – The Photon Oct 14 '17 at 17:50
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You've already got the standard answer regarding a fixed supply voltage and some known LED forward voltage. But it is also possible to design a circuit that will be relatively efficient (you write, "optimium level for the longest life span") with respect to linear regulation, generally, and where you don't really need to know the LED forward voltage and where it will work well with a battery whose voltage isn't at all fixed over its lifetime. (A \$9\:\textrm{V}\$ battery will actually have a lifetime voltage that is: \$6.5\:\textrm{V} \lt V_{BAT} \lt 9.5\:\textrm{V}\$.) If you don't mind using BJTs, that is.

schematic

simulate this circuit – Schematic created using CircuitLab

The value of \$R_2\$ sets the current. Making it smaller increases the current and making it larger decreases the current in the LED. The value of \$R_1\$ is somewhat less important. Note that there is no need for a potentiometer (which is a good thing, I think.)

The circuit does require that the battery voltage be at least \$1.5\:\textrm{V}\$ more than the LED requires. But in your case, there are few LEDs requiring more than \$4\:\textrm{V}\$ to operate and your \$9\:\textrm{V}\$ battery will be completely dead by the time it reaches \$5.5\:\textrm{V}\$. So this circuit is fully functional throughout the battery lifetime.


To design this, start by first estimating the lowest battery voltage. I wrote above that \$6.5\:\textrm{V}\$ is a good minimum to use for a \$9\:\textrm{V}\$ battery. I figured two standard diode drops for the two transistors, so the base of \$Q_1\$ should be \$1.4\:\textrm{V}\$. (In reality, \$Q_1\$'s \$V_{BE}\$ will be \$120\:\textrm{mV}\$ more than \$Q_2\$'s \$V_{BE}\$ to account for the two orders of magnitude difference in collector currents. But the average between them will probably be about \$700\:\textrm{mV}\$, so no harm done here.)

At an LED current of \$18\:\textrm{mA}\$, the base current for \$Q_1\$ will be near or less than \$180\:\mu\textrm{A}\$. I decided from that figure that I wanted at least \$280\:\mu\textrm{A}\$ flowing in \$R_1\$ when the battery voltage was at its lowest point. So I computed \$R_1=\frac{6.5\:\textrm{V}-1.4\:\textrm{V}}{280\:\mu\textrm{A}}\approx 18\:\textrm{k}\Omega\$. With a completely fresh battery, the current in \$R_1\$ will then be \$I_{C_2}=\frac{9.5\:\textrm{V}-1.4\:\textrm{V}}{18\:\textrm{k}\Omega}=450\:\mu\textrm{A}\$.

Knowing that \$280\:\mu\textrm{A} \le I_{C_2} \le 450\:\mu\textrm{A}\$, I can now estimate that \$Q_2\$'s \$V_{BE}\$ will be about \$60\:\textrm{mV}\$ less than the typical \$700\:\textrm{mV}\$, or \$640\:\textrm{mV}\$. So now I can figure out that \$R_1=\frac{640\:\textrm{mV}}{18\:\textrm{mA}}\approx 35.5\:\Omega\$. I went with \$33\:\Omega\$ for the standard value here (because I know that the \$V_{BE}\$ may actually be a little less than I estimated.)

That's all there is for designing this. You don't really need to know the required voltage for the LED. It could be most any LED (all of them I believe I've seen, which are just LEDs and not entire circuits.)


It should be quite stable. Regulation of current through the LED should be in the area of \$\pm 2\%\$ over the range of battery voltage variations. However, there will be probably another \$\pm 4-5\%\$ due to part variations with the BJTs (which can be most any small NPN BJT you have laying around.) So if you build several of these, they will still be fairly close in behavior to each other.

Current efficiency is good. It wastes at most \$450\:\mu\textrm{A}\$ in order to provide \$18\:\textrm{mA}\$ for the LED, and as the battery voltage declines this added requirement also declines with it. This works out to around \$97-98\%\$ efficiency (in terms of assigning and regulating the current through the LED.) This control overhead should be relatively hard to beat in a linear regulator.


Since efficiency has come up, I thought I'd add a "switcher" version of the circuit that should work over the same range of voltage inputs and will be relatively efficient in power consumption. (Perhaps 75% efficient?)

It's a relatively simple modification to the above circuit:

schematic

simulate this circuit

\$R_2\$ has to be lowered a little bit in order to increase the peak current so that the average will be about right. \$R_1\$ gets moved over into the emitter leg of \$Q_3\$ and must also be adjusted by about the same factor, plus a little smaller yet because of some added voltage drops. \$L_1\$ gets inserted into the LED leg and now \$D_1\$ is required to provide a current path for the inductor when the switches are off. \$R_3\$ helps start the circuit and its value isn't critical. \$L_1\$ also isn't critical and a variety of inductors within a factor of 10 in value can be applied, I suspect; including a variety of copper resistance values, too. \$C_1\$ was added here because \$9\:\textrm{V}\$ batteries have notorious internal resistance.


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  • \$\begingroup\$ Regarding the efficiency, is that compared to using a single resistor? With an LED forward voltage of 1.8 V and a battery voltage of 9 V, there is about 140 mW being dissipated in Q1 (9 V - 1.8 V - 33 Ω * 18 mA) * 18 mA. A 400 Ω resistor would dissipate 130 mW to do the same job at that battery voltage (obviously it would not provide current regulation, so it would be inferior in that respect). \$\endgroup\$ – Andrew Morton Oct 14 '17 at 18:35
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    \$\begingroup\$ @AndrewMorton No, that's an efficiency for the circuit's current, which will consume, in theory, about \$18.28\:\textrm{mA}\$ to \$18.45\:\textrm{mA}\$ in order deliver \$18.00\:\textrm{mA}\$ of it through the LED. I can't do anything regarding power efficiency in a linear regulator. \$\endgroup\$ – jonk Oct 14 '17 at 18:41
  • \$\begingroup\$ @AndrewMorton Thanks for the comment. I added a phrase at the very end to include your thoughts. \$\endgroup\$ – jonk Oct 14 '17 at 18:42
  • \$\begingroup\$ @jonk: You can control the imgur picture size by adding an 's' (small), 'm' (medium) or 'l' (large) before the filename dot. This makes the schematics with only a few components in them a more believable size. It works for all graphics files. See meta.stackexchange.com/questions/189635/…. \$\endgroup\$ – Transistor Oct 15 '17 at 0:39
  • \$\begingroup\$ @Transistor Thanks. I tried that before on a different post and the image just disappeared completely. So I gave up on the idea. In the case I tried, every single one of the letters of the image just coincidentally happened to be already lower case letters. Not sure if that was the reason. But it stopped me trying again. Thanks! \$\endgroup\$ – jonk Oct 15 '17 at 3:44
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General formula for LED resistor is ...

\$\huge R = \frac{V_{Supply} - V_{LED}}{I_{LED}}\$

Where:

  • \$I_{LED}\$ is your chosen LED drive current which should be less than the maximum value stated in the LED's spec sheet.
  • \$V_{LED}\$ is the minimum forward voltage of the LED at your chosen \$I_{LED}\$

However, it is not clear to me what this has to do with POTs and your interpretation of the lifespan on the LED. If you are intimating you want the LED to be driven with exactly 18mA, then you are better to drive the LED with a controlled current source instead of a simple resistor.

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  • \$\begingroup\$ What would the correct formula be to calculate the forward voltage of the LED ? \$\endgroup\$ – J.Blades Oct 14 '17 at 17:25
  • \$\begingroup\$ @J.Blades you need to look at the data sheet for the specific LED. Especially the graphs. \$\endgroup\$ – Trevor_G Oct 14 '17 at 17:29
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enter image description here

Figure 1. A graphical means of establishing the resistor value for an LED. Source: LEDnique.com

  • You can obtain the maximum continuous current from the LED datasheet. You then have the choice of running at that current or choosing a lower current to obtain a reasonable brightness or extended battery life.
  • Using the chart of Figure 1 you can work out the forward voltage drop of the LED. In this case we have chosen a green LED and decided to run it at 20 mA.
  • If the supply voltage is 5 V then the resistor has to drop \$ 5 - 2.2 = 2.8 \ \mathrm V \$.
  • The required value is \$ R = \frac {V}{I} = \frac {2.8}{0.02} = 140\ \Omega \$. The nearest standard value of 150 Ω will do fine.

You can use the chart to work out resistor values for infra-red (IR), red, orange green, yellow, blue and white LEDs but check the datasheet values too.


Your calculations:

V=IxR
V=Ix(R1+R2)        R2 is the potentiometer    
R2=(IxR1)/V
R2=(0.018x500)/9
R2= 1 Ohm 

You have a mistake on line 3.

V = IR1 + IR2
IR2 = V - IR1
R2 = (V - IR1) / I
R2 = V/I - R1
R2 = 9/0.018 - 500
R2 = 500 - 500 = 0

You don't need the pot according to your calculation. The 500 Ω resistor is right.

You have the problem that you forgot that there is 2 V dropped across the LED, however, so there is only 7 V across the resistor. The 500 Ω resistor, therefore, will only allow \$ I = \frac {V}{R} = \frac {7}{500} = 14 \ \mathrm {mA} \$ through the LED.

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