0
\$\begingroup\$

I am using a 10k digital potentiometer to adjust the brightness of an LED. However, I need to make sure that when I reduce the resistance on the pot, it will not exceed the maximum power ratings.

The max power the pot can handle is 473mW.

So using P = I^2R, we get Imax = 6.88mA.

The battery is 3.3V

From here I am stuck and unsure how to move forward with those values. I have a circuit like this (10k, not 1k): Circuit But according to this answer, that circuit seems to be called the 'pot smoker' How can I make sure that I do not blow out the pot? Is there a circuit I can create to put in between the pot and the LED?

Thank you!

\$\endgroup\$
  • \$\begingroup\$ Add a series resistor that sets a minimum bound on the total series resistance. \$\endgroup\$ – uint128_t Apr 8 '16 at 21:09
  • \$\begingroup\$ @uint128_t great, I was on the right track. I was thinking of putting a 5kohm resistor there, but is there a mathematical way of determining the correct size resistor to put there to ensure I would not exceed power ratings? \$\endgroup\$ – Julio Vasquez Apr 8 '16 at 21:11
0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Current limiting simplicity: add R2.

Let's check your maths: from \$ P = I^2R \$ we can get \$ I_{MAX} = \sqrt {\frac {P}{R}} = \sqrt {\frac {0.473}{10000}} = \sqrt {0.0000473} = 6.9~mA \$, as you correctly calculated. Let's be conservative and round down to 6 mA.

The maximum current for a visible LED will be in the case of a red one as red has the lowest forward voltage drop - typically about 1.8 V at a reasonable brightness. That leaves us with 1.5 V to drop across the series resistor at 6 mA. Mr. Ohm says \$ R = \frac {V}{I} = \frac {1.5}{6m} = 267~\Omega \$ which we will round up again to the nearest value of 270 Ω.

Simply pop this somewhere in the circuit to set the minimum resistance in the circuit to a safe value for the pot.

Thanks for the reference to my pot smoker.

\$\endgroup\$
  • \$\begingroup\$ the pot I am using is 10k not 1k which is where i got the 6.88mA Value. However, plugging that value into your calculations yeilds 1.5V/6mA = 250ohms. (6mA for to be conservative). Thank you so much for your help!! \$\endgroup\$ – Julio Vasquez Apr 8 '16 at 21:27
  • \$\begingroup\$ I missed that in the text and only read the 1 k in my original schematic. I'll edit. \$\endgroup\$ – Transistor Apr 8 '16 at 21:30
0
\$\begingroup\$

You should place a series resistor to limit maximum current. Eventually could add an emitter follower to increase maximum current.

\$\endgroup\$
  • \$\begingroup\$ great, I was on the right track. I was thinking of putting a 5kohm resistor there, but is there a mathematical way of determining the correct size resistor to put there to ensure I would not exceed power ratings? \$\endgroup\$ – Julio Vasquez Apr 8 '16 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.