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I am still beginner to design and so I would to confirm that the variable resistor setup is correct?

I have an LED that operates at 2.7V with 100mA so I added a 23 Ohm resistor. Vcc provides 5V and so I calculated by doing (5-2.7)/100mA = R giving R 23 Ohm.

Does that seem correct?enter image description here

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    \$\begingroup\$ If the 100 mA rating for the LED is given in the Absolute Maximum Ratings table, you should not operate the LED at that current - aim for 75 - 80 mA max to ensure a longer life for the LED. \$\endgroup\$ – Peter Bennett Aug 28 '19 at 15:41
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    \$\begingroup\$ If that 10k pot is rated at 1/4 W (most trim pots would be 1/8 W) then the maximum current it can handle is \$ I = \sqrt { \frac {P}{R} } = \sqrt { \frac {0.25}{10000} } = 25 \ \mu \text A \$ regardless of the wiper position. As you adjust the wiper to decrease the resistance the max power dissipation reduces accordingly. \$\endgroup\$ – Transistor Aug 28 '19 at 15:50
  • \$\begingroup\$ Are you wanting to manually adjust the "set current" for the LED, which will be pulsed by an MCU? If so, you want to use a constant current circuit which is adjusted by the pot and switchable by the MCU. \$\endgroup\$ – jonk Oct 8 '19 at 21:15
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A few potential problems:

  • Low resistance potentiometers are not a good idea, unless you pay $$$
  • You've forgotten saturation voltage of the transistor (perhaps 0.2V)
  • Base current is likely insufficient to saturate the switching transistor
  • A 100mA current surge may cause problems for a feeble +5V DC supply.

That transistor consumes 0.2V of your 5V supply when saturated. To saturate that transistor, 10mA base current should be available, if 100mA collector current flows.

The calculation for collector resistor, for 100mA current now becomes \$ (5 - 0.2 - 2.7)\over{0.1}\$. A 20 ohm resistor might be right.

The calculation for base resistor involves the transistor's base-emitter junction voltage around 0.7V. Fairchild's data sheet suggests it would be about 0.86V @ 25C, for a base current of 10mA. If you were to drive the base from a very stiff 5V logic signal, base resistance would be \$(5 - 0.86)\over{0.01}\$. That's 414 ohms. However, logic gates are not infinitely stiff suppliers of current, so you'll need to reduce this resistance.

Finally, the +5V supply surge current: A large-value capacitor placed close to the 20 ohm resistor, and close to the transistor's emitter may help to keep surge current from disturbing nearby logic run from the same supply.


Your base drive FLASH suggests a momentary pulse. Your transistor may survive a short current pulse without overheating, were it to be biased into non-saturation. This scenario opens up the possibility of controlling LED current by varying base resistance. A fixed base resistor in series with a potentiometer might be appropriate, if current accuracy was not an issue.
If FLASH lasts too long, the un-saturated transistor may flash instead of the LED.

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That's correct but the 10k potentiometer is not a good idea. You might really want a 10\$\Omega\$ or 100\$\Omega\$ pot.

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    \$\begingroup\$ You may not want a pot at all, unless it's a high powered one. Your 23 ohm resistor is dissipating about a quarter watt, so that's the order of magnitude of power you're dealing with, and a lot of pots won't handle it well. \$\endgroup\$ – Cristobol Polychronopolis Aug 28 '19 at 14:54
  • \$\begingroup\$ Thank you for the feedback but I do need one to dim the LED. What do you recommend? \$\endgroup\$ – M2T156 Aug 28 '19 at 14:58
  • \$\begingroup\$ It depends on how much you want to dim the LED. Dimming an LED predictably in this way is difficult...we usually use PWM for LED dimming. \$\endgroup\$ – Elliot Alderson Aug 28 '19 at 16:15

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