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I'm trying to design a dimmable LED circuit board. It's controlled by Arduino with its PWM signals. I'm new to electronics so I have some question regarding my design and knowledge on it. I've attached two diagrams I've made to help with the visualization. Vcc is 24 VDC and the 5V is the PWM pin from the Arduino.

The idea is to control the RGB LEDs (1W per channel) using a TIP122 Darlington transistor. From what I know, there is about a 1.4V drop from the base to the emitter in the TIP122 (Vb-Ve = 1.4V). The voltage at the collector and base should be the same (Vc=Vb). The base voltage is the voltage out of the voltage divider (R1 & R2). If R1 is equal to R2 (R1=R2), then Vb should be half of 5V or 2.5V (Vb = 2.5V). From that, the emitter voltage is 1.1V (Ve = Vb - 1.4V = 1.1V).

The emitter current is the summation of the base current and collector current. However, the base current is super small compared to the collector so we can assume the emitter current is the same as the collector current (Ie = Ic). By changing the value of resistor (R_Q) after the TIP122, we can control the current flowing through the circuit (Ie = Ve/R_Q = 1.1V/R_Q). This is assuming that current gain is not bigger than 1000.

From beginner LED circuits, I’m suppose to put a resistor (R_L) in series with the LED(s) to limit the current flowing through the LED and prevent the LED from being destroyed. But if the resistor after the TIP122 will limit the current, do I still need the R_L resistor?

As an example, let’s use the circuit in the second diagram. The usable voltage that can power the LED would be the difference between the supply voltage and the voltage at the TIP122 collector (Vcc-Vc = 24V-2.5V = 21.5V). Let’s assume the 3 LED in series each has a forward voltage of 6V (Vf = 6V) and a forward current of 750mA (If = 750 mA). Thus there is 18V of total forward voltage and 3.5V leftover (21.5V – 3*6V = 3.5V). Now normally, we’d put a 4.67 ohm resistor (R_L = 3.5V/750mA = 4.67 ohm) to keep the current flowing through the LEDs 750 mA. But with the TIP122 and R_Q, is the R_L still necessary? If R_L is NOT needed, is it still true even if the leftover voltage is a lot larger? So say 2 LED is in series instead of 3 thus giving a leftover 9.5V instead.

If anything I said is false, please let me know. I’m just explain my thought process in my design and would appreciate any feedbacks on it. Thanks.

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  • \$\begingroup\$ Things to think about: what is your PWM speed and can the darlington switch that fast? If I were to do something along these lines, I'd pick a N channel mosfet. They can switch faster than a darlington and are easier to drive and you will have less losses. \$\endgroup\$ – Aaron Jan 13 at 21:12
  • \$\begingroup\$ @Aaron If it's coming from a typical Arduino then the PWM frequency will be less than 1kHz. \$\endgroup\$ – Elliot Alderson Jan 13 at 21:14
  • \$\begingroup\$ I'm using an Arduino Micro and their website said it has a pwm frequency of 490Hz. I've used it with a Darlington before on a test circuit and it worked fine to me. There might be some tiny flickering if I giving it a blow 10% duty cycle but it's not very noticeable. \$\endgroup\$ – Agriculex Jan 13 at 21:44
  • \$\begingroup\$ @Agriculex What purpose is served by the PWM'd light from R, G, and B LEDs? What's the end goal? (It matters in selecting from among a variety of things I might say.) \$\endgroup\$ – jonk Jan 13 at 22:19
  • \$\begingroup\$ @jonk The PWM is to adjust the output/brightness of the LED. This is for a plant growth apparatus. It will be used to give plant samples different colors and intensity of light from the PWM'd light. \$\endgroup\$ – Agriculex Jan 14 at 19:57
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Without \$R_L\$, a much larger voltage may appear across the transistor, from collector to emitter. It is incorrect to say that the voltage at the collector is the same as the voltage at the base. The voltage at the collector will change as needed to control the current.

If the base voltage is about 2.5V then the emitter voltage will be about 1.1V. If the LEDs drop 6V altogether then the remaining voltage, \$24 - 6 - 1.1 = 16.9\text{V}\$ will appear from collector to emitter of the transistor. If 750mA is flowing through the LEDs then that much current is also flowing through the transistor, and it will be dissipating about 12.7W of power. You will need a sizeable heat sink to keep it from burning up.

There is no free lunch here. If your power supply is 24V and only 6V is across the LEDs at 750mA, then you will be wasting 13.5W somewhere. You can add a high-power resistor in series with the LEDs to reduce the power consumed by the transistor, but you are wasting 75% of your power no matter what.

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    \$\begingroup\$ That's insightful to know. I did notice the transistor getting a lot hotter on the red channel compared to transistor for the blue & green channel in the 3in1 RGB LED. Red does have a smaller forward voltage too. \$\endgroup\$ – Agriculex Jan 13 at 21:40

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