# Constant Current Drive for LED I am going through the circuit as shown above. In such LED drive design (constant current), I am not sure about the role of R16 (2 ohm) resistor. Is it for transistor to have less power dissipation since as current flows R16 will have some voltage across it, reducing Vce drop?

Thank you.

• Where did the circuit come from? Oct 15, 2019 at 10:26
• Its old legacy design Oct 15, 2019 at 10:29
• So, what does the design documentation say? Oct 15, 2019 at 10:30
• Unfortunately no document present. Oct 15, 2019 at 10:32
• Given the potential design flaws of the circuit as shown, is it possible the circuit has been copied incorrectly, more specially the value or component type of R16? With the non-sequential component reference numbering is it possible there are other parts of this circuit not being shown?
– Nedd
Oct 15, 2019 at 13:00

The QTLP690 LED has a maximum pulsed current of 160 mA however, the constant current driver circuit formed around Q6 and Q3 can sink a current peak of about 200 mA. This is based on Q6's base being about 2.65 volts (1.25 volts from the LT1634 and two lots of 0.7 volts from the diodes). The 2.65 volts will become about 1.85 volts at Q6's emitter and drop by a further 0.8 volts at Q3's emitter.

That leaves around 1.05 volts on resistor R10. Because R10 is 4.7 ohms, the implied peak current sink is 223 mA and this exceeds the maximum pulse current rating on the LEDs.

So, is there any real need to go further with this? The design (as presented) is flawed.

• My apology, I didn't get the exact LED used in design, so instead of that I took QTLP690. In actual C503B-RCN-CW0Z0AA2 LED is used which has 200mA max peak current and is used with 20msec over 1sec duty cycle. On the other hand, I am not worried about LED current / design flaw. I am trying to understand the use of R16. Oct 15, 2019 at 11:26
• The C503B-RCN-CW0Z0AA2 has a peak forward I of 200 mA for a pulse thinner than 0.1 ms. Operating at 223 mA for 20 ms is still a big problem. If it weren't a big problem (somehow?) then you need to calculate what voltage might be dropped across the resistor in order to calculate the voltage dropped across the transistor. Given that Vdrop and the Icollector you can calculate transistor power. Once you have power you need to estimate the thermal resistance of the transistor mounting. This allows you to estimate the transistor's junction temperature. This will tell you if R16 is needed. Oct 15, 2019 at 11:40
• @Micro is the question fully answered now or is there some follow up query? Jan 8, 2020 at 15:46

the 2 ohm resistor is for the inrush current protection of the LED diodes, and the 4.7 ohm is for stability to prevent thermal runaway.

Looks like a stuck clock fuse protection resistor where the resistor burns up and opens before the LEDs burn out. The resistor if rated for 0.5W and drawing 0.5W = 1/4A*2Ohms will run at max temp <150'C . But maybe they used a bigger power resistor.

Looks like a poor man's or old technology safety solution when you can get SMD CC chips instead now which you can drive with a FET.

• 2 ohm and 4.7ohm resistors are 100mW rated Oct 15, 2019 at 11:29
• Must be operated at low PWM then..flawed design as shown Oct 15, 2019 at 11:31