0
\$\begingroup\$

I'm trying to evaluate LED drivers for a project I am working on, and I came across this reference design for the TI LM3402/04: https://www.ti.com/lit/ug/snva342e/snva342e.pdf. The design employs a FET to shunt the LED current to ground using PWM as a means of controlling brightness, which is an approach I'm familiar with, but I encountered the following schematic for the dimming circuit on page 20 that has me stumped:

LED Dimming schematic from page 20 of the AN-1839 reference design

[edit] I think the way I presented this diagram is confusing. Including the full schematic below to make clear that Q4 is not the switching transistor for the LED driver, but rather a means of shunting the LED current to ground for PWM. The switching transistor is included in the LM3402 IC.

Full schematic of the driver circuit.

From what I understand, the dual diode D2 is some type of current control. I'm a bit confused about the two VDIM inputs, I'm assuming one is the PWM signal and one can be used an alternative to VCC for driving the dimming FET if JMP-1 is disconnected? The distinction there is extremely unclear.

The circuit with R4, R6 and Q3 is what really has me stumped. I'm guessing it's some sort of gate driving circuit, with R5 helping provide current?

If anyone can help me work out what's going on here I'd really appreciate it.

\$\endgroup\$

2 Answers 2

0
\$\begingroup\$

L1 and Q4 form a bosst converter to be able to use LEDs that need a higher voltage than we can provide. So we want to switch Q4 and we want to switch it fast (to keep switching losses low).

MosFET are controled by voltage on the gate. But they also hae a cpacitance between that cate and the channel. To change the voltage on a capacity we need current. That's why we use the Q3 as a gate driver to charge and discharge the gate of Q4 "as quickly as possible". BUT the resitance on the gate is really low. This can lead to undesired effects like oscillations (wires are inductors and we have a capacitance, that's a nice resonant circuit with a pretty low Q. To reduce the Q (dampen the circuit) we put in R5.

\$\endgroup\$
1
  • \$\begingroup\$ I think the way I presented the schematic above was a bit confusing: Q4 is actually a shunt transistor used to PWM the LED current. I've updated the question for clarity. \$\endgroup\$
    – flimsy
    Commented Jan 21, 2022 at 18:21
0
\$\begingroup\$

Q4 and L1 form a boost converter circuit along with the LEDs. R5 etc is just the gate drive for Q4.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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