- Joined
- May 8, 2012
- Messages
- 52
- Points
- 8
I think the title is pretty explanatory.
Background:
I'm designing my own ESP8266 based PWM 3 channel driver.
I'm using a 0-1500ma CC driver (LD06AJSA) that accepts a wide-band PWM input. I'll be using the ESPs PWM output pins for duty input to the CC board.
Question:
Let's say I'm giving the driver 6vcc.
The duty-adjusted square wave that it puts out is also equal to 6v (Vpp) with no load as per my oscilloscope. (Nice smooth PWM output, though)
If my diode (hypothetically) was rated with a Vf of 3v... I'd be exceeding the Vf of the diode by 100%, even if the output is current controlled, right? I assume this is super no bueno.
When I use this channel to control a diode with a Vf of 5.5v, my bench supply DOES show that the circuit is consuming less current, proportional to PWM.
Extras:
I can also buck the 6v down to 3.3v before powering this one channel's driver, but this still exceeds the Vf of the diode by 0.3v. If you can't tell, I'm trying to control a 638nm diode, with a very low Vf from the same Vcc that my G/B diodes will tolerate.
Please bare with me, everyone - I'm an electronics hobbyist, teaching myself this stuff via youtube... So, while I think I understand some basic concepts here, I could be extremely confused.
EDIT: I decided to trust my understanding of CC/CV driver topology and everything worked out fine.
I chose to buck the 6v input for the R channel to 3.3v before sending it to the CC driver. Which, honestly makes too much sense -_-.
Background:
I'm designing my own ESP8266 based PWM 3 channel driver.
I'm using a 0-1500ma CC driver (LD06AJSA) that accepts a wide-band PWM input. I'll be using the ESPs PWM output pins for duty input to the CC board.
Question:
Let's say I'm giving the driver 6vcc.
The duty-adjusted square wave that it puts out is also equal to 6v (Vpp) with no load as per my oscilloscope. (Nice smooth PWM output, though)
If my diode (hypothetically) was rated with a Vf of 3v... I'd be exceeding the Vf of the diode by 100%, even if the output is current controlled, right? I assume this is super no bueno.
When I use this channel to control a diode with a Vf of 5.5v, my bench supply DOES show that the circuit is consuming less current, proportional to PWM.
Extras:
I can also buck the 6v down to 3.3v before powering this one channel's driver, but this still exceeds the Vf of the diode by 0.3v. If you can't tell, I'm trying to control a 638nm diode, with a very low Vf from the same Vcc that my G/B diodes will tolerate.
Please bare with me, everyone - I'm an electronics hobbyist, teaching myself this stuff via youtube... So, while I think I understand some basic concepts here, I could be extremely confused.
EDIT: I decided to trust my understanding of CC/CV driver topology and everything worked out fine.
I chose to buck the 6v input for the R channel to 3.3v before sending it to the CC driver. Which, honestly makes too much sense -_-.
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