I'm trying to design a lighting circuit for an indoor garden, and I'm not understanding things that seem like they're really basic, so I need some help. The schematic I'm referencing is: Schematic

I have a 24V input source which I'd like to limit to 20V. LA is a series array of LEDs such that the total voltage drop across the series is ~19.8V. R1 and Z1 are designed to limit the circuit to 20V. The portion of the circuit after that (the area above the "Repeat 14x" label) is a constant-current circuit as adapted from this instructable, and will be repeated a total of 14 times. Each of those will draw approximately 0.02A.

My confusion stems from the presence of RA and ZA, that, in tweak 3 of the constant current instructable, limits the input to the gate of QB to about 5V. Doesn't it also limit the entire circuit to 5V, including the drop across the LED array? RB was calculated to be 27 ohm 1/4 W from the formulas provided in the constant current instructable; do I only need QA, QB, and RB in the circuit, and then connect the gate of QB and collector of QA to the R1/Z1 circuit?

Device value specifics:
R1 - 140 ohm, 1/4 watt resistor
Z1 - 20V 1W zener diode
RA - 100k ohm, 1/4 watt resistor
ZA - 4.7V 500mW zener diode
QA - NPN transistor (Fairchild 2N5088BU)
QB - N-channel MOSFET (Infineon IPP057N06N3 G)
LA - 6x3.2V LED in series (e.g Cree C512A-WNS)
RB - 27 ohm, 1/4 watt resistor

  • 1
    \$\begingroup\$ You accepted Oli's answer too fast here IMHO. Questions with an accepted answer get less attention, while other answers would be useful too. \$\endgroup\$ – Federico Russo Sep 19 '12 at 14:45
  • \$\begingroup\$ I agree with Federico here, thanks for the accept but generally it's best to leave it a while before you accept an answer, to see what other answers you get. Some people won't bother to look/answer if they see the answer has already been accepted. \$\endgroup\$ – Oli Glaser Sep 19 '12 at 15:13
  • \$\begingroup\$ You have too little "headroom" on the LEDs to guarantee regulation if your line voltage falls to 20V under load (or is limited to 20V purposefully). With 24V you are OK. You want say at least 1 V across Qb in normal operation to allow current regulation. | Rb ~= 0.6V/I_LEd as you probably know. P_RB =~ 0.6V x 20 mA = 12 mW so 1/4W very adequate. P_ZA =~ 5V x (24-5)/100k ~+ 1 mW so any available zener Wattage OK. \$\endgroup\$ – Russell McMahon Sep 19 '12 at 17:15

R1 and D1 will not limit the voltage to 20V. The resistor would need to be in series with the +24V supply rail for this to happen.
In any case, why do you want to limit the voltage to 20V if you are limiting the current through the resistors? The current limiting circuit will adjust the voltage across the resistors to the same value either way.

ZA and RA just limit the gate voltage to 5V (or not - I didn't look at the values, Steven is right there), in case you want to interface with a microcontroller according to the Instructable. The only effect this will have on the current limiting is setting an upper limit to the MOSFET turn on (which may be a good thing, but not necessary)
The circuit will limit the current through the LEDs just fine like this.

To show what happens with the circuit over a range of supply voltages, here is a quick circuit I created, without the unnecessary bits:

Current Limit LED


Current Limit LED Simulation

I set the current for around 20mA using 0.7V / 0.02A = 35Ω. The X axis is V+ (the supply voltage) being swept from 10V to 30V. The green trace is the voltage across the 3 LEDs, and the blue trace is the current through them.
You can see how once the voltage rises high above the LEDs combined Vf, any further increase has very little effect on the voltage across them or current through them (less than 1mA variance over 20V)

| improve this answer | |
  • \$\begingroup\$ I'm limiting to 20V out of an excess of paranoia about burning out the LEDs. since I won't be interfacing with a microcontroller in this part of the circuit, I'll leave out RA and ZA to save on cost. QB's G-S breakdown is 30V so I'm not worried about overloading that. You know, as soon as I read that R1 has to be inline with the supply rail, I realized I had read that several places. Now I feel kinda stupid. \$\endgroup\$ – John Sep 19 '12 at 14:31
  • \$\begingroup\$ @John: Your choice for R1 is inappropriate for a series resistor. 14 LED strings will total about 1/3 of an amp, to drop 4V in R1 the value must be no more than 12 ohms. And you're going to dissipate at least 4V*1/3A = 1.333W, so a 1/4W resistor won't cut it either. \$\endgroup\$ – Ben Voigt Sep 19 '12 at 14:46
  • \$\begingroup\$ Is it safe to assume that duplicating this circuit in parallel another 13 times will give the same results across each set of LEDs? \$\endgroup\$ – John Sep 19 '12 at 14:51
  • \$\begingroup\$ Yes, you can duplicate as many times as you like up to your supplys current capability. \$\endgroup\$ – Oli Glaser Sep 19 '12 at 15:05

R1 + Z1 don't serve any function. You will get 20 V, but at the zener's cathode, not at the high end of the resistor. That will still be at 24 V.
But it doesn't matter much: if you have a current source the current will be independent of the input voltage anyway.

Same for RA and ZA; the zener voltage will appear at the zener's cathode, but the 24 V remains untouched.
The value of RA is way too high; it will allow only 200 µA, and that's too little for a zener to regulate properly. The zener voltage won't be the full 4.7 V, and the FET may not conduct sufficient current. I would use a higher zener voltage to start with, like 6.2 V, and decrease RA to 1.8 kΩ, this gives you 10 mA zener current.

Oli eliminated the zener, but you'll have a too high voltage on the gate then. Maximum gate voltage is 20 V. I would keep the zener, or use a resistor divider to keep the voltage below 20 V. Advantages: a resistor is cheaper than a zener and also the divider needs less current, so that you could use two 10 kΩ resistors there.

RB will limit the current to 0.7 V/ 27 Ω = 26 mA.

| improve this answer | |
  • \$\begingroup\$ Q2 will make sure the gate voltage is not much above the Vth for the MOSFET, so the gate should never see more than a few volts. Adding another 10k resistor from gate to ground wouldn't hurt if you want to be extra cautious. \$\endgroup\$ – Oli Glaser Sep 19 '12 at 14:46
  • \$\begingroup\$ @Oli - I don't think the resistor is extra cautious. QA will only begin conducting after QA conducts, and that will be because of the 24 V on the gate. Granted, very briefly, but I don't like it :-) \$\endgroup\$ – stevenvh Sep 19 '12 at 14:58
  • \$\begingroup\$ Well I don't like you not liking it, so I'll add the parallel resistor in so we're both happy ;-) \$\endgroup\$ – Oli Glaser Sep 19 '12 at 15:07
  • \$\begingroup\$ @Oli - :-) \$\endgroup\$ – stevenvh Sep 19 '12 at 16:22

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