Driving 3 LEDs in Series @ 1000mA. Is this circuit ok? - Page 1
Forum section: electronics / beginners

author: made2hack
author_rank: Frequent Contributor
timestamp: February 10, 2014, 12:16:46 pm
content: I want to drive 3 LEDs in Series at constant current of approx 900mA - 1000mA.Will this circuit work? The LEDs are Vf 3.55V @ 1000mA.R1 = 0.25W 100k Ohm ResistorR2 = 2W 0.47 Ohm ResistorZ1 = 4.7V 1.3W 5% Zener DiodeQ1 = Fairchild KSP13 Darlington NPN 30V 500mA 625mW transistor- Datasheet:https://www.fairchildsemi.com/ds/KS/KSP13.pdfQ2 = Vishay IRF830PBF Power Mosfet 500V 4.5A 75W- Datasheet:http://www.vishay.com/docs/91063/91063.pdfThanks everyone!

author: made2hack
author_rank: Frequent Contributor
timestamp: February 10, 2014, 12:17:13 pm
content: What I meant to say is that have I ensured a constant current of approx 900mA - 1000mA?

author: tszaboo
author_rank: Super Contributor
timestamp: February 10, 2014, 12:29:09 pm
content: I've made a quick simulation of your circuit, with "close enough" components. The output current was parabolic dependent on your input voltage. Also it was very dependent on the temperature. When the temperature increased for 25 to 50 degrees, the current halved. Have you calculated the dissipation of the mosfet?If you need a very simple 1A current source/sink, you can make one with a LM317 and one sized resistor. That is reasonably stable for this application.

author: made2hack
author_rank: Frequent Contributor
timestamp: February 10, 2014, 12:37:45 pm
content: Yeah, the power dissipation I believe will be around 1W (0.9V drop on the Mosfet). The mosfet will be connected to heatsink and there are fans in the rest of the assembly.Source is 12VDC.

author: made2hack
author_rank: Frequent Contributor
timestamp: February 10, 2014, 12:50:14 pm
content: Should I switch the Darlington for a regular BJT NPN?Will it make things "safer"?

author: mariush
author_rank: Super Contributor
timestamp: February 10, 2014, 05:11:48 pm
content: Are you against simply using a proper led driver? They're around 2$ a piece, plus about maybe 1$ in the other parts required.For example:AL8807MP-13 :http://uk.farnell.com/diodes-inc/al8807mp-13/driver-led-buck-36v-1a-8msop/dp/2138313NCL30100SNT1G :http://uk.farnell.com/on-semiconductor/ncl30100snt1g/ic-led-driver-buck-cntrl-6tsop/dp/1830588ZLED7020-ZI1R :http://uk.farnell.com/zmdi/zled7020-zi1r/led-driver-buck-1-2a-sot89-5/dp/1898428You can look in the datasheets to see how they do the current limiting internally.

author: richard.cs
author_rank: Super Contributor
timestamp: February 11, 2014, 09:58:48 am
content: I've used this design before and it works well. What's the zener there for? Just to protect the mosfet gate during startup? The ~4V you'll get with that R1 will barely turn that mosfet on, a 12V zener might be better. I think that must be where your dependance on input voltage that NANDBlog saw comes from. You should select a zener that doesn't conduct under the normal operating conditions, or loose it entirely if you can be certain your supply will never spike over 20V.The key is to thermally bond the mosfet and the bipolar, then as the mosfet heats up the base-emitter junction voltage of the bipolar falls and the circuit regulates to a slightly lower current. I often build this one selecting the same package for both transistors so they can be mounted back to back.

author: tszaboo
author_rank: Super Contributor
timestamp: February 11, 2014, 03:00:09 pm
content: Quote from: richard.cs on February 11, 2014, 09:58:48 amNo... I've just realised, I did not use darlingtonIt is still temperature dependent, but only like 10%/25 degrees C. Also the voltage dependency is gone. Sorry, my mistake.

author: codeboy2k
author_rank: Super Contributor
timestamp: February 11, 2014, 03:02:02 pm
content: Q2 should be an NMOS, and make the zener be equal to the a typical gate bias voltage, say 15V so the zener biases the gate on, and the PNP BJT shunts current away from the gate to regulate the current at Vbe / R2. This also increases the voltage range of the supply.You can do away with the zener altogether and just drive the gate directly through a 100k from your input voltage and shunt that away through the BJT for current regulation. This only works if you know your input voltage will be less than the FET's Vgs(max). However, with 3 LEDs already at Vf= 3.55V (10.65V) you don't have much headroom, so using the zener is always going to be safer.

author: tszaboo
author_rank: Super Contributor
timestamp: February 11, 2014, 04:00:21 pm
content: Indeed, according the datasheet Q2 is NFET, but the schematic is PFET.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 11, 2014, 04:34:50 pm
content: Strictly speaking, the symbol isn't anything... it's nonsense. The closest thing to a FET with an emitter would be an IGBT, but that would be drawn with diagonal connecting lines, not horizontal. Some people have the bad habit of drawing a MOSFET that way, I suppose in reference to an NPN (the arrow represents the common terminal) or JFET (hence the horizontal and vertical lines).Incidentally, the conventional enhancement mode symbol (a gate, with three line segments parallel, and the arrow pointing into (N channel) or away (P) from the center segment) already shows the body diode (that's what the arrow means!), so any symbol with a parallel diode (usually zener) is amusingly redundant.Aaaanyway, the circuit is missing three things:1. Low gate bias (i.e., only a 100k) is fine, but the low voltage zener is a problem (or, concern at least). Low voltage zeners are rather conductive. Since this is just to protect the gate in dropout, anything from 6V (where the leakage isn't too bad) up to Vgs(max), 20V or so, is fine.2. Resistor value. This is Vbe/Iout. Something like a 2N3904 would give maybe 0.6V, so you'd need 0.5-0.6 ohms for that current. 0.47 is close, but a bit high in current.3. Vbe value. A darlington has double the Vbe of a plain BJT... so you've actually got a 2A+ current sink here.4. Optional, but good to have: if Q2 fails shorted, full supply current (give or take) can flow into Q1 base, nuking it as well. A series resistor, like 1k from source to base, prevents that.You didn't mention over what voltage or temperature range it has to operate, so I can't really comment on how suitable those aspects are. It won't be good: loop gain is low so it will be voltage sensitive; minimum dropout is Vbe + Iout * Rds(on), or over 2V; Vbe (for a plain BJT under linear conditions) varies roughly from 0.8V (0C or below) to 0.5V or less (100C or above), give or take bias (with the relatively low bias current through the 100k, this might be 0.1-0.15V lower still). It's somewhat temperature stable, in that as it heats up, current drops. But it's not a protective mechanism, because it will never actually shut off.If operating voltage is only 11-13V, efficiency won't be bad, but if it needs to tolerate, like, 15-20V for long periods of time, it might be worth considering a switching circuit to save the losses and heatsink. Needless to say if it's battery powered or something, you can also pretty much double battery life in that case.Tim