Methods for adding short circuit protection to a current-limiting LED driver. - Page 1
Forum section: electronics / projects

author: isthatme
author_rank: Newbie
timestamp: December 09, 2016, 01:43:17 am
content: Hi, so I'm designing a little LED driver with a current limiter. I've been thinking about ways to deal with potential shorts and stuff, because this is going to be put in a spot in our house, so I obviously don't want things catching fire. Also it's high(ish) power, maximum per channel is approximately 2 amps (absolute max).My current design is attached as a picture. The main idea is the LED (cathode to +12V, anode down into the circuit - the wire at the top of the picture) is connected to ground by the MOSFET. Rsense is a current sense, with the value going off to the opamp which multiplies that by a value I don't remember to drive a bjt transistor to pull the mosfet low if the current gets too high. The led is just an indicator.I've thought of a few things to add protection:1) Add a fuse to the 12V pin going to the LED, or maybe what mike uses here:https://youtu.be/cQWnLOCGpXM?t=6m. Or maybe a larger breaker earlier on or something. Anyways, this would protect against something shorting to ground on the positive pin.2) Add a transistor or something across the source pin of the transistor, so in case something shorts across the anode and cathode of the LED the MOSFET doesn't have to dissipate all the power of the LED. This is the one I'm having the most trouble with, I'd like to be able to turn the MOSFET completely so it doesn't keep burning off power. I'd also like to keep the component count down if I can.Is there anything more anyone could think of to add more board-level protection against shorts and the like?

author: Automobilie
author_rank: Newbie
timestamp: December 09, 2016, 02:39:03 am
content: Do you mean a short across the LED? If it's a current regulated driver, the current shouldn't exceed what the supply is set for.

author: isthatme
author_rank: Newbie
timestamp: December 09, 2016, 05:13:30 am
content: Yeah. I'd just rather not have the transistor dissipate that current, and have a way to shut off the channel so that I don't have to worry about the heat. Thermals might be kinda tight with where I'm putting this.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 09, 2016, 05:38:34 am
content: A properly sized fuse works fairly well.If the output is shorted, the transistor dissipates most energy. Die temperature exceeds absolute maximum rating (around 150 degC) quickly (before the transistor case is any kind of fire risk) - this blows the transistor as a short, stops regulating, giving unlimited current to the output, blowing the input fuse.You can use a simple thermal fuse as an added protection.

author: isthatme
author_rank: Newbie
timestamp: December 09, 2016, 06:34:47 pm
content: True, but will 12 watts bring the temperature up that fast? Because the current limiting is still happening at this point.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 09, 2016, 07:49:33 pm
content: I don't quite see what the circuit is doing, or is supposed to be doing...If you need a current source, but you can't afford the power dissipation under shorted-load conditions, then you can modify it in several ways:- Foldback. Rather than being perfectly constant, the current is made to be proportional to load voltage (a negative (incremental) resistance characteristic). The peak power point is in the middle of the load line (between X and Y axis zero-intercepts; draw it out). It's probably desirable that the circuit can withstand this power level continuously, in case of a rather unlucky partially-shorted load. Otherwise, another method can be used (with, or alone).- Timed fault (usually with auto-reset). A sense circuit charges a capacitor, at a rate (and to a voltage) proportional to the output voltage. When the capacitor voltage exceeds a threshold, the current source is turned off. The capacitor is allowed to discharge slowly, until a falling threshold is passed. The current source then turns on again. (The most basic timing method is simply charging the capacitor from a resistor, or resistor divider, connected to the output. Hence, it charges to a voltage proportional to the output voltage, and at a rate that's also proportional. The aim is to have the capacitor voltage be a surrogate for the transistor temperature, so the RC time constant should equal the thermal time constant.)- Thermal cutout. This can be tricky, because it's difficult to sense the transistor's internal temperature. If the circuit goes slowly enough (i.e., the maximum power dissipated by the transistor is not enough to burn it out before the circuit can respond), this works safely. It also addresses high ambient temperature: if it's an unusually hot day (or everything's become caked up with dust), it's better to dim or turn off the light, rather than create a hazardous condition.The simplest example is putting a PolyFuse (or other PTC device) in series with the transistor collector/drain, and physically attaching them, so the transistor heats up the fuse. A typical PolyFuse trips around 150C, and its trip current falls linearly as the ambient temperature rises. Thus, a 50mA fuse might trip at, say, 25mA when the ambient is ~125C. Because thermal stuff changes so slowly (over seconds), the transistor may overheat by the time the fuse trips -- but hopefully not by too much, nor for too long. A transistor will tolerate junction temperatures over 200C, for short periods of time, which should give enough overhead for the fuse to trip.You can also put a thermistor on the current setpoint, so the current output itself is reduced at high temperatures. Thus, dimming the light (and perhaps turning it off entirely), rather than causing it to fault out.More advanced methods include power regulation circuits, like the LM5069. This solves the problem of the foldback current limiter having a power peak in the middle of the operating range: it folds back much more aggressively, so that power is kept constant instead. The power might still be too high for continuous operation, so a timer is started at the same time; when that times out (hopefully, corresponding to the transistor die temperature getting dangerously high), it shuts off and waits a relatively long time.- Use a switching supply instead. Who needs power dissipation when you can control the total power output, period?Tim

author: isthatme
author_rank: Newbie
timestamp: December 09, 2016, 08:40:08 pm
content: Thanks for the advice. I guess probably the most I can do is limit the dissipation on the mosfet.