Current Regulator Circuit Power Dissipation - What Happens? - Page 1
Forum section: electronics / beginners

author: isthatme
author_rank: Newbie
timestamp: August 11, 2014, 01:55:53 am
content: Hi, so as a relative newbie to the world of EE, I've come across the need to use a current limiting circuit, but I'm not too sure how power dissipation and such will affect heat in such a circuit.Example:(where you could replace Q1 with an op-amp or some such to control the limit current, but that's unrelated to my question)When the load attempts to pull more current than the circuit is designed to limit it to (i.e. two amps where the circuit limits to 1 amp), does Q2 dissipate power? I'm not quite sure because in controllable dummy loads, for example, the transistor dissipates (most of, I think) the load.

author: c4757p
author_rank: Super Contributor
timestamp: August 11, 2014, 02:00:01 am
content: Q2's power dissipation will always be the product of the current into the drain and the drain-source voltage. Therefore, it's going to dissipate whatever power is necessary to drop the remaining supply voltage (after the load takes what it needs) at the current setpoint.Consider these equivalences:- A volt is the potential energy of charge, one joule per coulomb- An ampere is the flow of charge, one coulomb per second- A watt is the flow of energy, one joule per secondSo ifIamperes flow into something and loseVvolts for a period oftseconds, this means that you haveItcoulombs which have lostVItjoules of energy intseconds, or an energy loss rate ofVIt/tjoules/sec =VIwatts.

author: sleemanj
author_rank: Super Contributor
timestamp: August 11, 2014, 02:13:36 am
content: You have a supply, the supply (for our purposes) is a constant voltage.You have a load, we will assume it is a constant resistance.You want to limit the current of that load.Thus to reduce the current in the load we must reduce the voltage to the load.How do we reduce voltage to the load? We take some of it away.What component in the diagram can possibly take that voltage away? Q2 can.What does that mean? Q2 drops voltage, because we control the amount of resistance it presents by way of feedback to it's gate.We know the current passes through the load, and Q2 is in series with the load, so by definition it must pass through Q2 also.What happens when something drops a voltage and has a current passing through it?It is taking power, and for a MOSFET, what can it do with the power?It can only produce heat.

author: isthatme
author_rank: Newbie
timestamp: August 11, 2014, 03:16:39 am
content: Hrmm... I think I get it.So let's say we have a 12v supply, with the circuit limiting to 1 amp, but the load is drawing 2.The resistance doubles to keep the amperage in check, so Q2 has to sink the six volts at 1 amp, meaning 6W. (excluding voltage drop on the resistor).In formula form: P<dissipated> = ( ( V<source> / ( I<load> / I<limit> ) ) - V<resistordrop> ) * I<limit> )Is this right?

author: c4757p
author_rank: Super Contributor
timestamp: August 11, 2014, 05:02:59 am
content: Quote from: isthatme on August 11, 2014, 03:16:39 amWell, the current that the load "would" draw if not for the limiter doesn't factor in at all. Your equation would work if the load were a resistor, but only because you're basically calculating the theoretical un-limited load from the resistance, and then recalculating the resistance from that in your equation.It's more like this:P<dissipated> = (V<source> - V<load> - V<resistordrop>) (I<limit>)Where V<load> is whatever voltage the load requires in order for it to conduct I<limit> (assuming you can supply that much voltage). That's going to depend on the device: for resistors it's the product of I<limit> and the resistance, for diodes (including LEDs) it's proportional to log(I<limit>), but can generally be approximated as fixed at or a bit above the threshold voltage of the diodes.

author: IanB
author_rank: Super Contributor
timestamp: August 11, 2014, 06:58:07 am
content: Quote from: isthatme on August 11, 2014, 03:16:39 amA good way to look at many problems is to consider the limiting cases.One limiting case occurs if we replace the string of LEDs with a piece of wire. In this case 1 A is flowing and the whole 12 V supply voltage appears across Q2 and R3. If we ignore R3 then Q2 will be dissipating 12 W (= 12 V x 1 A).The other limiting case occurs if we replace the string of LEDs with a 12 ohm resistor. Now we have 1 A flowing through the 12 ohm resistor and therefore a voltage is developed across it of 12 V (= 12 ohm x 1 A). Now there is no voltage left, so there must be 0 V across Q2 and R3. (This can't happen in reality of course, but as a limiting case we can pretend it does.) With no voltage across it there can be no power dissipated in Q2.If we have the two limiting cases laid out, then the case with the LEDs must be somewhere in the middle. This tells us that Q2 must be dissipating somewhere between nothing and 12 W. Notice that to get this far we didn't need to know anything about what the load actually is.But to be more precise than this we need to know something about the LEDs.

author: isthatme
author_rank: Newbie
timestamp: August 12, 2014, 04:59:47 am
content: Thanks, I'm pretty sure I've got it.Quote from: c4757p on August 11, 2014, 05:02:59 amUsing this equation, with a 12V supply, 12V LEDs, with a 0.12ohm resistor or something (whatever, with a negligible voltage drop), Q2 would be dissipating negligible power, right? Even if the load was (trying to) draw more or less current.This would only change if the V<load> was lower than V<source> right? (and the resistor drop didn't make up for it) In which case Q2 would disspate more power?