Adjustable LM317 High Powered Current Source

all right here's close up view of my lm317 current boost transistor the transistors in MJ 2955 this will if done right with some proper heat sinking provide up to 10 to 15 amps the rating of the transistor there's the lm317 there's your current adjust as you see there let's move back out off to my left what you can't see is and is a power supply rated for fifteen volts I think it will do up to 25 amps here is my setup I'm using a 1 10 ohm resistor as a load I went ahead and cut on my voltmeter it's it will measure the voltage across the 10 ohm load if you read the vault if you read the voltage at say 10 volts divided 10 volts by 10 that's your current but this is going to go ahead and illustrate that I can in fact easily adjust the current it's already set for about 1 amp this resistor does not measure out to be exactly 10 ohms it's closer to about 11 to about 11 ohms which is why you see this note a couple of issues as I'm doing this this transistor is rated at a gain of between no more than 70 when I measure this is the connection from the lm317 as you saw in the schematic you will see in this command that goes on off the ground when I put a current meter and here measured it and I measured the current over there I was able to calculate the actual DC current gain of the transistor which turns out to be 116 for this particular transistor so listen to this lesson here just because you get a transistor MOSFET or whatever and you're reading the spec sheet and it says this particular reading or it's going to be a particular or whatever ninety-five percent of the time it won't be within that range a lot of the times it's not that's why when you folks use a circuit simulation software and then you run out and build it in the real world it doesn't work or it doesn't work the way you think it does anyway let's watch the meter as I adjust the current now I've dropped the current to approximately a half an amp let's notice something about the voltages as we measure them as explained by the way this resistor is getting quite hot but you'll also find out that this transistor is starting to get hot why is that remember your law of series circuits that is the actual input voltage from the power supply if I'm dropping around five volts on the load the other 10 volts will be dropped across the transistor in fact I will leave that connected and I will connect this to the emitter circuit on the transistor and there is where my other 10 volts disappeared - it's being dropped across the transistor which is why when I'm really doing some high current stuff I have this auxilary heatsink that will bolt on top of this heatsink and you will notice if you look under the board real closely that the transistor connections are offset from the board so what you're doing here let's okay I'm measuring the voltage drop across the transistor so what am i doing with a constant current source if I vary the load I am NOT going to see yeah what you're doing is you're having an adjustable voltage divider that when you adjust your current and your voltage drop across the load is solely dependent on the load if I change the load I'm going to change the voltage drop across the transistor lm317 circuit let's connect in a different resistor value here is the actual schematic of the circuit as you seen in the video here's our VCC of 15 volts coming in it comes in on the emitter of q1 which is a MJ 2955 PNP power transistor it's rated at 15 amps at a hundred and 15 watts we have a current flow from emitter to base called IB I be is set by this circuit the lm317 and these this very this potentiometer and I stuck in an extra 10 ohm resistor just in case I set the potentiometer all the way to zero I won't have a virtual-short from base literally to ground at any rate is nonetheless what you will find here is the current through the LM 17 is set by are set which is a combination of these two resistors your formula is 1.25 divided by our set all right these were actual measured voltages our set was adjusted to a hundred and forty five ohms this gave me an i set measured through this meter of eight point six two milliamps but i said is the same thing as ib which is also eight point six two milliamps the spec sheet for the m j 2955 says it has a gain between five and seventy that's your h fe or DC gain as i warned in the video and i'll warned to learn you many many more times what you read in the spec sheet and what you come up with on simulation software is not the real world it turns out that actually under test and that's after i let it run a while to heat up in the temperature to stabilize the DC gain turned out to be hundred and sixteen not seventy quite a bit off let's follow our current flow from emitter to collector and that current flow from emitter to collector is controlled by the current flow from emitter to base which is controlled by the lm317 and is set for eight point six two milliamps eight point six two milliamps name's 116 is approximately 1 amp and that's what I adjusted it for originally when I had an amp meter in series with my load in this case the load was a 10 ohm resistor basically 10 all these are big wire wound resistors all right cold if you measured the resistance through the original resistor I used while it's stamped 10 ohms it was reading more like 12 what happened is the resistor got hot and went on and as a circuit as it got hot and got to his expected operating temperature the resistance dropped closer to 10 ohms so I adjusted the amp meter that's in series with the load I adjusted our potentiometer over here to get a 1 amp current at a 1 amp current of course I was measuring 8.6 2 milliamps over here this is a voltmeter tied across it it was actually at 1 amp about 10.8 volts or something like that so that tells you that the load resistor is not exactly 10 ohms but it dropped down closer to what it should have been it was the voltage come out to a little bit over 10 and that's all there is to it I'm controlling this massive amount of current by a small account amount of current that I control through an lm317 this is far better than using just I could have used a potentiometer in series with the base but that doesn't work too well this is a much better way to do it at lm317 is cheap and easy to use as far as that goes with our current set-up of IC at one amp the circuit is using about 15 watts of power I want to disregard the few milliwatts used down here we're using 15 watts of power of which 55 watts is dissipated in the q and q 110 watts is dissipated in the load our load but they both at 1 amp at 15 volts input all right at our present 10 ohm load at a 1 amp current 10 watts will be dissipated by the load 5 Watts will be dissipated by Q 1 let's change the load to 5 ohms but leave our current at 1 amp we've sort of reversed the situation now 5 ohms times 1 amp is 5 volts over here though the remaining 10 volts is going to be dropped across Q 1 so Q 1 is dissipating most of the nth 15 watts in the circuit as a suggestion if you want to say have one amp and leave your load at 5 ohms or whatever current your circuit is drawing I would suggest that you drop the VCC if you can to say 8 volts that way with 8 volts VCC you will be dropping 5 watts in the load circuit and only 3 watts in the Q 1 circuit let me reiterate once again and this subject confuses people you can't just use any value for our load and expect to draw 1 amp let's say we're going to change this to 20 ohms at 1 amp that's 20 volts the only trouble is VCC is 15 volts the circuit is not working and it won't work so bear in mind your load resistance whatever the current is when you multiply them together they have to be a little less than they can't they have to be a little less than VCC or the circuit will not work now if I wanted to adjust my current for say a tenth of an amp a tenth of an amp times twenty ohms is going to be two volts then that's going to work fine of course two volts times a tenth of an amp is what two hundred milliamps and of course at fifteen volts your drop being 13 volts times 200 milliamps over here you're still dropping most of your voltage your voltage will divided according to the load and the current through q1 so keep in mind your input voltage versus the current out that you want our load times the current its voltage must not exceed VCC I hope that was some use to you thanks for listening you

Channel: Lewis Loflin

Views: 198,865

Keywords: LM317, constant, current, MJ2955

Id: n94xqPMhmTo

Length: 15min 23sec (923 seconds)

Published: Fri Aug 25 2017