Transistor - Voltage drop with less power loss - Page 1
Forum section: electronics / projects

author: vignesh_shan
author_rank: Contributor
timestamp: October 30, 2020, 06:11:33 pm
content: HiI have designed a LED circuit. Where input is 12V and LED forward voltage is 2V and current required to illumniate the LED is 20mA. So Adding resistor consumes lot of power and drop over that resistor. Is there any possibility to use Transistor to limit the current and also drop the remaining voltage across its Vce or acheive minimum power loss over resistor ??ThanksVignesh_shan

author: james_s
author_rank: Super Contributor
timestamp: October 30, 2020, 06:16:34 pm
content: What you describe is using the transistor as a resistor, no matter what you do, a linear circuit dropping the voltage is going to behave like a resistor and waste the extra power. The only option to get around this is to use a buck regulator, ie switching regulator where the transistor is toggled rapidly between fully on and fully off. For a single 20mA LED I doubt there will be any savings though. Do you need to run the LED at 20mA? A high efficiency LED can be more than bright enough as an indicator on less than 1mA.

author: Zero999
author_rank: Super Contributor
timestamp: October 30, 2020, 07:35:45 pm
content: Quote from: vignesh_shan on October 30, 2020, 06:11:33 pmYou need to use a switched mode power supply, which also requires an inductor. A circuit known as a buck regulator will do that.

author: Peabody
author_rank: Super Contributor
timestamp: October 30, 2020, 08:42:20 pm
content: 20mA is probably the maximum continuous current the LED will take, not the minimum to make it light up.How about putting 5 LEDs in series, and the highest value resistor that still lights them up. You will still drop the voltage, but almost all of that will go to producing light, not just heat. And it will give you the minimum current draw needed to produce the light you need.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: October 30, 2020, 10:13:24 pm
content: Instead of using a transistor, use an integrated LED driver buck converter.A few years ago I bought a few PT4115 chips from Ali out of curiosity and "just because they were cheap".It goes upto 1.2A and 30V input, but by changing the passives it can be modified for lower currents.It's also an example. These days there are many specialized LED driver IC's.

author: james_s
author_rank: Super Contributor
timestamp: October 31, 2020, 05:55:10 am
content: For a 20mA LED?I've used the PT-4115 and like it, but it's a power LED driver. For a little indicator I'd just use a resistor.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: November 01, 2020, 01:47:00 am
content: PT4115 can improve efficicency for a single 20mA led just as wel as for a 20W COB panel.From what I rembember it costs about 3ct, so that's also not prohibitive.And it fits with OP's original question of improving efficiency. (Which makes sence if it's fed from a battery for example)So why not?

author: Zero999
author_rank: Super Contributor
timestamp: November 01, 2020, 09:37:32 am
content: Quote from: james_s on October 31, 2020, 05:55:10 amHow do we know it's an indicator and isn't being used for illumination? If it's used as an indicator, then what sort of LED is it and under what conditions is it being used? Crappy yellow LEDs need a relatively high current and it needs to be very bright, if it needs to be seen in full sun.Quote from: Doctorandus_P on November 01, 2020, 01:47:00 amIt's not available from the main distributors, only from eBay & AliExpress, so how can one be sure they're getting a decent, reliable, genuine part?

author: james_s
author_rank: Super Contributor
timestamp: November 03, 2020, 03:36:29 am
content: Quote from: Zero999 on November 01, 2020, 09:37:32 amWhat it's being used for is irrelevant, perhaps I should have stated "indicator class LED" or just left it at "20mA LED". Either way at 20mA or less it is not usually considered cost effective to go with a switching regulator, you have to remember that the regulator itself will consume some current, and the inductor and freewheel diode also incur some losses, then there is the cost of the parts and the physical space they take up. If efficiency is of the utmost importance then give it a try, but run the numbers and see if it's actually buying you anything. I've never used the PT4115 for less than about 200mA so I don't know what the lower limit is for being practical. IIRC the datasheet has efficiency curves under various conditions.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: November 03, 2020, 07:24:39 am
content: LCSC also sells the PT4115:https://lcsc.com/search?q=pt4115

author: james_s
author_rank: Super Contributor
timestamp: November 03, 2020, 06:37:56 pm
content: I've bought them from several random Chinese sellers on ebay and never had a problem with the ones I got. You have to be careful about getting cheap clones/fakes of expensive Western parts but the risk is much lower for things that are cheap Chinese parts to begin with. Sure it could happen, but unless I was building a large number of something or it was a safety critical item I wouldn't worry about it, and if the latter I'd probably choose a part from a reputable Western company anyway.