Please Check Power Dissipation in Transistor. - Page 1
Forum section: electronics / beginners

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 20, 2012, 03:08:17 am
content: from what i've learnt, this is linear voltage regulator configuration. please check if i got the power dissipation calculation right.to regulate at 3V output/load voltage from 12V input supply, we need 3.7V at the transistor's base, assuming hFe so high that Ib is negligible. so..power dissipated by load (3ohm) = 3Wpower dissipated in transistor = 9Wtotal = 12W (12V x 1A)regardless of voltage regulation, the power dissipation is constant at 12W? (assuming the Rload is changable to be equal to Ve, say 3Ve = 3ohm, 1Ve = 1ohm, ie constant 1A through the load). the lower the output voltage, the greater the loss from the transistor? in other word, the lesser the output voltage, the more heat generated by the transistor?

author: IanB
author_rank: Super Contributor
timestamp: January 20, 2012, 03:29:59 am
content: Yes--you don't even need to worry about the transistor's base in the calculations.If your 12 V supply is delivering 1 A, then the total power output is 12 W, which must all go somewhere. If your load consumes 3 W, then the rest of the circuit (the transistor) must somehow dissipate the remaining 9 W. It is as simple as that. Load 4 W, transistor 8 W; load 1 W, transistor 11 W, etc.

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 20, 2012, 04:24:40 am
content: thanx IanB. now i realize this design is not power efficient. i've been wasting much of my time lately.

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 22, 2012, 07:26:37 am
content: another question. how about off the shelf regulator chip like LM317. is it the same power dissipation as the normal transistor analysis above? ie lower voltage regulation = more power loss in the LM317 (more heat)?

author: BravoV
author_rank: Super Contributor
timestamp: January 22, 2012, 07:30:25 am
content: Quote from: Mechatrommer on January 22, 2012, 07:26:37 amIt depends on the input voltage, or better term the drop-out voltage (input voltage - output), the larger drop out, the bigger power loss.eg. input 12 volt, output 2 volt @ 1 Amp -> drop out = 12v - 2v = 10v10 volt at 1 Amp = 10 Watts power dissipated at the regulator.

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 22, 2012, 07:44:40 am
content: thanks Bravo. so the simple answer is yes. it is the same class, whether power transistor, or voltage regulator chip, they are the same... "linear" voltage regulator which waste more power in larger voltage dropout (regulating at lower output voltage). so i know i cant get rid of it unless i go SMPS.

author: amspire
author_rank: Super Contributor
timestamp: January 22, 2012, 07:46:17 am
content: Quote from: Mechatrommer on January 22, 2012, 07:26:37 amThe LM317 power loss is the same as a transistor regulator.power = voltage drop x current, and so any linear regulator will have exactly the same problem. The cure is to go to a switching buck converter which would only draw 2.4 watts from the 12volts for the 1A output instead of 12 watts. The power loss in the switching converter would be around 400mW or less instead of 10W.If you have an old retired PC motherboard, it probably has some switching regulators for the processor that can do exactly what you want, if you want to salvage some parts for the job.Richard

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 22, 2012, 07:49:59 am
content: you got me at "salvaging things". thanx for the advice amspire.