Resistor heat in circuit - Page 1
Forum section: electronics / beginners

author: Thumper_LV
author_rank: Contributor
timestamp: December 11, 2023, 10:01:12 pm
content: I've been trying to understand some core concepts - one of these days I'll go ahead and start breadboarding to see what actually blows up. But I'd like to have some grasp of the theory.I'm considering using a resistor to drop from 40VDC to 32VDC. Yes, I know that's inefficient. But - the application is for pulses of 10 microseconds. The current will range from 0 to a maximum of...at the moment I'm going to say 0.5A although reality says 0.25A is going to be extreme. Typical use would be less than 0.1ASo, basic theory says switching in an 8ohm resistor will do what I need. But now I'm wondering about paralleling resistors to reduce thermal load and possibly serializing them as well to use common values. 10, 22, and 49.9 are simple building block resistor values. I can put 4 x 32 (10 + 22) which parallel calculation comes out close enough - or take it to 10 x 80 (10 + 22 + 49.9).Given the pulses will be 10 microseconds on, 5 microseconds off, and let's say continuous duty - it'll be switched on/off on a constant cycle - how much do I need to worry about thermals and power ratings of the resistors. It feels like 1/8W is actually more than enough - even though a never used "full load" of 0.5A = 4W? Do I have that right?

author: Ian.M
author_rank: Super Contributor
timestamp: December 11, 2023, 10:24:16 pm
content: Average power is I2R * duty_cycle. I calculate that as 1.333 W for your 8 ohm resistor passing 0.5A, and a negligible 53.3mW at 0.1A.Note that some power resistors are rated for full power operation at up to 200 °C surface temperature, far far hotter than you want anywhere near the rest of your PCB, so considerable derating may be required.However an 8 ohm resistor will only drop 0.8V at 0.1A and 4V at 0.5A (basic Ohms Law) so as you want to drop 8V (from 40V to 32V), your concept is fundamentally flawed. If you are proposing to shunt regulate the 32V to maintain an 8V drop, then the resistor will be passing 1A and dissipating 8W if the pulse switching is load side, or 5.333 W if the pulse switching is supply side.

author: Thumper_LV
author_rank: Contributor
timestamp: December 25, 2023, 03:38:15 am
content: I knew it was too simple to work...

author: MarkT
author_rank: Frequent Contributor
timestamp: December 26, 2023, 08:44:45 am
content: Quote from: Thumper_LV on December 11, 2023, 10:01:12 pmSo you want to be able to change the current value, but keep the 8V drop constant? That either takes a range of different resistors or a voltage regulator (in the latter case you'd switch after the regulator as regulators require decoupling capacitors and can't switch on and off rapidly.)Regulator will ensure the 32V is actually 32V and not load-dependent too...

author: EPAIII
author_rank: Super Contributor
timestamp: December 26, 2023, 09:12:06 am
content: The Voltage drop in a resistor is given by Ohms Law. E = I*R. Plain and simple. The Voltage drop is directly proportional to the current.So if you want to drop 8 Volts, you need to know the current in order to calculate the resistance value of the resistor. Reworking the equation above gives you R = E/I. You know E, that's the 8 Volts you want to drop. But you seem to be uncertain of the value of the current. You give values ranging from 0.5 A to 0.1 A and that is a 5 to 1 range. So calculating a resistor would give that same range of values, 5 to 1.8 V / 0.5 A = 16 Ohmsand8 V / 0.1 A = 80 OhmsUnless you can have a known and very constant current, then any fixed value of resistance would produce a wide range of Voltage drops. In short, a series resistor used to drop a Voltage is not a very good regulator circuit if the current is not very constant and KNOWN.If you want 32 Volts from a 40 Volt source, you are much better off using one of many Voltage regulator ICs. The companies that make them offer on-line circuit design facilities.