Maximum current for regulating voltage - Page 1
Forum section: electronics / beginners

author: sy
author_rank: Regular Contributor
timestamp: August 03, 2024, 01:17:04 pm
content: Hi guys, I wanted some help with the maximum current from a regulated output voltage using the scheme in the image below.I realize that I could just use a buck converter to achieve this, but I would like to modify this circuit so I regulate the 12V output down to 5V and keep at least a 3A current capability for the load.I believe there is something fundamentally wrong with my thinking, because the current is just kind of "set" by the load resistance. I'm not sure of other schemes that can achieve this.I would've thought the op-amp keeps the negative terminal at the same reference level as the positive terminal, and the transistor would open to allow the load to source more current, but that does not seem to be the case...Any suggestions would greatly be appreciated!

author: ledtester
author_rank: Super Contributor
timestamp: August 03, 2024, 01:35:30 pm
content: Quote from: sy on August 03, 2024, 01:17:04 pmThis is exactly what should happen with a constant voltage power supply -- I = V/R, so if V is constant I is determined by R.QuoteWhat exactly is not working with your circuit? What do you expect vs what are you measuring? What is Vcc and what part numbers are you using for U1 and Q1?Note that your circuit looks like a constant current sink but in that case R2 is viewed as the variable "load" and RL is a fixed shunt resistor.

author: BennoG
author_rank: Frequent Contributor
timestamp: August 03, 2024, 01:52:34 pm
content: Your output voltage is not a constant value.It is 0.418 times the input voltage so for 12V it it 5V and for 14V it is 5.85VSo no regulation but divided by a factor.Also at 5V 3A the R2 resistor is burning up 3V x 3A = 9W of heat.And the NPN transistor is burning up 3 * (7-3) = 12W of heat, so you need a big heatsink.Benno

author: sy
author_rank: Regular Contributor
timestamp: August 03, 2024, 01:54:57 pm
content: I think my main cause of confusion is say if the load resistance drops to 5Ohms, I would expect the negative terminal to stay at 5V so the current through the load is I=5V/5Ohm=1A.However, what I am seeing on the simulation is the negative terminal at 1V, and well the current though it would follow the same equation I=1V/5Ohm=200mA.I am not sure what parts I would be using yet, I thought the simulation would be general enough for this case. But to give something in the ball park say BUW12A? (450V 8A might be overkill) As for the opamp just need something to hold the emitter voltage at the same level, so I'm not sure if I should be looking at something in particular?

author: David Hess
author_rank: Super Contributor
timestamp: August 03, 2024, 01:59:53 pm
content: Q1 only has limited current gain, so at 1 amp the base current might be 40 milliamps which could exceed the maximum output current of the operational amplifier, allowing the output to sag.

author: sy
author_rank: Regular Contributor
timestamp: August 03, 2024, 02:03:36 pm
content: Thanks for all the suggestions guys. I just realized a regulator had to keep the output at a constant value despite the input, and there are some HUGE flaws with my suggestionThis is beyond the scope of what I know... So I might just op for the buck converterWould yous recommend any sources to get started with learning buck/boost converter design? I want to take a look into it when I get some time

author: BennoG
author_rank: Frequent Contributor
timestamp: August 03, 2024, 02:10:13 pm
content: I always use the LMR14030 on my PCB's almost exactly as the example in the pdf.https://www.ti.com/lit/ds/symlink/lmr14030.pdf

author: tooki
author_rank: Super Contributor
timestamp: August 03, 2024, 02:49:50 pm
content: Quote from: sy on August 03, 2024, 02:03:36 pmSome tips:1. Look at the evaluation boards offered by e.g. TI and AD, and read and carefully re-read the evaluation board documentation and the chip datasheets. They frequently explain why they designed the board a particular way. Read carefully, because key details are often given in passing, mentioned in a footnote or fleeting comment. You will likely encounter unfamiliar phrases, so invest time finding out what those mean and re-reading to get a full understanding of what the documentation is trying to tell you.2. TI and AD have great application notes, too.3. TI has a fantastic tool called WEBENCH Power Designer which lets you give it parameters for your PSU (like input voltage range, output voltage and current, and some design goals) and it will spit out configurations using their chips. You get a complete circuit with component values, a PCB layout, and you can run simulations. You can also change components and re-simulate to see the effect.4. DC/DC converter design is highly sensitive to PCB layout. Especially if you don’t really understand the layout guidelines, then closely adhere to one of the PCB layouts from the evaluation board or WEBENCH.

author: David Hess
author_rank: Super Contributor
timestamp: August 03, 2024, 04:33:13 pm
content: Quote from: sy on August 03, 2024, 02:03:36 pmNothing prevents using the input voltage to the regulator as a voltage reference, if the input voltage is constant. This is common in tracking regulators, where all of the regulated outputs rise or fall at the same time, and the higher output voltages must always be greater than the lower output voltages.In your case replace R4 with a 5 volt reference, which could be a 5 volt zener diode, to get a constant 5 volt output which does not depend on the regulation of the 12 volt supply.