12V -> 5 - 15V (8A) digitally controllable PSU - Page 1
Forum section: electronics / beginners

author: BitShard
author_rank: Contributor
timestamp: July 22, 2017, 09:14:47 pm
content: HiI'm trying to design a circuit that can take 12V supply voltage and produce an output voltage in the range of <5V - 15V at a maximum current of 8A. This output voltage would also need to be digitally controllable.The solution I'm currently exploring contains an LM723 adjustable voltage regulator with a digital pot (eg MCP4151) driven by an MCU. I'm not too concerned about high voltage transient frequencies, the most I would expect would be an increment change at 1Hz.Are there any better solutions that I'm missing because I'm too green?Thanks

author: Zero999
author_rank: Super Contributor
timestamp: July 22, 2017, 09:37:06 pm
content: The LM732 is just a linear regulator, which can't boost the voltage. The maximum output voltage is always a few volts less than the input voltage, so you'll be lucky to get anything greater than 9V, with 12V in.To get a higher output voltage, than the input voltage, you need a switched mode power supply. In your case, there will be occasions where the output voltage is less, as well as greater than the input voltage, you need a topology which supports that such as buck-boost, SEPIC etc.

author: BitShard
author_rank: Contributor
timestamp: July 22, 2017, 09:58:04 pm
content: Whoops, looks like I forgot to mention that I would need to solve that problem, too. Thanks for the info.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 23, 2017, 07:09:44 am
content: The digital pots are usually limited in there voltage range. So a DAC might be more suitable to set the voltage. Alternative to the DAC one might be able to use PWM and filtering - not as clean and fast, but tends have good linearity at low cost.To get a higher voltage one would need a kind of switched mode regulator. 8 A at 12 V is close to 100 W, but still in a range that can be handled by a linear output stage. So one might not need a sepic (or similar) switched mode converter that also goes down.The LM723 would likely need also auxiliary supply of more than 15 V and maybe a negative (to go all the way to 0).

author: Zero999
author_rank: Super Contributor
timestamp: July 23, 2017, 09:16:06 am
content: Quote from: Kleinstein on July 23, 2017, 07:09:44 amA agree: digital pots also tend to have poor linearity.QuoteThat's true. There could be a boost converter, followed by a low drop-out regulator. For higher voltages, the regulator could operate in drop-out mode with the boost converter providing all of the regulation. At lower voltages, the boost converter could be disabled, with the low drop-out regulator providing the regulation.QuoteYes it would need a higher voltage to get a low-dropout voltage. The minimum output voltage requirement is 5V, so it shouldn't need a negative rail.

author: BitShard
author_rank: Contributor
timestamp: July 23, 2017, 03:04:55 pm
content: Would it be best to implement this using an MCU to turn on and off the various PSU modules, if I were to take that route Hero?

author: Zero999
author_rank: Super Contributor
timestamp: July 23, 2017, 03:54:22 pm
content: Quote from: BitShard on July 23, 2017, 03:04:55 pmI don't see any need for that. When a boost converter's output is set lower than the input, minus a diode drop, it'll just sit there, passing current straight through to the output. The same thing will happen to a low drop-out regulator, when the output voltage is set too high: the output stage will just saturate in the on state, allowing current to pass straight through,

author: BitShard
author_rank: Contributor
timestamp: July 23, 2017, 06:11:02 pm
content: Oh, cool. Thankyou!