Powersupply Design Question - Page 1
Forum section: electronics / beginners

author: Zh4ng
author_rank: Unknown
timestamp: October 18, 2015, 11:39:32 am
content: Hello everybody,I've got a small design question. I need two voltages 3.3V and 5V.As a voltage regulator I chose the LD1117. The input voltage is 9V.Now my question:What's smarter:a) the LD1117-3.3Vand LD1117-5.0V parallel to 9Vorb) the LD1117-5.0V and LD1117-3.3V in seriesWhat makes more sense? Apart from the power consumption ...THX Zh4ng

author: Simon
author_rank: Unknown
timestamp: October 18, 2015, 11:43:46 am
content: You can do it either way. The only thing is that any current consumed by your 3.3 V rail will also be used on your 5 V rail. Many computer power supplies work like this in fact which is why there output specs are often dubious.It does mean that the 5 V regulator will take a lot of the heat and your 3.3 V regulator will not heat up hardly as much. If you only need a small amount of power from your 3.3 V rail then putting them in series could be an option but it does mean you need a slightly bigger 5 V regulator.Really you have to weigh up the options and for anyone to help you need to tell us what sort of power we are looking at

author: Zh4ng
author_rank: Unknown
timestamp: October 18, 2015, 12:02:59 pm
content: Thank you Simon,on the 3.3V rail a uC is supplied. He does not draw much power. On the 5V rail, a small motor (max. 500mA) and two LEDs.I have also thought of a greater voltage regulator on the 5V rail.I think I'll take the variant b with a bigger 5V regulator.....Thank Zh4ng

author: Simon
author_rank: Unknown
timestamp: October 18, 2015, 12:09:37 pm
content: Yes that should work. Obviously make sure your 3.3 V regulator has a low enough dropout voltage which is the voltage you need on top of your 3.3 V to make the regulator work. If for example your regulator has a dropout voltage of 2 V you would need 5.3 V to make it work properly and in fact preferably more. There are many low dropout regulators around which is what LDO stands for. Many regulators have a dropout of 1.7 V which is exactly what you would need and a bit too close really unless you are drawing only a tiny current so be careful you don't end up starving your microcontroller.

author: mij59
author_rank: Frequent Contributor
timestamp: October 18, 2015, 12:20:41 pm
content: I would go with option a, power both regulators form the 9V rail, it gives the most headroom for regulation.With option b you got only 0.7V headroom for the 3.3V regulator, electric motors are nasty loads, noise could affect the 5V regulator.

author: Zh4ng
author_rank: Unknown
timestamp: October 18, 2015, 12:22:02 pm
content: Hi Simon,that's why I selected the LD1117. LDO at about 1V -> 500mAI hope I have the datasheet correctly understood.Zh4ng

author: Simon
author_rank: Unknown
timestamp: October 18, 2015, 12:25:33 pm
content: As mig59 said you could get problems from that nasty motor load.I'd put a schotky diode in series wih your 3.3V regulator and a bypass capacitor after it if you really do it this way. It may be better to put the 3.3V regulator on the 9V rail, if it's only a few mA you won't have much heat to dissipate anyway and at the end of the day the totaldissipation will always be the same.

author: Zh4ng
author_rank: Unknown
timestamp: October 18, 2015, 12:26:33 pm
content: Quote from: mij59 on October 18, 2015, 12:20:41 pmHi mij59,good point. The noise could be a problem.Hmmmmmmmm, then i need some filtering.Zh4ng.

author: Zh4ng
author_rank: Unknown
timestamp: October 18, 2015, 12:30:46 pm
content: That's the nice thing about this forum. Many nice people to help.At the noise I did not think. I let the 3.3V rail at 9V with some filtering.Many thanks to Simon and mij59Zh4ng

author: c4757p
author_rank: Super Contributor
timestamp: October 18, 2015, 01:08:50 pm
content: 1117 is hardly an LDO by modern standards. If you're just powering a MCU and you want something small and compact, go for something like aTLV70233, 0.22V dropout at full (300mA) current and a nice SOT-23-5 package. Hard to beat.The comment about keeping the motors away from the 3.3V rail is a good one, but that's just as valid with the regulators in parallel. If your 9V rail isn't very low-impedance, spikes could just as easily backfeed through the 5V regulator, through the 9V rail, and up the 3.3V regulator. You'll need to pay close attention to power routing, filtering, and decoupling no matter how you go about this.

author: Ian.M
author_rank: Super Contributor
timestamp: October 18, 2015, 01:12:56 pm
content: Can you PWM the motor and run it from the 9V rail to avoid the need for a 5V regulator. The LEDs can surely run off either 3.3V or if they are blue or white, the 9V rail.