High side switching or enabling a voltage supply - Page 1
Forum section: electronics / projects

author: csreades
author_rank: Newbie
timestamp: December 16, 2021, 09:07:03 pm
content: I want to design a board with uC, interfacing etc all which I have done before and moderately comfortable with. What I am not sure about and struggling to find answer to is a question about supplying higher voltages and switching these. The board I am designing will be interfacing with a device that has a specific power on sequence, involving two high voltage 40V (0.6 A peak, 4 W average) lines that need to be enabled after some initial communication. The issue I have is that these two lines have the grounds tied together on the device, so a standard low side MOSFET switch will not enable me to control them lines indepdently. My other thought/issue is that I see a lot of the time devices that have a single 12V or 5V supply but then internally use 20/30/40 V so they must be using some design to up the voltage (whilst using increased current). I wonder if these two functions can be combined leading to a case where I have:Single 5/12 V input (VIN) that gets regulated down to my uC. This uC enables the device and checks everything is working, then it enables the high voltage supply lines which takes the VIN and steps it up to the required 40V.Is this the "correct" way of doing things? Currently I can only think of using a mechanical relay and a 40V PSU, but it just seems such a sticks and dirt solution I wanted something neater (and with less voltage bounce issues).As I think it will explain my questions my background is Mech. Eng. and I just wish I knew more about electronics. So there may be an obvious IC choice but until I know what the chip is called or where to look I am hitting a wall.

author: David Hess
author_rank: Super Contributor
timestamp: December 17, 2021, 03:53:56 am
content: It is better to enable a supply through the regulator than to switch the output because the transition can be much better controlled. The problem is that if the high side switch is too fast, then inductance in the wiring may cause destructive ringing, however there are high side switches designed to produce a clean switching waveform.

author: csreades
author_rank: Newbie
timestamp: December 17, 2021, 09:03:22 am
content: That is interesting to note thank you, would you know any of those high side switches that might be suitable for my application? Would something like thishttps://www.farnell.com/datasheets/2334167.pdfbe suitable for my applications?This would solve the switching of the high voltage line, but creating it is another challenge. Is it better to just use a single VIN of 40 V then regulate that down for the low voltage side of the circuit instead of using a low voltage in and generating the high voltage?

author: Siwastaja
author_rank: Super Contributor
timestamp: December 17, 2021, 09:10:13 am
content: If you are going to use a regulator anyway, just pick one with ENABLE pin. This has two purposes, first cost/area savings, because regulator already has a pass transistor it uses for regulation, so it can use the same transistor for switching the load completely off. Another is indeed better controlled rise time.If you are not going to use a regulator, i.e., the load uses "raw" voltage bus, then the best option is to buy an integrated high-side switch. Look up your favorite distributor, they have whole sections for such load switches. Plenty to choose from, relevant parameters are max voltage, max current and number of channels.The advantage in such ICs is, they offer current limiting, overtemperature shutdown, typically some sort of short circuit protection, can power inductive loads within reason. The disadvantage is limited switching speed so you can't PWM the loads (or need to do that at very low frequency, like tens of Hz).If you want a simple solution using discrete parts, a P-channel MOSFET on the high side works, but if the driving MCU works from lower voltage, then you need another N-channel MOSFET (or NPN transistor) to drive the gate of the P-channel MOSFET. This lacks current limiting and other protections so really only good for loads with small inrush current / small capacitance.

author: csreades
author_rank: Newbie
timestamp: December 17, 2021, 09:14:35 am
content: I was really hoping to use a single lower voltage input that can drive the whole system something around 12V and then generate the 40V on board. However if this is not the "done" way then supplying ~42V and regulating down is the way to go? An enable pin fitted regulator makes sense. That also opens the other possibility that I have electronically controllable voltage level?

author: MarkR42
author_rank: Regular Contributor
timestamp: December 17, 2021, 09:18:58 am
content: Right so if you want to step up the voltage from 12V to 40V, you'll need some kind of boost converter circuit. So choose one which has an enable pin and you can drive that from your microcontroller, job done? Or not?

author: Siwastaja
author_rank: Super Contributor
timestamp: December 17, 2021, 09:21:10 am
content: If the power needed for the 40V bus is small (say, maybe a watt or two), then it makes sense to supply from 12V and boost to 40V on board. On the other hand, if significant part of the total power goes into the 40V thing, then just supply more than 40V and regulate it down to 40V with a DC/DC buck regulator. Then you can further buck down 12V, 5V, whatever you need. Note that 42V might not be enough, linear regulators have minimum dropout requirements and DC/DC bucks have maximum duty cycle or minimum off time, often around 90% or some hundred ns.If you want to control the voltage using the MCU, this is a bit more involved. Fine-tuning is easy: typical solution is to connect MCU DAC output (or an RC filtered PWM output), through a series resistor, to the FB pin of the regulator, so that the normal FB resistor divider is still functional, but the DAC injects a tiny current offsetting the FB pin.40V, 0.6A peak is really quite some power to boost from 12V. If you can freely choose the input voltage, I might not do it, it's realistically at least 3A from the 12V input. Of course you decide, if youneedto run this off a car battery, then what else can you do but to boost. Forcing the end-user to buy an inverter to run a 48V mains supply would be a nightmare.And by peak, I assume it's longer than some tens of microseconds, meaning capacitors won't help supplying it, and you need to design the power supplies to deliver the full peak current "continuously". The only compromise you can make by this being "just a peak", is in heatsinking.

author: csreades
author_rank: Newbie
timestamp: December 17, 2021, 09:48:31 am
content: Quote from: MarkR42 on December 17, 2021, 09:18:58 amJob done... if I know what one of those circuits looks like, struggling a bit to find a good example online.Quote from: Siwastaja on December 17, 2021, 09:21:10 amIt is 2-4 W total for the 40V line, with peaks around ~100 usec range. This as you say might be something capacitors can take care of and then not have a large current draw on the low voltage line. However it does sound an awful lot simpler by just having 48V input which can then be much lower current. The MCU - DAC - Regulator control is interesting, I will have to see if I can breadboard that and play with it (I am practically minded).

author: David Hess
author_rank: Super Contributor
timestamp: December 18, 2021, 02:50:25 am
content: Quote from: csreades on December 17, 2021, 09:03:22 amIt says slew rate on of 1 volt per microsecond maximum so it should be safe. Just a MOSFET will switch much harder causing problems.QuoteIt could go either way. I don't know that one is better than the other. It just depends on the application.