help with potential voltage spikes - Page 1
Forum section: electronics / beginners

author: psubond
author_rank: Contributor
timestamp: October 04, 2018, 08:14:39 am
content: I saw Great Scott's boost converter write-up and video. He mentioned that there was a problem with voltage spikes from the inductor when power was turned off, link found herehttps://www.instructables.com/id/DIY-Boost-Converter-How-to-Step-Up-DC-Voltage-Effi/My question is, would a flyback diode across the inductor like one used for a motor or relay work with his schematic for eliminating the spike when power is turned off?

author: Mario87
author_rank: Frequent Contributor
timestamp: October 04, 2018, 10:36:12 am
content: From what I understand...yes it should work. However there are more knowledgeable people than me on here who may well be able to explain as to why it would not work. I myself cannot see a reason.After all, a relay is just an inductor (coil of wire that magnetizes when voltage is applied), and they use flyback diodes, but as above, there may be something I am missing that someone else can clarify better.

author: Zero999
author_rank: Super Contributor
timestamp: October 04, 2018, 10:50:25 am
content: Yes it will work, but I wouldn't expect it to be very efficient, as the inductor appears to be the sort used for EMI filtering, rather than switched mode power supplies.Also make sure you use the IRLZ44N, rather than the IRFZ44N. A logic level MOSFET is essential for decent performance with 5V of gate drive.

author: oPossum
author_rank: Super Contributor
timestamp: October 04, 2018, 01:45:55 pm
content: Towards the end of the video he mentions that the output voltage will rise when the load is disconnected. This is due to the microcontroller firmware not implementing a proper control loop. The easy hardware fix is to use one of the may switching regulator chips rather than a microcontroller.Don't put a diode in parallel with L1. It will no longer work if you do that.

author: Mario87
author_rank: Frequent Contributor
timestamp: October 04, 2018, 10:30:15 pm
content: Quote from: oPossum on October 04, 2018, 01:45:55 pmCan you explain why it would stop working?

author: oPossum
author_rank: Super Contributor
timestamp: October 05, 2018, 12:51:03 am
content: From Wikipedia:QuoteSo a diode in parallel with the inductor in either direction would pass current through the diode instead of allowing it to flow to where it is supposed to go.

author: psubond
author_rank: Contributor
timestamp: October 05, 2018, 04:57:48 am
content: ok ty, I have a spot on the board for the flyback diode but a simple fix if it will prevent it from working is to just not install it when i solder the board.

author: Mario87
author_rank: Frequent Contributor
timestamp: October 05, 2018, 08:12:44 am
content: Quote from: oPossum on October 05, 2018, 12:51:03 amThanks for that, just had a look at the diagrams on the Wiki page as well, all make sense now.

author: David Hess
author_rank: Super Contributor
timestamp: October 07, 2018, 01:06:22 am
content: D1 as shown acts as a flyback protection diode. Adding a diode across L1 will prevent proper operation.The problem he mentions in the video is a rise in the output voltage when the *output* is disconnected and that is a problem with the control circuit which has a minimum on-time.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 07, 2018, 02:09:57 am
content: Euch, once again no current protection or measurement or anything.But I expect no less from Instructables: where plans go to die for ad revenue.If you must have a basic circuit -- one not using an integrated controller -- consider this:https://www.seventransistorlabs.com/Images/555%20Boost.pdfThis is a peak current mode control, so the transistorwill not die*from transient conditions, whether startup, short circuit or anything else that happens.*I mean, I'm not guaranteeing anything. There are a lot of compromises to keep the parts count low. But it's a far sight safer than an open loopmicrocontrollerthat can't possibly see what's going on, or keep up with it.As you can guess from the number of support components (transistors, an op-amp..) needed, this is practically an abuse of the 555. The circuit is using it for the flip-flop and timer functions, to replicate most of the operation of a UC3843 -- which is no harder to understand and uses far fewer parts, with far better performance. I do suggest reading up on it, it's very good despite its age.I wish people would forget about the 555 and learn the 3842 series instead; they're possibly as universal as the 555 to begin with. It's also not hard to understand that there are two variables in a switching converter, not just one: the output voltage (and the compensation required due to change in that voltage, drawing current through the output capacitor), AND the inductor current (and the compensation required for it).Tim