Author Topic: help with potential voltage spikes  (Read 689 times)

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Offline psubond

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help with potential voltage spikes
« on: October 04, 2018, 08:14:39 am »
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 here https://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?
 

Offline Mario87

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Re: help with potential voltage spikes
« Reply #1 on: October 04, 2018, 10:36:12 am »
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.
 

Offline Zero999

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Re: help with potential voltage spikes
« Reply #2 on: October 04, 2018, 10:50:25 am »
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.
 

Online oPossum

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Re: help with potential voltage spikes
« Reply #3 on: October 04, 2018, 01:45:55 pm »
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.
 

Offline Mario87

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Re: help with potential voltage spikes
« Reply #4 on: October 04, 2018, 10:30:15 pm »
Don't put a diode in parallel with L1. It will no longer work if you do that.

Can you explain why it would stop working?
 

Online oPossum

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Re: help with potential voltage spikes
« Reply #5 on: October 05, 2018, 12:51:03 am »
From Wikipedia:

Quote
The key principle that drives the boost converter is the tendency of an inductor to resist changes in current by creating and destroying a magnetic field. In a boost converter, the output voltage is always higher than the input voltage. A schematic of a boost power stage is shown in Figure 1.

(a) When the switch is closed, current flows through the inductor in clockwise direction and the inductor stores some energy by generating a magnetic field. polarity of the left side of the inductor is positive.

(b) When the switch is opened, current will be reduced as the impedance is higher. The magnetic field previously created will be destroyed to maintain the current towards the load. Thus the polarity will be reversed (means left side of inductor will be negative now). As a result, two sources will be in series causing a higher voltage to charge the capacitor through the diode D.

So 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.
 

Offline psubond

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Re: help with potential voltage spikes
« Reply #6 on: October 05, 2018, 04:57:48 am »
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.
 

Offline Mario87

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Re: help with potential voltage spikes
« Reply #7 on: October 05, 2018, 08:12:44 am »
From Wikipedia:

Quote
The key principle that drives the boost converter is the tendency of an inductor to resist changes in current by creating and destroying a magnetic field. In a boost converter, the output voltage is always higher than the input voltage. A schematic of a boost power stage is shown in Figure 1.

(a) When the switch is closed, current flows through the inductor in clockwise direction and the inductor stores some energy by generating a magnetic field. polarity of the left side of the inductor is positive.

(b) When the switch is opened, current will be reduced as the impedance is higher. The magnetic field previously created will be destroyed to maintain the current towards the load. Thus the polarity will be reversed (means left side of inductor will be negative now). As a result, two sources will be in series causing a higher voltage to charge the capacitor through the diode D.

So 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.

Thanks for that, just had a look at the diagrams on the Wiki page as well, all make sense now.  :-+
 

Offline David Hess

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Re: help with potential voltage spikes
« Reply #8 on: October 07, 2018, 01:06:22 am »
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.
 

Online T3sl4co1l

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Re: help with potential voltage spikes
« Reply #9 on: October 07, 2018, 02:09:57 am »
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.pdf
This is a peak current mode control, so the transistor will 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 loop microcontroller that 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
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
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