Diode on light load keeps failing

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Ferrite beads are now available that offer appreciable inductances. If the Schottky diode was failing for the reason I suggested the current pulse wou ld be very brief, and even a ferrite bead might add enough inductance to sp read it out over a tolerable - though still short - period.

Getting over 20A when the average current is 0.2A implies a 200usec on time (in a 20msec mains period). That's 4x10^5 amp/sec for a triangular wavefor m. 4.3uH (which you can get out of a big ferrite bead) would generate a vol tage drop of about 2V, which might be enough to smear out the peak.

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Bill Sloman, Sydney
Reply to
bill.sloman
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A big ferrite bead will saturate on the order of 10uVs.

But I don't know where you get mains frequency from? This is a switching converter.

At switching frequencies, the ferrite bead helps -- on the ns scale, as Kevin noted -- to reduce peak voltage and current. Often a ferrite bead is used with a diode to provide damping on turn-off ringing (which is between transformer LL and diode CJO), and introduce a delay to turn-on (where a hard switching event might lead to more EMI).

It's turn-on delay because the bead saturates after a few microwebers (so, applied voltage for some nanoseconds, usually), then full load current passes. The same delay occurs at turn-off, but it's towards zero current so it just looks like normal ringing.

The downside to turn-on delay is more peak voltage on the primary side switch, but it probably needs an RCD snubber anyway.

Tim

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Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
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Reply to
Tim Williams

Care to tell what manufacturer / type (or any other identifying criterion usable in a search on aliexpress) these lights are?

Thx

Reply to
Dimitrij Klingbeil

The ones I use now don't have a brand and are sold by all kinds of vendors, often as a package with battery and charger:

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However, don't ride around with those without first installing a diffusor lens, else you'll blind other traffic participants:

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There is an efficient switcher inside and thus the brightness remains the same as the battery voltage drops. Gets a bit hot at full brightness on 105F days but ok in the wind while riding and so far I didn't have a thermal shutdown. I usually dim down when stopped for longer than a traffic light phase. I always ride with daytime lights.

The other older (and weaker) light is no longer made and I cannot recommend the new version because of a flimsy handlebar holder. Looks similar to this but mine is without the bulge around the head and has a removable 18650 Li-Ion cell:

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As for handlebar holders nowadays you have to often make such stuff yourself. The industry is largely no longer capable of producing ones that are sturdy enough. Same goes for tail lights, racks, panniers MP3 players, batteries and other gear. So ...

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--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

Why do you think that? Pimpom just descried it as a 5Vstandby supply.

My point was that the peak current might be a problem, and - if it were - it wouldn't take much to get the peak current below the level where it might be dangerous. A resistor would be even cheaper.

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Bill Sloman, Sydney
Reply to
bill.sloman

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Pimpom just called it "a 5V standby supply".

But - thinking about it - the power supply itself is going to be an off-lin e switcher, and the standby supply is going to driven by narrow spikes comi ng out of the high-frequency inverter that turns high voltage floating DC d ervied from the mains into an isolated low current supply.

Those spikes are going to have a frequency of the order of 100kHz or highe r.

If peak current is what is wrecking the diode that is failing it could be a t lower than the 20A that seems to be the limit for 60Hz repetition, so a a 0.2A output, the mark to space ratio has to be 100:1.

On a 100kHz waveform, that's a 100nsec on time every 10usec.

The current is turning on and off a lot faster than in the 50Hz case I went through, but the voltages are still much the same, so a 4.3uH inductor fer rite bead/chip could significantly smear out the peak current.

If it saturates at 10uVsec, you'd need to put an average of 100V across if for that 100nsec to get it into saturation, which would be difficult to man age.

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On the primary side.

Pimpom's diode seems to be on the secondary side, where the output voltage across the capacitor is 7.7V. A 10 microvolt.seconds capacity in the ferrit e bead means that it's going to take more like a microsecond to go into sat uration.

If we are talking about narrow spikes coming out of a 100kHz or faster swit cher merely to provide standby power, the charging period in each cycle is likely to be shorter than a microsecond.

You are making the same mistake as Kevin - thinking about the wrong side of switcher.

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Bill Sloman, Sydney
Reply to
bill.sloman

Thanks. I think I've seen these things advertised as helmet lights and wondered whether their construction and thermal properties would really make them a better design for use on a bicycle.

As for improvising holders and soldering together replacement structural parts for a luggage rack out of copper pipes, I'm no stranger to that :)

Reply to
Dimitrij Klingbeil

Perhaps I should have mentioned that it's a switching supply and that the rectifier is on the secondary side. I thought that the fact that it's a Schottky with a low-voltage output in a fairly modern TV would make readers assume a switcher with the diode on the secondary.

Reply to
Pimpom

f the

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ing

-line switcher, and the standby supply is going to driven by narrow spikes coming out of the high-frequency inverter that turns high voltage floating DC dervied from the mains into an isolated low current supply.

igher.

be at lower than the 20A that seems to be the limit for 60Hz repetition, so a a 0.2A output, the mark to space ratio has to be 100:1.

went through, but the voltages are still much the same, so a 4.3uH inductor ferrite bead/chip could significantly smear out the peak current.

if for that 100nsec to get it into saturation, which would be difficult to manage.

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age across the capacitor is 7.7V. A 10 microvolt.seconds capacity in the fe rrite bead means that it's going to take more like a microsecond to go into saturation.

switcher merely to provide standby power, the charging period in each cycle is likely to be shorter than a microsecond.

e of switcher.

It is pretty obvious - when you think about it. I was initially mainly reme mbering the large current pulses we got with small power supplies with just a diode charging a capacitor, and was more interested in registering the t hought that diode might be being asked to handle a higher peak current than it could accommodate.

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Bill Sloman, Sydney
Reply to
bill.sloman

f the

se

nce to

n time

ut the

ing

-line switcher, and the standby supply is going to driven by narrow spikes coming out of the high-frequency inverter that turns high voltage floating DC dervied from the mains into an isolated low current supply.

igher.

be at lower than the 20A that seems to be the limit for 60Hz repetition, so a a 0.2A output, the mark to space ratio has to be 100:1.

went through, but the voltages are still much the same, so a 4.3uH inductor ferrite bead/chip could significantly smear out the peak current.

if for that 100nsec to get it into saturation, which would be difficult to manage.

s

ad is

ween

a

(so,

age across the capacitor is 7.7V. A 10 microvolt.seconds capacity in the fe rrite bead means that it's going to take more like a microsecond to go into saturation.

switcher merely to provide standby power, the charging period in each cycle is likely to be shorter than a microsecond.

e of switcher.

One approach that would be quick & easy would be to use 4 diodes in series parallel. Or perhaps 2 in series plus a 1ohm R.

NT

Reply to
tabbypurr

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ff-line switcher, and the standby supply is going to driven by narrow spike s coming out of the high-frequency inverter that turns high voltage floatin g DC dervied from the mains into an isolated low current supply.

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d be at lower than the 20A that seems to be the limit for 60Hz repetition, so a a 0.2A output, the mark to space ratio has to be 100:1.

I went through, but the voltages are still much the same, so a 4.3uH induct or ferrite bead/chip could significantly smear out the peak current.

ss if for that 100nsec to get it into saturation, which would be difficult to manage.

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ltage across the capacitor is 7.7V. A 10 microvolt.seconds capacity in the ferrite bead means that it's going to take more like a microsecond to go in to saturation.

r switcher merely to provide standby power, the charging period in each cyc le is likely to be shorter than a microsecond.

ide of switcher.

s parallel. Or perhaps 2 in series plus a 1ohm R.

Quick and easy, but hard to support with rational arguments.

--
Bill Sloman, Sydney
Reply to
bill.sloman

The construction is good but sometimes the quality in production is a bit lacking. Forgot to apply thermal paste, screws not torqued down well and such. This is easy to fix though.

I never wear anything on the helmet because many of my riding sections are forested singletrack. The helmet is already bad enough in catching overhanging branches and I don't want to add to that risk. I kept the straps though in case I need it for emergency work or rescue particiaption at night.

Neither am I but it is still sad to see that you can get almost anything for automotive including a sturdy mount yet for bicycles most stuff is flimsy. Even expensive gear such as a B&M light on my road bike vibrated off and then almost fell into the front wheel spokes.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

As far as current is concerned, I have schottky diodes rated the same as or higher than the original, so there's no need to parallel them. The problem is that I don't have any rated higher than 45V.

If this were a low frequency circuit, I could just place two or more PN diodes in series and parallel each with a high-value resistor to distribute the reverse voltage evenly. But this isn't very practical with Schottky diodes that could leak several mAs when they get hot under load. I don't have any of the newer low leakage types.

Reply to
Pimpom

re diodes failing

fair enough, if you're confident high current pulses aren't the cause

I'm not seeing how leakage would be a problem. It means you wouldn't need the resistors.

But why use 2 schottkies when you can use one cheaper PN diode. It would need to be larger to handle the extra heat dissipation.

NT

Reply to
tabbypurr

I have diodes rated 10 times or more than the original.

An ordinary PN diode will be too slow and, as I mentioned before, the only ultrafast ones I have are high voltage types with a high forward drop. I haven't completely rejected the idea of using that.

Reply to
Pimpom

A friend uses a home-built 3000LM chest-mounted LED lighting system for night hiking, complete with digitised voice to report battery levels.

Might a lighting vest work for you on a bike?

Reply to
Clifford Heath

Nah, I use something similar to what has worked for motorcyclists for around a century.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

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