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I need clarification on an error on my part

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Ian Rogers

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I don't normally need help.. I actually don't need it now but!!

I have used the LM2567t-5 for years now.... No problems, it served me well.
I used the design in the datasheet as I needed up to 3A at 5V.
upload_2018-1-19_13-25-50.png
Proven and solid!!

Several years ago I couldn't get the 1n5822 so I replaced with a 21DQ04... Still the best thing since sliced bread..

But!! there is always a but! This superfast diode went south... So I went back to the 1n5822 ( so I thought ) I bought 1n5819's the difference has become apparent, as the last 5 units have been a disaster... The inrush current is @ 1.5A so a 2A fuse was deployed.. To my horror I realised that the 1n5819 is a 1A device and not as fast as the 1n5822..

So... The ADC in my micro just goes high, in my view the ground rail and been lost to the Avdd, also the loadcell amp is fried and the decoupling cap on my ICL7660 is swollen and the ICL7660 doesn't convert!!

The clarification:- Just by using the smaller diode is not preventing enough shoot through on the chip above? Thus all immediate chips are getting too much negative voltage? All devices after the swollen cap seem fine...
 

MrAl

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Hi,

Running the diode at full rating probably isnt a good idea.
And yes, if the diode is rated less then it's voltage drop at a given current would normally be higher because the lower rated diode has less cross sectional area, just like what would happen with a resistor of less cross sectional area but made with the same material. a resistor with 1/2 the area of another resistor made from the same material and same length would have 2 times the resistance and thus 2 times the voltage drop for a given current.

The 1N5822 isnt the best choice either though, even though i have used it in this very same application. That's a 3 amp diode as you know, and with a 3 amp output the diode rating should be more like 4 amps. That's if you use the full 3 amp output rating though. For my app, the current was LIMITED via a secondary transistor circuit that limits the current to 1 amp so the output can never go above 1 amp. I dont know if you have that kind of circuit but i suspect that you dont. Thus your circuit depends entirely on the behavior of the chip alone.

I cant remember if this chip has slow start built in or not. If it doesnt, then the initial current could be high because the cap is discharged when the circuit is first turned on.

Interesting that you should bring this 1 amp diode question up here because i've often thought about what would happen with just a 1 amp diode in a 1 amp application. To be more accurate though, i would either find a diode that is rated for 4/3 amps or limit the current to 3/4 amps if i used the 1 amp diode.

The diode normally has to deal with the max current in the inductor, and we think that is 3 amps, but it can actually be higher and that pushes the RMS current higher and thus the power dissipation. I think that is the reason for the 4/3 rating factor requirement although it's been a long while since i looked at this circuit now.

If you throw a 5 amp diode in there it will probably run forever.
 

kubeek

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The clarification:- Just by using the smaller diode is not preventing enough shoot through on the chip above? Thus all immediate chips are getting too much negative voltage? All devices after the swollen cap seem fine...
Hi, a slow diode will just waste more power in heat. The figure is reverse recovery time, therefore how long the diode stays conducting after it should have shut off. This will burn more power in the diode, but it will not cause any negative voltage on the output.
However when the diode finally dies and if it dies open circuit (which is likely since the diode in this case is weak and the transistor in the LM2576 is rated for higher current and has better cooling, so the diode will sooner or later be burnt to a crisp), then you will have pulsed high voltage on your output caps, which has the potential to kill any subsequent circuitry.
 

Ian Rogers

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I have re-called the five offending units... I have a VERY decent load tester that can load the circuit to 3 amps...

If I load to 1A and check the output on a scope then check the that on the offending diode, that'll be conclusive..
If you throw a 5 amp diode in there it will probably run forever.
The MBR1060 ( which I use on a sister project ) has never failed..
 

ericgibbs

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hi Ian,
I used the Vishay devices in my SMPS, never a problem,
E
 

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Beau Schwabe

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We use this regulator all the time in industrial applications.... for 5V logic we usually place an additional EMI filter on the output. You can also place a forward biased diode in series with the inductor between the cap and the inductor... this way the FB (feedback) still regulates properly.
 

Ian Rogers

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We use this regulator all the time in industrial applications.... for 5V logic we usually place an additional EMI filter on the output. You can also place a forward biased diode in series with the inductor between the cap and the inductor... this way the FB (feedback) still regulates properly.
This will require a PCB mod... The design is sound!! Has worked for over 10 years...
 

be80be

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Thermal Shutdown and Current Limit Protection My bet is the Current Limit Protection kicks in way before the diode pops if there both 3 amp rated.

Remember them days when you could get lots of glass diodes I had so much fun lighting them up like leds it took lots of juice lol.
But i don't think a 1 amp in your case would hold at 3 amps those old glass ones would light up at twice there rating and die.
 
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Beau Schwabe

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What is the purpose of that diode?
The purpose would be to minimize any negative spikes from the inductor that could make it's way into the load. The positioning of the diode would create a "perfect diode" since the feedback to the regulator would compensate for the diode drop... IOW the output would still be at the correct regulated voltage.
 

kubeek

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The purpose would be to minimize any negative spikes from the inductor that could make it's way into the load. The positioning of the diode would create a "perfect diode" since the feedback to the regulator would compensate for the diode drop... IOW the output would still be at the correct regulated voltage.
I dont think there is any possibility of a buck converter producing negative output voltage, and the diode there only makes additional unnecessary losses.
 

MrAl

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The purpose would be to minimize any negative spikes from the inductor that could make it's way into the load. The positioning of the diode would create a "perfect diode" since the feedback to the regulator would compensate for the diode drop... IOW the output would still be at the correct regulated voltage.
Hi,

What negative spikes? When do you think this occurs?

The extra diode would drop efficiency significantly also, in a circuit that is not exactly modern too.
At 1 amp and 5v output in a 100 percent efficient converter we have 5 watts in 5 watts out.
Add a diode, and we have to have 5.7 watts in and 5 watts out, which means we loose 12 percent.
That on top of a converter that is only 80 percent to begin with, we drop down to around 68 percent efficiency.
So we go from 80 percent to 68 percent. There better be a darn good reason for that diode :)
 

MrAl

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The whole purpose of the schottky is to prevent negative spikes... BUT with an inrush of 2A mine was spitting out it's dummy as it was a 1A device..
Hi,

Which diode are you talking about. I was talking about the "extra" diode in series with the inductor, which i dont think does anything but drop efficiency at best.
That would be a second diode not the original Schottkey.

Are you perhaps saying that you saw negative spikes on the very output, across the cap?
 

Ian Rogers

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Which diode are you talking about. I was talking about the "extra" diode in series with the inductor,
I'm talking about my original.... As you say! A second diode is burden, not required... I have ( however) seen a rectifier connected back to Vin... Again, I don't know what that does either...
 

MrAl

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I'm talking about my original.... As you say! A second diode is burden, not required... I have ( however) seen a rectifier connected back to Vin... Again, I don't know what that does either...
Hi,

Maybe to keep some input voltage present in the event the input power is disconnected and there is little load current. The output would have full voltage for a while while the input had nothing if there was no diode.
We see something like this in some LM317 circuits.
 

crutschow

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Maybe to keep some input voltage present in the event the input power is disconnected and there is little load current. The output would have full voltage for a while while the input had nothing if there was no diode.
We see something like this in some LM317 circuits.
Yes.
Many regulators, such as the LM317, are sensitive to any reverse voltage from output to input, such as can occur when the input voltage is suddenly lost and the output capacitor still has voltage.
The diode from output to input prevents that reverse voltage from being more than one diode drop.
 
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