Hello,

I've got a problem that's been plaguing this circuit for a while. I think the answer is some kind of circuit protection but I need to understand the problem.

The symptom is that the LM7805 5V regulator gets destroyed sometimes and then it reads a voltage just a few mV less than the input voltage, ~12V in this case. Once this happens I need to replace the regulator.

I'm not sure what causes this issue. Today I tied the PANEL_V and BATT_V nodes together and after that, I read the high voltage on the 7805 output pin.

I know that I need to add reverse connection protection. I've googled this but have not come up with an answer. So if anyone can point me at some good resources that would be appreciated.

Any ideas on what is amiss with this circuit and why I keep killing 7805's?

Thanks.

 

Just read this in the datasheet: "Each type employs internal current limiting, thermal shut down and safe operating area protection, making it essentially indestructible."

So how do you do it?

Does this IC get hot?

considering the protections it has built-in there are only two more things left I guess, either you put a higher voltage than 5V at the output causing a current to flow back into the IC (don't know if it has a protection against reverse current) How come you don't damage the PIC with 12V power supply?
Is there anything tied to the 5V that can cause a reverse current?
Or you have a higher than maximum rating input voltage, also not very likely, because it's a battery...

I'm asking myself how fast is the current limiting function of this particular chip, can the maximum peak output current be exceeded by the discharged cap at the output when powering up the circuit and therefore damage the regulator?

 

You also need to find the right matching capacitors. I think C1 and C2 should be more like .33uF and .1uF respectively. 100uF and 10uF are rather high, and could cause the output to become very unstable. This could cause the regulator to not regulate properly. Try adjusting the cap values and see if it helps.
Der Strom

NOTE: It is also always a good idea to put a heat sink on the 7805, especially if you are dissipating 1 watt or greater. What is your battery voltage and the current output of your regulator? You can find the power dissipation by using the formula, (Vin - Vout) * Iout

 

I noticed you have fairly large value caps
on input and output of your regulator.

Look at page 11 of attached datasheet especially fig. 1.

You may also want to refer to page 1, upper left
for recommendation on cap size, unless you selected
the caps you have for a specific reason.

Just a thought.

Edit: DerStrom 8 replied while I was typing.

 

Edit: DerStrom 8 replied while I was typing.

Sorry about that, Pencil. I tend to do that a lot

 

How come you don't damage the PIC with 12V power supply?

I have the same question.

 

related to what has already been stated, I found a website saying "With the LM7805, the output capacitor should not exceed 1mf, as larger values could damage the 7805 due to backfeeding of current when power is switched off." the attached schematic shows 1uF,though. so maybe they meant 1uF max and "der strom 8" and "pencil" are right?

 

related to what has already been stated, I found a website saying "With the LM7805, the output capacitor should not exceed 1mf, as larger values could damage the 7805 due to backfeeding of current when power is switched off." the attached schematic shows 1uF,though. so maybe they meant 1uF max and "der strom 8" and "pencil" are right?

I have seen a few cases in which people think "mF" is the same as "uF". It may just be a mistake like this, as you suggested

 

These chips will self-regulate their power dissipation when necessary. You can run them without a heatsink and they will keep themselves quite safe.

The thing that kills them dead dead dead instantly is a reverse bias, when Vout < Vin. That can happen if the load is light, the input goes to zero fast, and their output cap holds up their output. I do not see that in your schematic (assuming the schematic is 100% correct) but something to try is put a diode (1N4001 is good) across Vout to Vin.

That should keep them alive forever.

 

Bingo. Ernie nailed it. The backwards current protection diode mentioned in the datasheets is my opinion of the right thing to do. See also diode protection for the adjust pin when you use a 317 or similar and have a cap on the adjust pin.

 

My best guess is what ernie said. When I first saw it I didn't really see a need, but install a simple diode where it is reversed biased between Vin and Vout. It will allow the caps to discharge without putting voltage on the regulator backwards.

 

Why don't they just implement a diode in the chip if it's such an obvious point of failure?

 

Why don't they just implement a diode in the chip if it's such an obvious point of failure?

Well, there are LOTS of good reasons:

It's a well documented failure mechanism that every vendor is susceptible to. If Simon Says "Don't do this" and you do it anyway then you're out, not Simon.

This device has been in production since... the 70's maybe? Do you really want to pull engineers off some new product to redesign one of your oldest longest lasting devices, loose valuable resources that could be designing new devices that keep you competitive, and perhaps induce a different fault mechanism?

Finally, and probably most importantly, this ain't no teeny tiny small signal diode but a real current handling rectifier that needs to pass a large current from a large capacitor. Exactly how much really depends on the end user, not the chip designer. Plus that diode takes space to make, using more space on a wafer means fewer chips on that wafer, making them more expensive.

I see no reason to fix what ain't broke.

 

Finally, and probably most importantly, this ain't no teeny tiny small signal diode but a real current handling rectifier that needs to pass a large current from a large capacitor.

Thanks Ernie, actually I wasn't really aware of this failure mechanism. I've seen tons of circuits where they put 47uF or more at the 78xx series output. (they never blew) Although the mechanism of reverse current may be documented somewhere, I've not seen a recommendation of how much is TOO much in terms of capacity.
I just opened a datasheet from Fairchild and at no point it's saying something like: "maximum peak reverse current". There is clearly something missing in the datasheet. It's kind of obvious that someone could try to put a bigger capacitor than in the typical application notes.

Well , I just learned something.

edit: well of course, you are right. from another datasheet "When a surge voltage exceeding maximum rating is applied to the input terminal or when a voltage in excess of the input terminal voltage is applied to the output terminal, the circuit may be destroyed.".

I guess there is no way to know how much is too much, because the internal circuit parameters are unknown...

 

Thanks all for the responses. I use a 12V battery and a bench supply that I change the voltage on when I'm ready to test power regulation the solar panel (bench supply in this case) to the battery. I could see the input getting disconnected (the 12V battery) and it dropping lower than the 5V on the output.

Sometimes the PIC survives the 12V treatment but mostly I replace it along with the regulator. I will examine the caps and the diode solution. Thanks again!

 

Something doesn't sound right there looking at your schematic. You have a 12v battery, into a 100uF Vin cap (at 12v) and 10uF Vout cap (at 5v)? So when the battery is disconnected you say Vin drops much lower than Vout?

The diode from Vin->Vout is always a good idea but I'm not sure that is your problem here.

The schematic shows a FET driving a motor, and what looks like a "disconnect" type solar panel regulator FET QZ? The disconnect type solar regulator will let a 12v panel float to 22v or so when open circuit so when regulating it is constantly attaching a 22v source to your battery, which may have a slightly high impedance if it is SLA type.

I'm not sure what the motor is doing but I think you might have 22v supply spikes from the solar regulator. I would add a resistor before the 100uF Vin cap C1 on the 7805, to kill the spikes. Make the resistor as high as possible so that probably means dropping (say) 2v at normal operating current. You could also increase C1 from 100uF to 1000uF which will reduce the spikes further and also further reduce the possibility of Vin being < Vout.

Also you might want to change from a disconnect solar regulator to a shunt type regulator that shorts out the solar panel when the battery >X, this will give you a much reduced wattage and heating in the panel and increase panel life. Obviously you need to add a series diode so when you short the panel you don't short the battery.

 

I'm not sure what the motor is doing but I think you might have 22v supply spikes from the solar regulator.

spikes maybe a good point, but 22V is not enough to kill your 7805, right?

 

Yes it's a disconnect type solar panel regulator. So voltage can swing like that. The thing I don't like about the shunt method is that I have to burn something like a 1/4 - 1/2 watt and do something with the heat. The device runs outside so weatherproofing and ventilation are competing goals. I've read about designs that short the panel when regulating which sounds interesting, as you're not also making a radio antenna out of your panel transmission lines with the regulation switching, but you'd have the same voltage swing problem I think.

The motor drives a small 12VDC pump (from 800mA to 2.5A), or drives banks of LED's (12VDC from 1A - 4A).

I definitely see problems when it's switching high current panels, 7A, so fixing those spikes needs to be done. I will try out your idea by adding a series resistor and large filter capacitor in front of C1, the 7805 input cap.

Here's the updated power section. I am currently running the 100uF and 10uF caps. I'll change to what's in the datasheet. I got those values off a microcontroller how-to a couple of years ago and it's worked until now.

 

adding a series resistor and large filter capacitor in front of C1, the 7805 input cap.

I think a 47ohms and a 1000mfd will do that.

 

So help me out with calculating the power dissipated in that 47 Ohm resistor. Assume max of 200mA for circuit (PIC, status LED, etc.)

So voltage drop would be: .2 * 47 = 9.4V. Something must be wrong because I need at least 7 to run the 7805.