Simple question. What is the purpose of diode across 7805

[William At MyBlueRoom]

I've seen this in a few schematics, but not all. I do like to employ a diode inline with external power but I would like to know what the purpose of a diode across a regulator is.

 

[justDIY]

it is meant to protect the transistors inside the regulator when it is being used in a high capacitance circuit

for example, lets say you have a huge output capacitor for some reason, and then your Vsup disappears for some reason (crowbared) - now the output side if the reg is at a higher potential than the input side, so current will want to flow from the output cap, through the reg, reverse biased ... the datasheet explains how to calculate wether or not you need that diode, as the reg have some built in protection.

so, with the diode there, if the output somehow reaches a higher potential than the input, the diode becomes forward biased and conducts the excess voltage safely around the reg.

 

[Russlk]

if the input in shorted to ground, the charge on the output capacitor could damage the 7805. The diode discharges the output cap.

 

[William At MyBlueRoom]

Thanks, I'll have to consult the datasheets to see if I require it. Are only 78xx regulators affected?

 

[justDIY]

all linear regulators are affected, in varying degrees ... the datasheet is the best source for specific information

 

[hyedenny]

but.... isnt that diode backwards???

 

[Russlk]

but.... isnt that diode backwards???

Of course! Read the discussion!

 

[mramos1]

If you flip the diode, lots of power will go to the circuit around the regulator.

The way it is now, voltage on the output will run back to the input (with a diode drop) and keep the regulator happy.

You probably looked at it like I did the first time, I do everything left to right.
Power is on the right.

 

[Nigel Goodwin]

You probably looked at it like I did the first time, I do everything left to right.
Power is on the right.

Yes, it's a confusing way to draw a circuit!.

 

[hyedenny]

oh whoops - I DID look at it backwards.... and they dont even drive on the wrong side of the road in Canada!

 

[Nigel Goodwin]

and they dont even drive on the wrong side of the road in Canada!

NO?, I thought they did?.

 

[mramos1]

Well we know he did not mirror it for a PCB my mistake because the text is still correct.

I stared at it. I went, OK power on the right. Do it that way. You can do it.
Only after I knew what the diode was for.

So he took the schematic from people that drive on the wrong side of the road... Or draws backwards.

Hey I learned something as I have never used a diode to feed the power back to the regulator (never had a big cap on that side I guess). I have used diodes to keep the user from attaching the power backwards many times.. So it all worked out.

So, what side do they drive on in Australia, just wondering? In Canada, it is on the "RIGHT" side..

 

[justDIY]

The way it is now, voltage on the output will run back to the input (with a diode drop) and keep the regulator happy.

under normal circumstances, no current will be flowing through the diode at all ... since the input voltage will be of a higher potential than the output, the diode will be reverse biased.

if something happens to cause the input voltage to drop below the output will the diode will become forward biased, and then do its job, and provide an low resistance path for current to discharge from the circuit, rather than forcing its way backward through the regulator

I wonder if there is any benefit to using a high speed schottky barrier instead of a gp silicon in this application, or would that depend on the power levels involved?

 

[Nigel Goodwin]

I wonder if there is any benefit to using a high speed schottky barrier instead of a gp silicon in this application, or would that depend on the power levels involved?

No, there's no benefit, fast recovery diodes are only really required for switchmode power supplies, or supplies fed from line output stages. If you replace them with normal rectifiers you soon find out why! 8)

 

[DigiTan]

I've got a question. Why does the diode go from to the output to input? Couldn't you just have the diode in series with the output to block reverse current and drop more voltage than the regulator in this scenario?

Also, if the diode in the diagram allowed too much leakage current, couldn't that cause a positive feedback loop and potentially make the regulator loose stability?

 

[ljcox]

If you put the diode on the output side is series with the load, it would drop 0.7 or more Volt and reduce the regulation.

The leakage current of the diode will go into the load so the output current of the reg will be that much less than it would be if the diode was not there, for example, if the load current was 20 mA and the leakage current was 1 mA then the reg output current would be 19 mA.

Assuming that the output current is >> than the leakage current, it should not cause any problems.

 

[mramos1]

So with the diode backwards, the regulator will take all the voltage through it before the diode? I have never tried it, and will not try it. I do not use a diode there anyway. But I will remember that. Thanks.

 

[ljcox]

So with the diode backwards, the regulator will take all the voltage through it before the diode?

I don't understand what you mean. I assume you are referring to the case where the diode is across the reg as suggested in the data sheets.

 

[Russlk]

A diode in series with the output will drop 1/2 volt which is not desirable, and the diode resistance will reduce the regulation.

Leakage in the diode will reduce the amount of current that the regulator has to supply to the load. There is no feedback and no harm to the circuit as long as the leakage does not exceed the load current.

 

[mramos1]

Russlk: Thanks for reading my comment and explaining.

I have never tried it, but figured if the diode "was in backwards" which it is not in this case, but if it was would conduct and bypass the regulator.

As a 78xx would have positive voltage coming in and the diode would allow DC to pass through it to the circuit.

ljcox, sorry just re-read yours too..

Anyway, he had it right to start with...