Protection diode placement

[sentry7]

Hey guys,

I need to put a safety diode in the power section of my design; below is a diagram of the power section. Since the 7805 regulator is voltage sensitive, should the diode go before the regulator?

[Zero999]

Yes, put the diode before the regulator.

If the diode were connected after the regulator, then not only would the regulator be no longer protected, but the voltage drop due to the diode would ruin the regulation.

[sentry7]

Thanks for the help guys. So this is the correct placement? I've also been thinking about the switch placement. SW10 is supposed to be a main power switch, but with it being where it is, even if it's in the "OFF" position, there's still power going to the regulator, right?

[sentry7]

SW10 in down position will short out the regulator and possibly destroy it. When switched off, the switch should connect to nowhere, not ground.

Sorry, I think my diagram is confusing (or I'm confused). The up position is OFF, and it is connected to nothing. the down position is ON and it is labeled as +5v (for the load). The skinny triangle is the way that the program shows voltage source.

[DanielS]

If there is any chance the input side of the 7805 might get shorted to ground or powered down while there is still any voltage on the output, you may want to add a diode from the 7805's output to its input to protect the 7805 against reverse biasing - many linear regulators are ridiculously sensitive to that too.

[mariush]

You don't need two 0.1uF capacitors at the input ( C20 and C21), one will be enough. However, most datasheets recommend an input capacitor of 0.33uF or higher if there's long input power cable.

I'd suggest going for a 10-22 uF ceramic or electrolytic capacitor rated for 25-50v (even if your input is only 7.5-9v) besides the 0.1uF ceramic capacitor. It wouldn't hurt to have some capacitance at the output as well, even if it's not required in most cases. I typically use anything between 33 and 100uF electrolytic, rated for 16-35v even if the output is 5v, because there's little difference in diameters and volume of capacitors at this specs.

As for reverse voltage protection, you may consider using a p-channel mosfet instead of diode, it's quite easy and just a bit more expensive. See Afrotechmods video for a nice tutorial :



ps. I'd also move the on/off switch right at the input jack, otherwise your linear regulator will be always on and waste a tiny amount of power (maybe less than 1mA).

[sentry7]

Considering the suggestions, here's is the final circuit; is it sound?

@mariush,

I didn't go with the MOSFET this time; for this project, the power loss due to the diode isn't critical. Thanks for that design tip though!

[Alex Eisenhut]

Why is the barrel positive? And put a reversed diode across the in and out of the regulator.

[eneuro]

As for reverse voltage protection, you may consider using a p-channel mosfet instead of diode, it's quite easy and just a bit more expensive. See Afrotechmods video for a nice tutorial :

What if reverse voltage will be outside mosfet gate VGS limits? This simply addon doen't look as good protection no more and 1000V 1N4007 looks much better than +/-20VGS only 

Maybe putting 12V zenners througth a 1k resistor could be fine to protect p-mosfet gate?

I didn't tested it practice, but probably I will, since overall concept is fine but it might be issue at some transients etc since low power mosfets will have very small total gate current, so they are more sensitive to ESD, etc...

[dom0]

If there is any chance the input side of the 7805 might get shorted to ground or powered down while there is still any voltage on the output, you may want to add a diode from the 7805's output to its input to protect the 7805 against reverse biasing - many linear regulators are ridiculously sensitive to that too.

Not necessary at 5 Volts. Except if your input can become more negative than a Volt or so (in non-automotive applications: how?), then you need it.

[tggzzz]

You have to define what you are protecting against - i.e. threat analysis. And then you must alway remember that you can't make something foolproof because fools are so damn ingenious.

I learned that early on when i built a simple solenoid pulser test box that was protected againse reverse voltage, short circuits and open circuits. "Bigfoot", for that was what all his colleagues in the production department called him, set the PCB on fire. Turned out there was a problem with a solenoid and, rather than draw the obvious conclusion, he decided there was a PSU problem. Since the solution to all PSU problems is a large capacitor, that's what he connected across the output.

[singapol]

Thanks for the help guys. So this is the correct placement? I've also been thinking about the switch placement. SW10 is supposed to be a main power switch, but with it being where it is, even if it's in the "OFF" position, there's still power going to the regulator, right?

The usual reverse voltage protection is a diode across input and output ( pin 1, 3 ). Download the 7805 datasheet and note the polarity of this diode. See pg. 11 ,12.

http://www.ti.com/lit/ds/symlink/ua7815.pdf

[sentry7]

OK guys,

Here is the final (I think) revision. Just so that you all know, there will be 9V applied to the jack.

[MarkF]

I think you want this.  Diode D10 where you have it doesn't make sense.

[sentry7]

I think you want this.  Diode D10 where you have it doesn't make sense.

Thanks. I'm not familiar with this technique; in fact, maybe someone can clear some things up for me. Let's assume there is no protection or shunt diodes in the circuit, just the voltage source and regulator with capacitors on the in and out. If I understand correctly, the capacitors clean up the voltage so that there is no (or little) sudden spikes. However, if the source drops quickly and there is still voltage on the output cap, the discharged current will go through the out of the regulator and through to com (damaging the regulator). To stop this from happening we put a shunt resistor across the in and out of the regulator to divert the current around the regulator to ground.

If we want to keep the current from reverse biasing the regulator, why can't we just put a forward diode at the output so that the voltage remains on the capacitor?

[tautech]

If we want to keep the current from reverse biasing the regulator, why can't we just put a forward diode at the output so that the voltage remains on the capacitor?

You can, but the output voltage beyond the diode will be one diode drop down ~0.6V.

Protection diodes specifically as detailed in datasheets is for protection against remote bulk capacitance discharging through the regulator after power down.

[rs20]

Why is the barrel positive?

This. Barrel = ground is the standard; you shouldn't go against this standard unless you have a specific reason.

[sentry7]

Why is the barrel positive?

This. Barrel = ground is the standard; you shouldn't go against this standard unless you have a specific reason.

I thought the two prongs above the "J" was the barrel, and the long rectangle was the center?

If we want to keep the current from reverse biasing the regulator, why can't we just put a forward diode at the output so that the voltage remains on the capacitor?

You can, but the output voltage beyond the diode will be one diode drop down ~0.6V.

Protection diodes specifically as detailed in datasheets is for protection against remote bulk capacitance discharging through the regulator after power down.

  Of course. Now it makes sense. Can someone explain the change that MarkF made?

[rdl]

The idea is that when the regulator is turned off, the charge on C22 and any capacitance present on the output is diverted around the regulator through D10 instead of into the regulator's output which could damage it. Seems like either way of connecting the diode does that. I guess if the charge was large enough it could cause damage when passed back to the regulator's input, and connecting D10 ahead of D9 would avoid that.

[DanielS]

I think you want this.  Diode D10 where you have it doesn't make sense.

Why does it not make sense? Directly across the regulator is the usual way of protecting linear regulators against input shorts or reverse bias.

If you connect the protection diode from the output to the input jack terminal, the input jack becomes live if there is voltage present on the output, which would be potentially unsafe. Connecting the diode after the input polarity protection diode prevents that.

[rdl]

 Anything is possible, but the odds of a voltage appearing on the output that is higher than the input seems pretty small.

[sentry7]

Wow. I have learned a lot from your responses guys. I have been doing regulator circuits all wrong for a long time.

May as well have all my questions answered now.

1. As far as I can tell, D9 does not protect the regulator (or the load) if the 9V battery is put in backwards. Even though there won't be any current through D9, there will still be a negative voltage at the input of the regulator.

2. What happens with the regulator when the battery starts to lose voltage and it falls below 5V?

[Siwastaja]

Anything is possible, but the odds of a voltage appearing on the output that is higher than the input seems pretty small.

This is completely normal, happens in many situations all the time, and (after burning a few, and they burn as shorts!) my design principle with 78xx regulators (and LM317) nowadays is that I always put the diode there (directly from output to input, as recommended in datasheet - why couldn't they just integrate this protection!), unless I conduct proper analysis that it is not needed in that particular case.

Most typical cases are:
- Large enough output capacitance. Every power-off stresses the device over the limits, and it may fail randomly, sometimes after months of use and several power-offs.
- Input bulk capacitance smaller than output capacitance.
- Input short circuit, which discharges the input bulk capacitance quickly, or even just other unregulated loads on input side!

TLDR; Always connect a diode directly from 78xx output pin to its input pin, unless you know you don't need to.

[Gyro]

I think you want this.  Diode D10 where you have it doesn't make sense.

The idea is that when the regulator is turned off, the charge on C22 and any capacitance present on the output is diverted around the regulator through D10 instead of into the regulator's output which could damage it. Seems like either way of connecting the diode does that. I guess if the charge was large enough it could cause damage when passed back to the regulator's input, and connecting D10 ahead of D9 would avoid that.

Surely if you connect the return path of D10 ahead of D9 then you're guaranteeing that if anyone connects a wrong polarity input to the jack it will immediately pull the output of the regulator negative - exactly what the polarity protection diode D9 is there to prevent! 

It also increases the likelihood of the output getting discharged via D10... any time the input jack gets accidentally shorted. If it is connected after D9 then it in extremely unlikely that it would ever happen in the first place (it would need a specific short on the board at the input to the regulator).

 

[mariush]

2. What happens with the regulator when the battery starts to lose voltage and it falls below 5V?

The linear regulator needs an input voltage up to 2v above the output voltage to output a stable 5v. The exact amount depends on the output current and temperature of the chip itself. The 7805 datasheets don't usually have detailed information about this as it's just a plain boring old common component with lousy specifications for today's chips, but other linear regulator chips will have graphs with the dropout voltage depending on the current output and possibly even temperature.
With the currents a 9v battery can produce, let's say up to 100mA, most 7805 regulators will have an internal dropout voltage of about 1.5v, maybe a bit less.  That means when the battery's voltage drops below about 6.5v (but don't forget the voltage drop on the reverse voltage protection diode as well, which would be about 0.5-0.8v for a regular 1n400x diode), the 7805 will no longer be able to output stable 5v. It will output less than 5v and will continue to basically output  input voltage - dropout voltage up to a threshold, let's say maybe 2-3v at the output.

For a slightly better linear regulator with decent amount of output current (800mA) and still very cheap, see LM1117 or a 1117 made by other manufacturer, here's an example search at Digikey.

It's a linear regulator similar to 7805 but has a dropout voltage of less than 1.2v at low currents and temperatures as you can see on the datasheet (see page 7. figure : http://www.ti.com/lit/ds/symlink/lm1117-n.pdf .. It also doesn't need protection diode unless you go crazy with capacitance on output (like 1000uF or more) and the only thing you have to be careful about is that the pins on the regulator may be in another order compared to 7805 and that it needs 10uF or more capacitance on the output with some esr (no ceramic capacitor on output).

If you're sure your product won't use a lot of current, you can go even further and pick a linear regulator optimized for low current and low dropout voltage, like LP2950 for example which has a maximum output current of 100mA: http://www.digikey.com/product-detail/en/LP2950-50LPRE3/296-20933-1-ND/1216905

As you can see on the first page in the datasheet ( http://www.ti.com/lit/ds/symlink/lp2951.pdf ) at about 60-80C (the regulator will heat up inevitably from regulating voltage because it's a small chip) and up to 100mA, the regulator will have a dropout voltage of up to 0.45v , which means excluding the reverse voltage protection diode, your device will get 5v even when your battery goes as low as 5.5v