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I'm adding the power circuit to my design. There's going to be a lot of chips (about 19) so I haven't calculated the total draw yet.

I'd like to implement some basic reverse polarity protection along with a fuse. Some examples I've read about and seen recommend using a diode and fuse.

But I definitely could use some advice on what values to use for them. I figure this thread might help some other people too.

Anyway, my schematic is attached below. Any and all advice welcomed!

Thanks!

The old fail-safe for this is to use a standard full-wave diode bridge. Granted, you'll always the voltage drop of two forward-biased diodes (~1.4v) but you'll never have to worry about DC polarity or frying a circuit.

From the design, I'd assumed that "External Power"meant that this header was a supply for additional peripherals which would be turned on and off with the main board. Just goes to show that it's important to think carefully about what you're communicating with the labels on your schematics. (And even moreso, boards, where a poor choice of label on the silk screen might lead someone to plug something in the wrong way!)
Oh, that's a brilliant idea! ("Why didn't I think of that?") The extra diode drop doesn't strike me as much of an issue: if you're using a 5 V regulator, a 6 V wall wart would borderline (not that they're all that common anyway), and once you've taken the step to 9 V or higher there's plenty of room for a couple of diode drops. I guess it might even accept an AC PSU, if you had enough capacitance after the bridge?

The other thing I wonder about is protecting against overvoltage. I've got a friend with a video upscaler that came with a 5 V wall wart; the board regulates its input down to the 3.3 V it uses with a small surface-mount regulator. She's blown up the 3.3 V regulator at least twice due to plugging a very similar-looking 12 V wall wart into the device. (The regulator is soldered to a big pad on the board, but even with the extra heat sinking that provides, I guess it's just not enough to handle regulating down from 12 V.) Is there a safe and reasonably cheap and compact way of dealing with that problem, preferably one that doesn't just short the PSU itself?

A single Schottky diode will suffice for reverse polarity protection, developing a forward drop about 0.5 volts, compared to the 0.8 volts typical of a conventional diode, such as a 1N4001. Be generous with the diode's current rating to handle the inrush when the capacitors in the circuit charge at power-on. An SB230TA is a good choice for this application and is in the same package as the 1N4001.

The 7805 regulator is self-protecting against overloads and outright shorts. However, I'd use the 78S05 to gain a higher current rating. The fuse's primary worth will be in protecting the power source should the capacitor at the regulator's input go south. No fuse is fast enough to protect the silicon in the circuit from damage.

The 78xx series of regulators can tolerate a wide input voltage range if adequately heatsinked. For example, the 78S05 I mentioned in my previous post can tolerate up to 35 volts at the input. Hence almost any commonly-available wall wart that produces DC output is suitable for this application.

1N4148 is too small, it only can take 100mA of continous current.
1N4001..1N4007 can take 1A of continous current, but to be on the safe side it's better to use a diode which is able to handle twice the current to be expected.
The price of Schottky diodes has dropped over the years, so I think that SB230 (or SB560) would be the best choice.

Would suggest to put a 330nF ceramic capacitor (or 3* 100nF in parallel) between the input of the 7805 and GND to prevent the 7805 from oscillating.
The 7805 datasheet gives some hints about what capacitors to use at the input and output of the 7805.

When inserting a diode into a power supply line like that, it becomes a RF demodulator,
means if your circuitry becomes part of a machine park where there is some RF interference to be expected
(and when your cable between wallwart and 7805 isn't too short for working as an antenna),
professional tinkerers put a 220pF..1nF ceramic capacitor in parallel to the diode to get rid of that effect.
(Because that capacitor in parallel to the diode plus the capacitor at the 7805 input would shunt RF signals to GND).


The problem with wallwarts is, that for some wallwarts the center pin of the output connector is VCC,
and that for some other wallwarts the center pin of the output connector is GND,
so to be on the safe side we can't get around using a diode.

If the diode is inserted into the power supply line, it has a voltage drop, and Schottky diodes have a lower voltage drop.
Another approach would be putting the diode in reverse between the output of the fuse and GND,
so when a wallwart gives out a reverse polarity the diode becomes conducting and blows the fuse.


Note, that the 7805 needs more than 7.5V at the input for generating 5V at the output,
means when a 7.5V powered 7805 gives out 1A, it burns more than (7.5V - 5V) * 1A = 2.5W,
so for the sizing of the heatsink it's better to be safe than sorry.


The average hobby shop is supposed to have the 7805 on stock.
Another option would be buying the Microchip MIC2940A-5.0WT low drop voltage regulator,
because it's more rugged than the 7805 (from the datasheet it can survive -20V...+60V at the input),
and because it's able to generate 5V from a 5.35V input voltage at 1A.

Also, there are switchmode DC-DC converters like the TR2024S05 which are more power efficient than linear regulators.

If quality of wallwart (output) power jack connectors becomes a topic, switchcraft connectors might be useful.
Sometimes we had to replace unreliable wallwart power jack connectors with switchcraft locking power jacks at work...

LT1084 certainly is very picky about the capacitors at low output currents.
MIC2940A had caused me less trouble.

I agree with Garth that the easiest for the beginner is something like a wallwart with 9V output plus a 7805 with adequate heatsinking.

Thanks everyone for the input!

I have revised a few things such as dropping the fuse, beefing up the diode and putting a 0.33uF cap near the 7805.

This design is already 10x better than my last so I think I will stick with the 7805 with a large heatsink. Mainly because I already have a lot of them in stock.

Thanks!

It’s still worthwhile having a fuse. Very, very occasionally, a 7805 or similar regulator can fail in such a way that it becomes a short circuit. A fuse may not blow in time to protect your circuit, but it should blow before the mains PSU / wall wart sets the house on fire (assuming you have used an appropriate rating).

A bonus of having a fuse, is that can also be used as a test point to measure the current flow of the power circuit.

Mark

Hi guys,

Attached is my power circuit (pdf) designed from the information gathered within this thread.
I'd like to have a mobile 65C02 computer that I can take around with me. Size doesn't matter here (within reason).
What I'm trying to do is to have a circuit with two power sources, a Raspberry PI power supply (5.1V, 2.5A), and a 9V battery.
E.g.
On the go: flick the switch to the "Battery" mode
At home: flick the switch to the "Mains" mode

1. How is my circuit? Are there any critical design flaws? Excuse the crudeness, I am quite the newbie.
2. I have duplicates like the polyfuse and the pairs of capacitors (100uF, 0.1uF) one for each power source where it enters the board. I think there's no way around this, which is fine.
3. I'm not sure if I'm using the most appropriate polyfuse (Mfr. No: 06R075BPR) for my cct:
- Hold Current: 750 mA
- Maximum Voltage: 6 V
- Trip Current: 1.3 A
- Current Rating - Max: 40 A

I'd like to use a 78S05 as recommended in one of the posts above, but I already have a 7805 and a heatsink. Plus running in Battery mode wouldn't use much current (I'd imagine).

I used 1117s a lot (hundreds of boards) without ever having problems - even when sourced from China and using questionable capacitors. If you go with a couple of tantalum caps, I doubt you will have any problems.

Unless I'm missing something, you have the battery and the Pi +5V supply coming in and going thru a pair of caps in parallel. Caps won't pass DC, so neither position of the switch will result in having +5V to power your circuit.

The National Semiconductor datasheet for LM1117 has this reference circuit with a battery: