SOLVED: Protection against external power supply

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mr.bojangles
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Posted by mr.bojangles: Wed. May 27, 2015 - 12:03 PM
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Hello everyone!

 

I am designing a board with a uC to read data from a sensor and send to pc via USB.

 

I am using the 5V from USB to power the board and the sensor.

 

The board has 3 cables that are connected to the sensor: Data, 5V and GND.

 

My question is: I would like to protect the board against stupid users, in case someone decides to use an external power supply.

I would like to know if there is a simple solution for that, like a diode. The problem with this is the voltage drop, which is too big for the sensor, since the USB voltage is not always perfect 5V.

Is there something I am missing here? Or should I aim for something a little bit more complex, like amp-op?.

 

Is there anyone out there that has done something similar?

 

Any tip will be appreciated.

Thank you very much! 

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Learning and Information, General Electronics
Last Edited: Mon. Jul 20, 2015 - 12:05 PM
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Kartman
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Posted by Kartman: Wed. May 27, 2015 - 12:17 PM
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Amp-op? I gather that's dyslexic for op-amp.

You could use a transorb diode like SA05. It will attempt to protect from reverse polarity and overvoltage. You might want to add a polyfuse in series to protect the transzorb and your circuit.

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ki0bk
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Posted by ki0bk: Wed. May 27, 2015 - 12:41 PM
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You could just remove the external DC connector! 

 

Jim

 

 

FF = PI > S.E.T

 

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DocJC
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Posted by DocJC: Wed. May 27, 2015 - 06:55 PM
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How many PCB's, 1,  100,  100,000?

 

How expensive is the board your are protecting?

 

What are you protecting against?  12V, Mains?, AC?, Lightning?

 

With an in-line fuse and a "crowbar" circuit you can certainly pop the fuse.

That makes the User, hopefully, have to tear open the box to replace an internal fuse, and perhaps thing twice about what they were doing.

 

A reverse biased diode on the circuit side of the fuse will again pop the Fuse, in needed.

 

The issue is that popping a Fuse takes fractions of a second, so you need a true protected power supply if you really want to protect the downstream circuitry.

 

Using a true USB connector ought to go a long way to keeping people from zapping the circuit.

One has to ask, who is the target audience?

 

Most people that could circumvent a USB connector for connection with a power supply ought to know at least enough to be very careful and not plug it into AC or Mains.

 

JC

Last Edited: Wed. May 27, 2015 - 06:56 PM
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gchapman
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Posted by gchapman: Thu. May 28, 2015 - 04:41 AM
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mr.bojangles wrote:
I would like to know if there is a simple solution for that, like a diode. The problem with this is the voltage drop, which is too big for the sensor, since the USB voltage is not always perfect 5V.
USB also has a suspend current limit of 500 microamp.

An encapsulated zener diode and PTC is a PolyZen but it's not rated for automotive usage and would need an added TVS for lightning.

TE Connectivity

PolyZen Devices in Circuit Protection

http://www.te.com/usa-en/products/circuit-protection/overvoltage-overcurrent-devices/pptc-zener-diode-devices.html

"Dare to be naïve." - Buckminster Fuller

Last Edited: Thu. May 28, 2015 - 05:11 AM
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mr.bojangles
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Posted by mr.bojangles: Thu. May 28, 2015 - 08:27 AM
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Hello Everyone! Thanks for the fast reply.

 

So, answering some questions…

 

The DC connector is necessary to power the sensor. The board will be later in a case and I intent to use 3 banana connectors (Data, 5V and GND), where the user will connect the sensor. So the 5V connector is an output and should not be connected to an external power supply.

 

The board is not expensive. There is only a uC and some resistor/capacitors/diodes… And that’s why I would like a simple/cheap solution, to not increase the costs…

 

The audience is also other engineers. So if I tell them not to use an external power supply, they should not do it. But I think there are people capable of things you don’t imagine, so if possible and simple, I would try to protect the board.

 

I liked the idea with the zener and a fuse… I will try to work on that…

 

I am also thinking that the best solution might be a label on the case written: “Do not connect an external power supply!” haha

 

Anyway, thanks again!

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Gold Markclawson
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Posted by clawson: Thu. May 28, 2015 - 08:43 AM
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So the 5V connector is an output and should not be connected to an external power supply.

I know this is not totally foolproof but why not give it some kind of connector never normally found in power supplies ?

So if I tell them

A label saying something like "Apply 5V here to release the smoke" should keep most engineers entertained enough to not actually try it cheeky

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meslomp
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Posted by meslomp: Thu. May 28, 2015 - 10:53 AM (Reply to #7)
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clawson wrote:

A label saying something like "Apply 5V here to release the smoke" should keep most engineers entertained enough to not actually try it cheeky

 

Cliff,

For all we know he is building a smoke generator with external smoke density sensor.

Then this would be a very bad idea.

 

How much voltage drop are you allowed to have?

If you are going to use 5V USB the voltage can actually already be 4.4V minimum. If the sensor does not already like that you might already be in trouble (worst case).

You could add a schottky diode then the drop is only 0.3V. 

 

I also wonder how the schematic is made. It might well be that your power supply can already handle the 5V on its output and thus there is no problem at all.

 

 

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sparrow2
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Posted by sparrow2: Thu. May 28, 2015 - 11:44 AM
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what are the needed current?

 

perhaps something like a 0.1 ohm resistor on 5V and measure the voltage drop, if negative take action!  

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mr.bojangles
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Posted by mr.bojangles: Mon. Jun 1, 2015 - 08:54 AM
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Haha Yes, the label mentioned by Cliff is funnier than mine… laugh

 

This circuit will be used to test different kinds of sensors, so I can’t say right now what is the max voltage allowed. And that’s another reason why I would like to have banana connectors, and not a specific one… Most of the sensors that will be tested have a different connector, but they all “speak the same language”.

 

The  schottky diode was also a good idea, with a smaller voltage drop… But as I said, not sure what is allowed frown

 

I think I will stick with the label idea… If the user is stupid enough to do it anyway, so there’s nothing I can do… But I will continue thinking… 

 

Thanks everybody for all the replies!

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meslomp
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Posted by meslomp: Tue. Jun 2, 2015 - 05:49 AM
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if you want to make things more or less fool proof there will be some cost impact.

 

If you add a diode on the incomming USb line that will drop the voltage a bit. then add a SMSP to ensure that your system always has 5V.

put the outgoing supply on a seperate SMPS and there a schottky in series to make sure that when the user puts power on it nothing happens. Then you can compensate for the schottky and in future if you need higher voltage sensors connected all you need to do is adjust the output of the SMPS.

 

should not be a big cost adder and keep your system save.

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mr.bojangles
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Posted by mr.bojangles: Mon. Jul 20, 2015 - 11:55 AM
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Hello everyone!

 

So, I’ve found a solution for my problem. It has some impacts on the cost, but nothing very significant.

 

It’s a voltage regulator with reverse current protection… I tested the circuit and applied an external power supply to the 5V output and no smoke came out of it laugh

 

The name is TPS737 from TI.  And with ultralow dropout voltage (130mV).

 

Thanks again for everybody that contributed to the problem!

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Kartman
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Posted by Kartman: Mon. Jul 20, 2015 - 11:11 PM
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We were trying to solve the problem, not contribute to it!!!

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mr.bojangles
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Posted by mr.bojangles: Tue. Jul 21, 2015 - 05:57 AM
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haha yeah, that's true!

I meant "... contributed to find a solution to the problem!" smiley

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