Pololu S7V8F5 Shutdown pin

[codysmith105]

Hello EEVBlog,

First post here. I'm Cody, I love the videos and what is being done here, I just didn't want to clutter up your forum until I had a real question to ask. My background is in programming and 3D animation, but I've always wanted to learn electronics so when I found the channel I fell in love with it pretty much instantly.

OK my issue, I have two microcontrollers, the VEX Cortex and a Raspberry Pi which I am hoping to use together, so I developed this enclosure to mate the two:

So for the electronics bit, there is a little switch on the Cortex that turns it on (it's powered by a 7.2v battery), and I essentially want the Raspberry Pi to turn on when that switch is pulled as well. Now I cannot modify the Cortex, this is for competitive robotics and we aren't allowed to mess with the Cortex. But there are a host of 5v pins that go hot when this thing turns on, a bunch are part of PWM connectors but there are UART ports that have both ground and +5v pins as well, so for simplicity reasons I want to use them.

Now the Pi will be powered by another battery (a 9.6v battery), the reason for this is current related. The Cortex has a 1A breaker in it, and we need all of that 1A to power motors so the Pi with a 200-400 mA draw needs to be powered separately. Our competition lets us use either another 7.2v battery or a 9.6v battery to power sensors and co-processors so this worked out well.

The original idea was to use a relay, but I've had issues sourcing one that fits and plays nice so I was looking for alternatives when I found this. It's a power regulator from Pololu with a shutdown pin. Now we already needed a regulator for the Pi so my thought was that this shutdown pin on this regular might just save the day, so I started reading the product page and I found this:

The SHDN pin can be driven low (under 0.4 V) to power down the regulator and put it in a low-power state. The quiescent current in this sleep mode is dominated by the current in the 100k pull-up resistor from SHDN to VIN. With SHDN held low, this resistor will draw 10 µA per volt on VIN (for example, the sleep current with a 5 V input will be 50 µA). The SHDN pin can be driven high (above 1.2 V) to enable the board, or it can be connected to VIN or left disconnected if you want to leave the board permanently enabled.

So I found myself confused on that part at the end there. Basically in a nutshell, I don't understand the difference between not connected at all, and being driven low. I talked to my dad about it (he has some old college background with electronics) and he tried to explain it, but my understanding is really fuzzy.

So EEVBlog, what is the difference between not connected and being driven low? Also, since the switch on the Cortex probably kills the connection to it's battery, will a +5v pin from the Cortex, connected to this shutdown pin on the regulator be considered "low" or not connected?

Thank you so much in advance! -Cody

[Jon86]

Not connected means the pin is what's called 'floating'. It's not connected to any reference voltage which means if it's high impedance then the voltage tends to bounce around due to noise. On that chip, it's probably connected to a positive voltage internally, and adding a resistor takes the voltage down below the threshold to 0v or ground. The whole 'floating' concept depends on what kind of input/output you're working with, but in general if it's floating you're not really giving it a specific voltage, it's just all over the place unless something is pulling it up or down.

[Jon86]

Gorgeous enclosures by the way. 

[mmilejski]

Hi,

When the SHDN pin is not connected it is driven high through a pull up resistor to VIN and it is at VIN. Driving it low means connecting SHDN directly to GND.
When you connect your Cortex 5V to the SHDN you will have the same situation, it will be high with Cortex turned on and basically unconnected (and thus pulled high by mentioned pull up resistor) with Cortex turned off. You just need to find a way to connect the SHDN pin directly to GND when you power your Cortex off and it will work.

[dr_p]

here, read this:  http://en.wikipedia.org/wiki/Pull-up_resistor

[codysmith105]

Gorgeous enclosures by the way. 

Thanks very much! The internals are not finished yet but I did 3D print a "test-fit" one, I had Shapeways make it. They have really good printers there. There was an issue, but I'll be addressing it with the second print. It basically amounted to the CAD I was using of the Pi having variances in the USB port between manufactures. My Pi has a slightly different USB hub/port/thingy which made it not fit sadly. The plan is to mount everything needed and have room for prototypal stuff if need be, we are deploying this on 2-4 robots this year at the VEX World Championship.

Depending on how things go we will either open source the enclosure CAD and BOM or we will sell it to promote our teams growth.

When the SHDN pin is not connected it is driven high through a pull up resistor to VIN and it is at VIN. Driving it low means connecting SHDN directly to GND.
When you connect your Cortex 5V to the SHDN you will have the same situation, it will be high with Cortex turned on and basically unconnected (and thus pulled high by mentioned pull up resistor) with Cortex turned off. You just need to find a way to connect the SHDN pin directly to GND when you power your Cortex off and it will work.

Could I maybe just pull off / destroy said resistor? I mean it would sort of "solve" our problem? Could anyone here identify it without a schematic?

[nardev]

Gorgeous enclosures by the way. 

Agree!!

p.s. GDESI BRE SRBINE PUSKA TE NE UBILA

[dr_p]

You just need to find a way to connect the SHDN pin directly to GND when you power your Cortex off and it will work.

[codysmith105]

I don't entirely understand this

I get that it's a transistor used as a switch, but I don't see how it would connect the regulator to ground when the Cortex is off.

In fact as far as I can tell it would do the opposite? When the Cortex is on wouldn't the base become positively charged thus attracting the negatively charged electrons from ground to the emitter and through to the collector? And of course when the Cortex is off, no voltage would be applied to the base so no current should be able to pass through the transistor?

If I'm wrong, I'm very sorry. I'm still learning my electronics ABC's but after a few hours of scratching my head on this I'm convinced that this does the opposite of what I need. Maybe you read it as on instead of off, or maybe I am fundamentally misunderstanding something here.

Some clarification would be great, I feel it's important to know if I'm getting this wrong and why. -Cody

[Jon86]

Trying to think of current/voltage and all that in terms of indivudual electrons and charge is just going to confuse you...

[codysmith105]

I'm sure you are right, and actually it is.

Am I right about the diagram? It's driving me crazy.

[dr_p]

Cortex ON   ->   NPN transistor "ON"   ->   SHDN pin pulled low   ->     fuck, I misread what was needed 


Yes, it does just the oposite.


edit:





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