Hi!
So I am preparing to get some prototype PCBs made from my schematic. I have been testing on bread board and seems to work, I have however a few questions before laying out the board and sending gerber files to production.

I have 4 mosfets (Q100, Q101, Q102, Q103) on the board to switch 12V DC from pins on my uC (electric imp). However If i measure on the output from mosfets (i.e. P100), when the i/o pin is low, I read like 5-6V even thou I would expect to see 0V.
If i put a load on, even small like 10kOhm, i will read 0V as expected.
Given that I will all ways have a load on, it is not really a problem, but I was just wondering why this is happening?

Another thing: look at the block "12V control for electric lock". Is the CARD_PIND sufficiently protected and does D201 and R202 make sense?

And last: if you see anything else that seems off please comment =

 

I read like 5-6V even thou I would expect to see 0V.

Result of diode leakage current flowing through your meter?
IMO R202 would offer more protection if it were between R201 and the zener.
BTW, if this lock is a car door lock the mods may close this thread as being against the Terms of Service.

 

HI Alec thanks for the reply.
Im not sure about the diode leakage thing, I could try with a better meter (this one was cheap) or a scope and see if i get same readings?

Ok so moving the R202 seems like a good idea.

The lock is not a car lock, its a small cabinet lock, this one
http://www.aliexpress.com/item/Free...h-file-cabinet-storage-cabinet/702300265.html

 

The diode leakage would give a higher voltage reading if you used a higher-resistance meter. But even a cheap DMM might have a higher resistance than the typical 1 megohm resistance of a scope.

 

OK thank you, I will look into it, do some testing.
I have another question: How do I invert my feedback on CARD_PIND so that it is default low and high when button is pushed or mosfet is on?
I want to do this so I can wake up the uC on its interrupt pin

Was thinking about doing like this, dont know if it could be done more simple thou:http://www.transtutors.com/university/images/1889_What is Transistor NOT Gate.JPG

 

That will do the job.

 

That will do the job.

Ok updated the schematic with discussed changes. Does the transistor inverter look to be configured right?
I only get 2V on the lock status pin (CARD_PIN1), its enough to be seen as digital high, however why is this I only get 2v?

 

Config looks ok.
What 'low' voltage do you get at Q200 drain?

 

Result of diode leakage current flowing through your meter?

That's what I was thinking, but he data sheet for the diode specifies a max of 25nA at 20V (and more at higher temp). If the TS's meter is 10MΩ and he's seeing 6V, that's 600nA. Curious.

 

Config looks ok.
What 'low' voltage do you get at Q200 drain?

When gate voltage is low i measure:
2V without load between drain and ground on Q200 and 9.8V with load (80mm PC fan).


The meter im using is a TENMA 72-7770 rated at 10M ohm input impedance. Later today I will try with high quality meter instead

 

load between drain and ground

?? The load should be between drain and 12V.
What drain voltage do you get when Q200 is turned on?

 

?? The load should be between drain and 12V.
What drain voltage do you get when Q200 is turned on?

Yeah sorry, poorly written by me, what i meant to say was:

Measuring from drain to ground:
Results with load connected (between 12V and drain, as in schematic):
Q200 ON: 0V
Q200 Off: 9.83V

Results with no load:
Q200 ON: 0V
Q200 Off: 2V

Measuring between drain and 12V rail:
Results with load connected (between 12V and drain, as in schematic):
Q200 ON: 12V
Q200 Off: 2.1V

Results with no load:
Q200 ON: 12V
Q200 Off: 10V

 

So, if with no load there's 2V across a 10meg meter, the diode+circuit-board leakage current is 200nA. Seems reasonable.

 

So, if with no load there's 2V across a 10meg meter, the diode+circuit-board leakage current is 200nA. Seems reasonable.

I just put a 33kOhm resistor in parallel with D200 and measured without load. I still measure like 9V across D200 (which would be parallel with the load if there was any).
Does it still seem reasonable?

 

I still measure like 9V across D200 (which would be parallel with the load if there was any)

You said 2V. Why now 9V? The load shouldn't connect in parallel with D200; it should go in series with D200.

 

Something still makes no sense and is probably a measurement artifact we aren't aware of.

Measuring from drain to ground:
Results with load connected (between 12V and drain, as in schematic):
Q200 ON: 0V
Q200 Off: 9.83V

The on resistance of Q200 is in the mΩ range, so there should be a tiny voltage across it when ON. Maybe not enough to see on the 20V scale of the meter.

Results with no load:
Q200 ON: 0V
Q200 Off: 2V

Perplexing unless there is sufficient diode leakage.

Measuring between drain and 12V rail:

Results with load connected (between 12V and drain, as in schematic):
Q200 ON: 12V
Q200 Off: 2.1V

This is a puzzle, since you are showing a 9.9V drop across the load with no current flowing through the load. You would expect rail voltage on the low side of the load, if no current is flowing because Q200 is OFF.

Results with no load:
Q200 ON: 12V
Q200 Off: 10V

If all the current is coming through the diode, you have 10V driving a 10MΩ impedance, or 1µA. That's out of spec for the diode.

 

Oh wait a minute: There's a direct path from the drain to 3V3 through R201!

 

What is Q210 collector voltage with button pressed/not-pressed ?

 

Hi guys, thanks for the patience

Oh wait a minute: There's a direct path from the drain to 3V3 through R201!

Oh shit yeah, true. Im not really sure how to fix this. I was thinking about putting a diode between Q200 drain and button "1" but then I cannot detect on the CARD_PIN1 if mosfet was opened by uC. Any suggestions?

Also I just realised I do not need the whole transistor logic inverter circuit (R203, R204 and Q201), since I do not need to use the interrupt to wake up anyways

 

Im not really sure how to fix this.

What are the implications of not fixing it?