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NPN circuit not behaving...

I'd thought I knew how to use an NPN but my circuit does not work...

A momentary switch S1 is connected to a 3.7V battery. Pressing SW pulls U3 Enable high turning it on. A microcontroller is powered which latches U3 Enable high through D2.

When VUSB is applied I want to force U3 Enable low to turn the Microcontroller off (maximize charging current to battery). But applying 5v to VUSB does not drive U3 Enable low.
I put a LED in series between R3 and Q1 and that does turn off when VSUB is applied but U3 stays on....?

Any help greatly appreciated....

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KrisBlueNZ

Sadly passed away in 2015
Your circuit should work OK.

What type of transistor are you using for Q1?
What is the voltage on U3 pin 3 in each state?
What is the minimum threshold voltage specification for U3 pin 3?
 
Hey there, thanks for asking!

the USB is connected to the common ground of the circuit.

I'm using a 2n3904 transistor.
Output Enable (active high), U3 Pin 3 voltage is zero before button press and 3.3V in the latched state.
The max 'off' voltage is 0.25V, the min 'on' voltage is 1.6V

Thanks for your help,

p.s. the circuit was my best idea at how to override the latch on the enable when USB (5V) is applied. If there's a better way I'd much appreciated suggestions.
 

KrisBlueNZ

Sadly passed away in 2015
You say the CE pin voltage is 0V before the button is pressed, and 3.3V when the MCU has latched it ON. So what is the CE pin voltage when VUSB is present?

That circuit is very straightforward and can be diagnosed step-by-step pretty easily. You understand how it works, because you designed it. So you should be able to figure out the problem with just a multimeter.

Regarding how to unlatch the regulator enable, the way you've done it is fine, but you could do it by having the MCU drop its control output when it detects the presence of USB 5V (assuming it can detect this somehow). That would also allow you to impement a short delay while USB 5V is present, if you want to, so that the MCU can be confident that the USB connector is properly plugged in before it turns off the secondary supply.
 
Hmmm that's another idea.
My reason for shutting the micro down was for two reasons.

1. Bi-LED Status. The battery charger has a tri-state output which is a 1 pin output to a bi-led. The micro also uses 1 pin to drive the same LED so i needed to make sure the charger and micro did not drive the pin at the same time (i.e. one has to be high Z).

2. Hard reset. Should the micro crash and leave power latched the user would need wait for the battery to discharge to recover. My design is supposed to force the micro off (by removing VCC) so that a hard reset occurs on USB insertion.

Will take some measurements of CE pin when USB is present. My guess is that it's not being pulled low enough to turn the reg off..
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
The BAV70 is two switching diodes in a single package. They share a common emitter pin only. I mention this because it's not properly drawn on your print.

Chris
 
The BAV70 is two switching diodes in a single package. They share a common emitter pin only. I mention this because it's not properly drawn on your print.

Chris

Indeed it is, and works nicely here as it prevents flow from the latch through to the button and vice versa.

I could have drawn the diodes separately connected by a common cathode but wanted to represent that it was a single 3 pin package on my schematic. I guess I could have drawn two diodes with a box around them? How would you have drawn it?
 

KrisBlueNZ

Sadly passed away in 2015
Yes, I would have drawn it as a box containing two diodes, internally commoned at the cathode. That's how it's shown in the data sheet as well.

The symbol you are currently using seems clear enough in meaning - at least, when shown in the context of the whole diagram. In fact I didn't consciously realise that it was wrong until Chris pointed it out - I immediately understood what it represented when I saw it. But it is not really explicit, and I believe that diagrams should emphasise clarity and avoid any possible confusion to the maximum possible extent. So I would never draw it the way you did.

Edit: You don't need a double diode; a single diode from the microcontroller output is enough. You don't need a diode from the pushbutton as long as you have a diode from the microcontroller output.
 
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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
I guess I could have drawn two diodes with a box around them? How would you have drawn it?

Yes, that would be a way.

The other way would be to just draw it a 2 diodes and label them as (say) D1a and D1b which tells us they're part of the same device.

The way you suggest would be the clearest.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
I'm sorry but our industry doesn't accept schematic freestyle. We're not going to let what happened to the English language, do to texting habits, migrate to our field. Besides that point are you aware that a Programmable Unijunction Transistor is drawn very similar to what you drew?

Yes, two common cathode diodes in a box; just like the manufacturer drew it.

Chris
 

KrisBlueNZ

Sadly passed away in 2015
I'm sorry but our industry doesn't accept schematic freestyle.
On that subject, what do you guys think about my Darlington transistor symbol? It's a standard transistor with two arrows on the emitter - see Q1 and Q2 in the diagram on https://www.electronicspoint.com/help-amp-t262454.html#post1565548.

I haven't seen it before; I just use it to save space, because I think the standard symbol is disproportionately big. Do you think its meaning is obvious? Do you consider it wrong?
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
Kris, have you been texting? :p
After beating up the new guy I'd be accused of nepotism if I said that was OK. ;)
Send it off to the IEEE to see if they'll accept it. Hey, if they say it's OK who am I to argue?

Chris

PS. Yeh, Darlingtons do take up real estate.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
On that subject, what do you guys think about my Darlington transistor symbol?

I'd probably wonder if it was some strange sort of transistor I hadn't seen before.

However, I'm not too much against drawing it as a normal transistor and then adding a note somewhere that it's a darlington.

Another approach I've seen used (with some special MOSFET-like devices) is to place them in a square rather than a circle.

Drawing a darlington is probably just as complex as drawing a mosfet with a body diode. And we often don't draw the body diode (to simplify the drawing?)
 

KrisBlueNZ

Sadly passed away in 2015
Thanks Chris and Steve for the feedback.

I'll create a new symbol that shows both transistors but doesn't take up a huge amount of space.
 
Edit: You don't need a double diode; a single diode from the microcontroller output is enough. You don't need a diode from the pushbutton as long as you have a diode from the microcontroller output.

Hey that's a good thought! I used a double diode as I imagined it would protect each source from the other, but I guess the latch will never be higher than the button source (battery).

I did solve my BJT problem (not grounding CE). It turns out the 2N3904 left 39mV on the CE pin when the base was at 5V, this was enough to keep the Vreg on. I played around in Multisim with other diodes and ended up with a DTC114EET1G which did properly ground CE and has pre-bias resistors build in saving on soldering and baord space.

Thanks to all for comments, much appreciated and learned along the way.
 

KrisBlueNZ

Sadly passed away in 2015
I did solve my BJT problem (not grounding CE). It turns out the 2N3904 left 39mV on the CE pin when the base was at 5V, this was enough to keep the Vreg on.
But you said earlier:
Output Enable (active high), U3 Pin 3 voltage is zero before button press and 3.3V in the latched state. The max 'off' voltage is 0.25V, the min 'on' voltage is 1.6V.
(Highlighting mine.)

39 mV is pretty low - low enough to be considered "low" by every discrete semiconductor and IC that I've ever heard of. And it's a lot lower than the minimum low threshold of the CE pin on that IC. So you haven't explained what the problem was.
 
Shoot... your right, I had read that as 25mV. So it's still a mystery why the NPN wont ground that pin against a 10K pull up...
 
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