Q: Transistor as a switch with pull down ?

For what purpose ? The transistor's leakage current won't be enough to turn the relay on.

Graham

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You might be right. I just wanna make sure that the transistor doesn't turn on because it is floating as the microcontroller is defaulted to an input pin upon startup.

So to be short: Even if the micro controller does nothing else than an infinit loop after startup and the IO pis is defined as an input there are no chance of the base of the transistor can go beyond 0,7v and thereby setting the collector emitter in an on state, even so that no other resistor is on the base except the one from base to the IO pin (defined as an input) ?

I hope you don't mind my exact question. I am just trying to make sure i understod everything.

Regards Benjamin

the

If the uC defaults to an input on start up is there an internal pull-up ?

Maybe it's wise to have a resistor from base to ground anyway. Say 4k7.

Graham

the

I think he's basically saying that internal leaking in the BJT plus any pin leakage (input pins still leak, though not all that much) shouldn't be enough to cause the transistor to pass 100's of mA. Even for a moment. Some folks may also worry about you rubbing your feet on a carpeted floor on a cold dry day and static-zapping the higher impedance base node of the BJT and doing something untoward. But it's not exactly floating as there are protection diodes and other impedances present on the node. So although I'm not an expert on the details, my experience would say it's just fine without the pull-down. But it won't hurt anything, either. And if this were pinned out to a connector, I'd add the pull-down.

Jon

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Exactly my point, but looking at schematics around the net i usually only see a resistor between base and IO pin. And i was surprised so i just wanted to know if this is normal/good pratice.

Regards Benjamin

to

turn the

Hi

The uC (PIC12F508) has internal pullup but you must enable it first as i understood from the spec.

Ok so it looks like it is safe to just have the base to IO resistor.

Say I remove the IC from the socket (to program it) and accidentally powers on the circuit, nothing is now connected to the base except for a resistor with one pin floating. Will it be able to turn on in any way (without the pull down resistor) ?

Again, I am just trying to learn something from you guys :-)

Regards Benjamin

turn the

--
Good practice dictates that you know what your circuit is doing all
the time, and even though it\'s tempting to think that a floating
base won\'t affect anything downstream, it might.

Consequently, it\'s a good idea to pull the base down (or up,
depending on what your circuit\'s supposed to do) in order to take
the uncertainty out of it.

Be careful with attaching too much importance to what the donkey
states. 

Note that he\'s back-pedaling in order to try to fix his earlier post
and hasn\'t stated whether the pull-down should be on the µC side or
on the transistor side of the series-connected base resistor.

turn the

Even without that enabled, there is leakage. The spec for the part will disclose that value, too. The programable pull-up (weak mosfet) just adds to that.

I can't tell you professional practice. My knowledge is as a hobbyist only. I have used pull-downs before, myself, just to make myself more comfortable. And if I were sending this line to a connector, I'd definitely use a pull-down. I want to KNOW where the input signal is going to be at, in that case. The extra resistor gives me comfort, if nothing else.

But I'm not a professional, so others will have to weigh in on that score.

If this is to an important relay, let's say, and you are likely to have folks replacing microcontrollers while powered up, I'd add the resistor. Yes, having a nice, low impedance to ground to hold the base is a 'nice to have.' In hobby stuff, I do it sometimes when it is convenient (I'm not wire-wrapping or point-to-point soldering it all up, in other words.)

Well, if you are just asking if tethering the base to ground via a nice resistor will cover more unintended ground, then yes it will.

Jon

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Hi JF

:-)

I think I would add the pulldown between base and ground and a current limiter between base and IO. It seems to be the safest way for me.. as a hobbyist as well :-) A 10K to ground and a 1K to IO (or perhaps a 4.5K)

Would you agree on that ?

Regards Benjamin

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I will add the pull down then from base to ground.

Thank you all for your help.

Regards Benjamin

turn the

Couldn't resist sniping eh ?

Exactly what part of " base to ground " is unclear ?

Graham

That's the safer/surer side. Best of luck (though it seems you are going to be fine without luck in this case.)

Since you are searching for various approaches here, you might also consider inserting a zener in series with your protection diode across the relay coil. The rate at which current declines is proportional to the voltage across (from dI/dt = V/L) it and adding a zener lets a somewhat larger voltage to develop which allows a faster dI/dt. That has some benefits, at times. Use something reasonable (most any common value.) If you don't have the diode or diode+zener, I believe what happens is that your BJT has its own breakdown voltage and the coil just dumps through the BJT after achieving its own breakdown voltage. That may or may not be okay, which is part of why diodes, zener+diodes, or RC snubbers are added.

I don't recall reading what the coil current needs to be, but there is another consideration here in sizing your base drive resistor (not the pull-down.) If you plan to use the BJT as a switch, where the voltage between the emitter and collector is minimized and below say 0.3V or so, you should imagine that the BJT is well into saturation and the effective beta is then low. I usually figure about beta=20 for modern general purpose transistors and beta=10 for the old ones I may find in an old box of parts. That means you need to figure on about 1/20th or

1/10th of the collector current into the base. Your PIC can handle a fair amount on its own (worst case is 25mA, I think, but that is an absolute maximum if memory serves and not something to rely upon for normal use.) And as yuo pull more current from the PIC, the voltage drops at the PIC's I/O pin. I figure about 80-90 ohms driving low and about 120 ohms driving high. Since you would be driving high, keep in mind that there is some voltage drop going on inside the PIC. Use that to help calculate your resistor.

For example, let's say that you wanted the collector to handle about

100mA for your relay and you are using a 2N3904. I'd say beta=20 and figure that the base needs 100mA/20 or 5mA. If the PIC is running on 5V (I think you said as much), then the I/O pin will be at 5 - 5mA*120 or about 4.4V (rough guess.) I usually figure a base-to-emitter voltage of about 0.7V when passing 100uA into the base. 5mA is about 50 times that much. With about 60mV per decade of current change, this suggests about an additional 100mV or so (close to two decades worth.) So make it 0.8V. With 5mA, small general purpose transistors like this are getting a bit non-log-linear though after about 1mA of base current. So add another .1V just to be safe. Call it 0.9V. Your resistor is then (4.4V - 0.9V) / 5mA or 700 ohms. A 10k would obviously be too weak in a case like that. On the other hand, if you only need about 10mA for the collector current, that would drop the base needs to 10mA/20 or about 500uA. We are talking now about a base voltage of say 0.7V + 40mV or .74V. Your PIC will only drop about 60mV, so now we compute (4.94 - 0.74) / 500uA or 8400 ohms for the base resistor. That's more like the 10k you are talking about, earlier.

If you can accept a volt or more drop across the collector/emitter of the BJT, then your gain is more like beta=100 to 200. And you don't need as much base current to drive the collector current. But then you lose some of the voltage applied to the coil, too. And you may need to consider power dissipated in the transistor.

Jon

--
Commenting on your back-pedaling is hardly sniping.