Hi

When people use a resistor from an output pin of a microcontroller attached to the base of a transistor, which is acting as a switch on a circuit; what is the resistor for exactly?

http://www.thebox.myzen.co.uk/Workshop/Motors_1_files/shapeimage_2.png

What might happen if it was not present?

Thanks

 

The resistor limits the current into the base and prevents overvoltage on the base. Without the resistor driving the base would probably blow the device due to overvoltage unless you were using a high impedance drive to the base.

 

Thanks

So the transistor would get damaged? That would lead to the microcontroller then getting damaged?

In that order?

 

I would think it depended on the transistors power rating.... Generally I'd think the microcontroller/IC would go first.... I've killed a 555 similarly (and they are pretty robust), that had a beefy power transistor connected to it...

 

Thanks

So the transistor would get damaged? That would lead to the microcontroller then getting damaged?

In that order?

Chips are pretty robust these days and often limit their output current so the processor or perhaps even both might survive but you don't want to design that way with everything heating up and running close to death.

 

I have been looking at some postings and I am trying to gain a better understanding of the issues.

Does anyone have pointers to tutorials that might help me get a better understanding?




I saw some comments from a thread on another forum here:
http://electronics.stackexchange.co...the-diode-and-capacitor-in-this-motor-circuit

but when I read through the posts its quite hard to understand.

The resistor is to limit current the digital output must source and the transistor base must handle. The transistor B-E looks like a diode to the external circuit. The voltage will therefore be limited to 750 mV or so. Holding a digital output at 750 mV when it is trying to drive to 5 V or 3.3 V is out of spec. It could damage the digital output. Or, if the digital output can source a lot of current, then it could damage the transistor.


1 kΩ is again a questionable value. Even with a 5 V digital output, that will put only 4.3 mA or so thru the base. You don't show specs for the transistor, so let's figure it has a minimum guaranteed gain of 50. That means you can only count on the transistor supporting 4.3 mA x 50 = 215 mA of motor current. That sounds low, especially for startup, unless this is a very small motor. I would look at what the digital output can safely source and adjust R1 to draw most of that.

 

Digital output "gives" 5V at the output. Transistor base-emitter needs max 0.8V or so.
If base voltage is force higher then this 0.8V the huge amount of current will start to flow. And this current will start generates the heat inside transistor. This heat in transistor will increase the current even further. And this increase in current will again increase the heat in transistor and this will lead to a destructive result (transistor will burn). Also this heat is also grow inside microcontroller, so the microcontroller can also be damage.
So the prevent this disaster to happen we add RB resistor. This resistor will limited the base current and drop the voltage ( From 5V to 0.8V) also according to the Ohm's law.

but when I read through the posts its quite hard to understand.

Which part you don't understand?

 

use the base resistor and put in another transistor following the first one to handle more current. this will help isolate the chip from the load better. spikes might blow one transistor, but probably will not get through two.

 

Makes more sense now thanks

Its all the terms I am unfamiliar with, but I read a bit before so things are slowly making more sense. I mean a minute ago I didn't really understand sink and source... thanks google got it now.

All the simple things when you don't know much, just add up and its almost like reading Chinese, for a non-Chinese speaker.

Where is that 750mV from? Is that a standard base emitter saturation voltage?

 

Is that a standard base emitter saturation voltage?

Yes. Driving more current through the b-e junction will increase the voltage only slightly.

 

Where is that 750mV from? Is that a standard base emitter saturation voltage?

From data sheet. But most BJT will have Vbe around 0.7...0.9V.
But don't forget that in saturation base current is more important than Vbe.
http://forum.allaboutcircuits.com/showthread.php?p=666860#post666860
http://forum.allaboutcircuits.com/showpost.php?p=577296&postcount=7

 

If the I/O pin has max current 40mA

Would I use a resistor to the base of the NPN with a value between 15 - 22 Ohms?

 

a 22 ohm resistor in series with 5 volts applied would pass 190 ma or so to drop 4.3 volts. a little high? use a 220 to 1000 ohm resistor.

 

If the I/O pin has max current 40mA

Would I use a resistor to the base of the NPN with a value between 15 - 22 Ohms?

No, why such low values?

 

I used a blooming online calculator and its still wrong!

 

I used a blooming online calculator and its still wrong!

Can you post the link? And what value you have put into this calculator?

 

Ok I was doing it wrong. Trying to 'will' the voltage down as if it was an unknown quantity.... when I already had a voltage value.... tut

 

I have a voltage: 5V, but I am trying to get the voltage to drop to fit this 0.75V

I was using the calculator in an insane manner. The problem being... it made sense to me.

So now I put in the correct values...
5v
20mA
results = 250 Ohms

but I want the voltage dropping to a value, so what do I use.... voltage divider calculator?

These questions might be criminal... please bare with me

 

I might have it.

I do what I was doing, and the subtract results from 5V?

so work out a resistance with 20mA that will give me 4.25V

leaving 0.75V?

 

but I want the voltage dropping to a value, so what do I use.... voltage divider calculator?

Why you want specific voltage drop? As I said earlier the Vbe voltage in not so much important here. The base and collector current is important here.
For example if your load connected between collector and VCC consume 60 mA of current. The base current need to be higher then
Ib > 60mA/20 = 3mA. So

Rb < (5V - 0.7V)/3mA = 1.4k so you use 1K resistor.