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# I'm struggling to understand the function of transistors

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Hi,
I hope this is the right sub-forum.
One of the SMD devices that I thought was just a PNP transistor for supplying 5v to LEDs turns out to be a BCR185W - PDF datasheet - which is a digital transistor. New to me.

I wonder if anyone can explain things below to me in a way that my awkward brain can understand please?

I did wonder why the faulty microcontroller on the control board had no resistor between its output pin and the base of the BCR185W and from what I can glean after reading a webpage all about understanding digital transistor datasheets (it helped a bit, but still don't understand a lot of it) that is because there are resistors built into the transistor. Am I right that this has been done so that logic level outputs (0v and 5v) can be directly connected to the transistor, with the inbuilt resistor being tailored so that when the 5V is applied, the base current is just right to turn the transistor on in saturation mode so it really does act like an on/off switch?

I'm struggling to understand the functioning of transistors in general which doesn't help despite reading umpteen webpages on them. Any help much appreciated:

Assuming NPN, I understand that a transistor has a voltage requirement to the base to cause current flow C to E. I now realise after much reading that the current is important as the current at the base alters the current C to E flowing through the transistor. I don't know what part of the datasheets for transistors tell me what base current I need to apply to get the transistor into saturation so that it passes the full allowable current C to E.

Going back to the base voltage. I gather that most transistors need about 0.6v to 0.9v to the base. I've found most transistors in circuit I've tested with a voltage meter show more than that to the base. I've done lots of reading to educate myself and wonder if I've got this next bit right (I'll write it as if it's fact because I am writing what I think is the case, but want to be corrected please!):

A transistor only uses whatever base voltage it requires, so let's say in this instance 0.7v. If 5V is put on the base then still only 0.7v is travelling from base to emitter, being used as it were.

It needs a certain amount of current to saturate, below that it is in linear mode and the amount of current it allows to flow C to E (gain) is proportional to base current. So, if you have a 5V supply and you choose 1mA current to the base, then you need to disperse 4.3v at 1mA, so you use a 4.3K base resistor. If you want 2mA at the base, you use a 2.2K base resistor.

Because of the relationship between voltage and current, would you measure voltage (assuming emitter to ground NPN) from ground to base and read 0.7v, or would you read a different voltage that depends on what current you are supplying to the base, even though only 0.7v is flowing from base to emitter? If we measured either resistor used above, you would still read 0.7v at the base, then why do I find higher voltages on bases on so many transistors I check in working circuits?

I'm really confused. Will the base draw whatever current it can pull, and the resistor will cause the voltage to drop as more current is drawn, so you'll always read 0.7v on the base?

If anyone has the time, could someone please look at the BCR185W datasheet I've linked to and tell me, using a PIC, what base resistor I'd use to turn it on in saturation mode, and most importantly, how you arrived at that, what parts of the datasheet you looked at, how it was calculated, etc? (even if the built-in base resistor means there's no calculating to do, it can just be connected straight to the PIC I/O pin.) It'll help me understand the calculations and how to read the necessary parts of the datasheet.

Thanks.

Sherldonnnn

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The datasheet tells you that for this little transistor to saturate with a saturation voltage drop of maximum 0.3V when its collector current is 10mA then the base current must be at least 0.5mA so the ratio of collector current to base current is 10mA/0.5mA= 20. Most American little transistors (2N3904 for example) specify a ratio of 10 but most European transistors (BC547 for example) specify a ratio of 20. This is saturation where hFE (current gain) is not used. hFE is used for a linear amplifier with plenty of collector to emitter voltage so it is not saturated.

It seems that you do not understand that the base-emitter of a silicon transistor is a silicon diode. It conducts a very low current when its voltage is about 0.5V and it conducts a normal operating current of about 0.5mA when its voltage is 0.7V. Some transistors have a base to emitter voltage of 1V at a fairly high current. The base to emitter voltage changes when the temperature changes. You talked about applying 5V directly to the base of a transistor. The current would be so high that the transistor will explode (if the power supply can produce the very high current).

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