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formerlyme
Posted: December 31, 2015 08:55 pm
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Hi,

I hope this is the right sub-forum. Following on from a question I asked before about using a PIC to get a dehumidifier control board working, I was finding out what parts are on the original board which the PIC will have to control.

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

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Sch3mat1c
Posted: December 31, 2015 11:56 pm
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QUOTE (formerlyme @ December 31, 2015 02:55 pm)
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?

Yes. AKA "prebiased" transistors.

QUOTE
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.


1. Just assume. Use Ib = Ic/10.
2. Check min hFE at the desired Ic. Use an actual hFE(sat) == Ic(max) / Ib(on) somewhat less than this (perhaps half).
3. hFE drops precipitously near Ic(max), so also choose a transistor large enough for the job.

You'll run afoul of #2 if you use #1 without checking. There are few common examples, but high voltage, "high speed" BJTs tend to have poor gain. These are usually found in CRT televisions, for the horizontal output transistor: typical ratings are Vceo = 800V, Ic(max) > 5A, hFE ~= 5 (with switching times measured at hFE(sat) = 2.5).

Conversely, some types have particularly high hFE: low-Vce(sat) types (such as PBSS303NX), superbeta (2SD1273), and Darlington (TIP120). These do not need to be operated at such high base currents (and indeed, Darlingtons can be damaged with that much current), which saves not only the bias current required to drive them, but has the added benefit of reducing turn-off time (storage time).

QUOTE
Going back to the base voltage. I gather that most transistors need about 0.6v to 0.9v to the base.


Indeed, the pure BJT is more fundamentally a voltage-input device than a current-input device.

The trouble is, both base and collector currents depend on the exponent of that voltage, so the current increases dangerously fast for a rather small change in voltage.

This is excellent news when you want a high gain amplifier. It's not very practical to try driving a device with a constant voltage source though!

The base and collector exponents are different functions. They happen to have the same exponential slope, so we can take the ratio of them and find that it has a fairly stable value -- the hFE. But there's really no reason for hFE to be any particular value at all, and indeed, it varies widely from device to device, and with respect to temperature, and of course with collector voltage (because that's what saturation is: the collector voltage is so low that additional base current won't draw additional collector current, so hFE drops sharply).

So amplifiers are always arranged so that bias currents in a given stage are greater than the base currents required in the next stage, and use negative feedback to adjust for the variations automatically.

QUOTE

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!):


Now, be veeeery careful what you're saying here.

If you're measuring more than about 1V on the 'input' pin, it's not the base. (Or, it was, and isn't anymore...)

Those prebiased transistors don't have the base connected to a pin, so you cannot measure the base voltage.

The datasheet claims the pins are "B", "C" and "E". They lie. There is no such "B" pin on that part.

There are always three kinds knowledge:
1. What the source tells you about something.
2. What you think the thing is (based on #1, and other sources, and your own research).
3. What it actually is; complete truth. (Given that the philosophical notion of "knowing" is a dubious and incomplete matter to begin with, but supposing that were possible, I mean.)

Through #2, you can approach #3. Sometimes, #1 is sufficiently close, at least for a narrow purpose (like using prebiased transistors). Often, it is blatantly wrong, sometimes on tangential things (it doesn't REALLY matter what a pin is called), sometimes on direct things too.

Not that this helps you yet, barely having #1 in your set of "what you think", active working knowledge... But perhaps the perspective will help guide you. wink.gif

So anyway. If you measure a circuit with, say, an MMBT3904 in it -- you will notice the base-emitter voltage is relatively stable in the 'on' region, because Vbe is logarithmic to Ib (which is simply what I said earlier: the inverse of, Ib being exponential with Vbe).

When you measure a circuit with a current limiting resistor, then the excess voltage drops across the resistor, and base current is set by Ib = (Vin - Vbe) / R. Which, if Vin >> Vbe, basically means Ib is fixed, which is pretty handy!

QUOTE
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.


Yup!

QUOTE
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?


Yes, 0.6V or so for those kinds of conditions (when 'on', of course).

QUOTE
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?


Yes. Like I said, two exponentials -- the base and collector are largely independent, so that the B-E junction looks for all the world like a lone PN diode, and the collector mysteriously draws current, when its voltage is reasonable to do so, and at a (more or less) proportional amount (i.e., hFE being the proportion, given that it isn't the most stable property of a transistor or anything).

In reality, when the collector voltage is very low (under 0.1V, say), Vbe drops slightly. This can be seen as beginning to forward-bias the B-C junction (which, due to doping, usually has a higher forward drop than B-E, so isn't extreme or anything), and can be seen as diverting base current.

As collector voltage continues to drop, going negative, B-C current flow takes over, and the B-E junction voltage begins to reverse. But C-E current continues to flow, this time reversed -- because an NPN transistor reads N-P-N no matter which way you read it, after all. Practical devices aren't very efficient when reversed (due to the difference in C and E doping, hFE is much higher in the normal direction, and typically quite small when reversed, like ~2 for a MMBT3904), but it's interesting to remember that, yes, it does work both ways.

(MOSFETs are rather more commonly used in the reverse direction. In that case, when turned on, the D-S channel really does act like a resistor, no matter which direction you use it. Though it acts in parallel with the normal body diode action, so this is only useful for very low voltage drops.)

Tim


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Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.
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