Transistors

[Nowhere-Man]

I read the Beta of a transistor can be calculated if you know either the emitter or base currents, or the collector base currents. The collector base example was easy enough so you just divide base microamp current into collector current. Then, using the emitter and base had me a bit confused. Do you subtract the division sum from 1? That's what it seemed to suggest.
Anothe question is how many transistor radios would opt for just a diode as the detector? Or was it more common to use a transistor for detection.

[reset]

Beta is the current gain. The ratio of collector current to base current. Emitter current is divided between the base and collector or another way of saying it is the emitter current is the sum of collector and base current and alpha is the emitter current minus base current. So beta should always be greater than one and alpha should always be lower than one. 
If that make since then it would follow the ratios can be simplified by referencing a current of ONE. If Ie=1 then 1-alpha = base current. if I=1 then Ib = 1/beta.
I would recommend this book:
https://ia802607.us.archive.org/28/items...niques.pdf
Page 19 discusses current gain.

The diode vs transistor as a detector would probably be more an economics question. The manufacturer would but in bulk and get great pricing. If they used a 100 transistor for every diode the difference in cost would make using the transistor as a diode more cost effective. You can buy a JFET for 3 dollars each at dealer A or 2 cents each minimum of 1000 at dealer B. Someone will argue an operation advantage and it may be true but my take on it it economics drives business more than performance.
73

[Mark Hennessy]

Beta is the ratio of collector and base currents. Simple.

The emitter current is the sum of the base and collector currents. Also simple.

Given that, can you work out beta from just emitter and collector currents? Perhaps. Subtract Ic from Ie, the difference is Ib. Then go to line 1 above.

What about emitter and base currents? Subtract Ib from Ie, the difference is Ic. Then go to line 1 above.

But that assumes that your measurements are accurate, and given that Ib is usually measured in microamps in a typical small-signal circuit, that's by no means certain. Remember, beta is often much greater than 100 for small-signal transistors.

Ultimately, who cares? Beta is an irrelevance for most properly designed circuits. It varies so much from sample to sample, with temperature, with collector current, with planetary alignment, etc. A designer might assume a certain minimum value, or a (wide!) range of values, but would never assume a certain value. That would be commercial suicide.

If you cast your mind back to that transistor amplifier exercise, you'll remember that we didn't use beta at all. Which reminds me, you still haven't done the second exercise. Hmm...

In over 30 years of using transistors, I've never knowingly used alpha.

Unfortunately, old text books place the wrong emphasis on transistor theory (and plenty else besides). Also unfortunately, you seem to be unwilling to accept this. As a teacher, it's really painful for me to watch...

[ppppenguin]

I think alpha mattered a lot in the very early days of transistors where beta was very low and you couldn't usefully approximate alpha as unity. Leakage current mattered too in the days of germanium devices and made it more difficult to design circuits that were sensibly immune to temperature variations etc.

[Nowhere-Man]

I just worked out a scenario with 30 microamps base and 6 milliamps collector. Then adding that total to get emitter current. Yet what I was really trying to apply was Ohms Law, using a base current and load resistor with a Beta. It is still different to valves and when transistors came to be, circuits got more complex.

[Mark Hennessy]

OK, so that's a beta of 200. Emitter current will indeed be the sum (6.03mA).

To apply Ohm's law, you need to know the voltage across each resistor. So your scenario needs a bit more detail adding. To help, I'll give you an example:

Let's assume a supply voltage of 9V, and a silicon transistor. I'm also assuming that emitter is connected to ground.

To get 30uA in the base, you need a 287k resistor. That one is pretty easy to work out - I'll leave that to you to figure out and explain to us.

The collector voltage is more tricky because the scenario hasn't stated the desired collector voltage. In which case, I'll assume 4.5V. Having added that assumption, calculating the load resistor is so easy that I won't do it... Over to you.

Does that help? Do you want to outline your scenario, complete with the missing details that I've just mentioned, so we can double-check your understanding?

[Nowhere-Man]

It will be useful to keep a transistor thread going. When possible I can upload a diagram I'm using. However, my radio work at the moment is snatched while walking the dog. By day, I'm plumbing and wiring the boat that very soon I'll be living on. No soldering work yet or any electronic repair. That will come towards Winter.
What you said about emitter being the sum of collector and base current was useful. While sat with the dog in the drizzle I worked through a few sums and all was we'll.
Maybe later I'll share a simple PNP circuit I have where a voltage divider sets the base voltage and imput impedance. It's a late 50's diagram.
Work on boat tomorrow could be in heavy rain.

[Nowhere-Man]

30 microamps I use as 0.00003 but I know some people use divisors. I do the same when converting picofarads into, say, nanofarads.
For collector Beta I subtracted base current from 0.006 amp and that was the Beta. For emitter, I summed up the emitter current and then pretended it was just a fresh figure. I then used unity. It helped.

[Craig]

I think alpha mattered a lot in the very early days of transistors where beta was very low and you couldn't usefully approximate alpha as unity. Leakage current mattered too in the days of germanium devices and made it more difficult to design circuits that were sensibly immune to temperature variations etc.

That is absolutely the case. Germanium transistor beta was typically in the range 40 to 100, and leakage current from 50uA to 300uA, and highly temperature dependent. Finger heat will easily triple leakage current. And when trying to measure beta, you have to wait for the transistor temperature to stabilise, measure leakage and then subtract that from the collector current when measuring beta.

Got heavily into that when building a fuzz pedal for my guitarist son. Why Ge rather than Si? Well using the "correct" Ge transistors such as AC128 gives a soft edge to the distortion because of the truly dreadful frequency response. Si transistors have much higher frequency response, and hence produce faster rise times to hard fuzz, and hence produce a harder and edgier tone.

First time I have ever built a circuit whose purpose is to distort!

[ppppenguin]

Go back to the 1950s and a beta as high as 40 was a luxury.

Reiterating Mark's point: Beta is highly variable and you can't use it as a basis for circuit design. All you can do is used the published data to give a minimum beta and design your circuit to work properly over a wide range of beta above that.