ВЭФ 12 (Astrad VEF 12 Project) - including a transistor tutorial

[Crackle]

The VEF circuit I posted further back but it's a bit blurred.

There is a reasonably clear schematic on the radio museum link which I put in post 6

Mike

[Nowhere-Man]

0.3 would be the right number for germanium. However, this is silicon. So try again

I'm not sure about the "magic number" sentence - I won't think about it any more right now, but perhaps we'll come back to it. I'm trying to keep you focussed on the problem - nothing else.

Let me give you a bit more: if that transistor was germanium (perhaps an AC127), can you see how the 0.3V would result in 1.3V at the emitter?

Yes, it is understood. And the silicon potential hill is 0.6. I'll be sure to refer back to these comments. If it ever seems I am not digesting the info, don't worry, I do file and use these posts. In fact the other day I re-read Miguel's comments on lead identification or may go back. Lately I work at an obsessive level and have been analysing a 2.4+1 superhet as well as the other work I have. Can't seem to switch off. Still in a bit I'll be watching the rest of Planet Of The Apes 1969.

[Mark Hennessy]

Good morning.

So, we have 1.6V at the base, and 0.6V less at the emitter.

We know the value of the emitter resistor, and we know the voltage across it. So what does that tell us?

[pwdrive]

Mark, first off, ignore my last PM...pressed the wrong button! Second, no need to reply to my previous PM as I can see now that you were using 0.6 volts, I was using 0.7 volts, hence the PM's prior to that, anyways, no problems now, cheers.

Lawrence.

[Mark Hennessy]

Naturally, the voltage is nominal as there is a distinct "knee". But 0.6V is a good bet for a small-signal scenario like this - it might well be closer to 0.7V when significant base current flows, such as in a power application. But let's not worry too much about that right now...

[pwdrive]

Yes, when I was using 0.7 volts I hadn't come across the data sheet, I since have, it's a proper Philips one:

http://www.futurlec.com/Transistors/BC109.shtml

Lawrence.

[Nowhere-Man]

"Let me give you a bit more: if that transistor was germanium (perhaps an AC127), can you see how the 0.3V would result in 1.3V at the emitter?"

Yes, I follow you. You ask me to subtract from the 1.6 divider supply and the silicon (is it passive bias) digit is 0.6 as opposed to 0.3 for germanium. So we get 1 volt. I think we can keep this circuit on file and we can finish off all the voltages and dividers and we can print it out for reference. I will go back to the schematic over time and we'll develop it. Tomorrow I'll print out the diagram because I only have my small phone.

[pwdrive]

"Let me give you a bit more: if that transistor was germanium (perhaps an AC127), can you see how the 0.3V would result in 1.3V at the emitter?"

Yes, I follow you. You ask me to subtract from the 1.6 divider supply and the silicon (is it passive bias) digit is 0.6 as opposed to 0.3 for germanium. So we get 1 volt. I think we can keep this circuit on file and we can finish off all the voltages and dividers and we can print it out for reference. I will go back to the schematic over time and we'll develop it. Tomorrow I'll print out the diagram because I only have my small phone.

NM, while you're at it try to find the reason why the bias voltage is obtained from the potential divider, here's a clue, there's a similarity with certain stages of some valve radios.

Lawrence.

[Nowhere-Man]

Will do. It will be better to print out Mark's diagram at the library because I just have an old Blackberry phone on the boat. I should add out of interest apart from voltage dividers, the 1940's engineers often simply took voltage difference between anode and HT supply, then simply divided the known current into the voltage to be dropped to get resistance.
Just to repeat I am hopelly ordering several P or MP type transistors from Russia so the set gets repaired. In the future I also have a dead Radiomobile rumour has it are tough to repair.

[Terry]

It's easy enough to draw it for yourself.

You've already established that the base of the NPN transistor is fed from the 47k/10k potential divider between Vcc and earth.

The emitter is connected to earth by RE (1k) which has a bypass capacitor CE (no value stated).

The collector is connected to the 9V supply rail by RL (4k7) with the output being taken from the collector.

The input is to the base via C1 (no value).

That's all there is to it - you should be able to draw it in less time than it's taken me to type this!