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

[Mark Hennessy]

OK - this all sounds very encouraging. It sounds like you're not going to fall into the usual trap that I see

So, what's the collector voltage?

Hello again, NM,

Given that you're so very close, are you going to finish this for us all? One number is all we ask

I might start hiding the off-topic posts until we've got there.



By "off-topic", I mean the posts that aren't relevant to our transistor tutorial. And by "hide", that means I'll restore them all once we're done. I'm not suggesting this to be cruel, or to put pressure on you, but I'm struggling - even with 15+ years experience of teaching electronics - to find ways to keep you focussed on the task. It would be much easier if you were in the room with me - I could throw chalk at you

(Obviously, we don't use chalk. Whiteboard pens are far better projectiles that chalk ever was. Not that we'd ever throw whiteboard pens at students. Probably )

Joking aside, I'm actually very impressed with the bits of knowledge that you've displayed. However, we can all see that it's very fragmented. What you need is someone to bring it all together for you - to make the links that lead to understanding. There is a big difference between knowledge and understanding. And many of us on here would love to help - most engineers are passionate, and very evangelical. However, you need to play your part - you need to be willing to listen to the advice, and play along with exercises like this.

Some might ask why I'm bothering. Because I see potential. You can do this. Your biggest challenge will be concentration and focus. But, however your brain is wired, practice will help, I promise.

Do remember that your attitude to learning will influence other's willingness to help you. In other words, if everyone sees you trying your best, you will have free access to some of the best brains in the community. But think about how the off-topic posts might seem to others - might it look like you're not willing to learn? In other words, my constant hints to stay focused were not just for this thread - they were for your wider benefit.

Mike's request for your first name is a good one. I understand that many on the internet like anonymity, but we're all friends on here, and really, what harm can result? But I won't push that - whatever you're comfortable with. But while I think of it, do you travel much? If you drive or can get a lift, you're not all that far from me - a couple of hours max? Given a bit of planning, I'd happily spend a day with you, doing practical exercises with real circuits. And I'm sure I can find some spare bits of test gear for you to take away. I promise not to throw whiteboard pens at you

Anyway, please tell us the answer to the question. I'm 99% sure you know the answer, but just humour me, so I can relax

All the best,

Mark

[Murphyv310]

Take mine out Mark not a problem.

[Mark Hennessy]

It's OK, Trev - don't worry

[Nowhere-Man]

I just noticed something that's relevant to this thread so I'll share it. I was just looking at the Saalburg diagram and something very simple struck me. I will share it now: Voltage dividers simply split the supply voltage but need to then be viewed distinctly in the contex of load and current. So if we have a supply of, say, 230 volts with 22K and 11K, the split potential will be 153 volts and 76 volts. Yet with loads placed across each divider, a wattage and current will create a drop across the load.
So, to calculate an anode potential we need a total series voltage plus the current. Thus in series the. 22 and 11K are just the required resistance.
My my point is the voltage divider is just a way to split volts but loads and current are going to have an effect.
So, if you know the supply voltage and the anode voltage the difference is the drop required. Divide the known current into the sum to drop. If it's say 100,000 R you can use three 33's.

[Nowhere-Man]

Looking at Mark's diagram probably I'm just not used to transistor diagrams but I think what throws me is the resistances aren't on a common rail but via the transistor. So looking at the 9 V supply the current runs in series through the 47 k and the 10 K back to negative. That's series. Without any transistor at all we could have 57 K and 5700 K in parellel across the supply rails. Yet the 4700 and the 1K are not common to ground. Do you see the transistor load in series with the emitter load??? Alternatively what would happen if you used the total parallel resistance?
Going with what I know so far, the norm would be you'd have the HT current and voltages in the valve schematic and it would be easier I think.

[Nowhere-Man]

4.3 would be 9 volts minus the 4.7 dropped by the 4700 load. There's another observation. Ok I got it. Let's assume we know the "tables" gave us 4.3 collector volts and supply is 9 volts. So we need to drop the difference between 9 and 4.3 which is 4.7. Divide the known 1 milliamp into the 4.7 to drop and you get 4700 R. That will be the "trick" question Mark aimed at me. I think it only confused me since normally the voLtage would already have given for a tube anode. It's almost 3.30 am so nobody can accuse me of not doing my bit. I am going to put the kettle on and watch a movie.

[Murphyv310]

There is no trick in the question, just look at the current flowing in the the emitter resistor you know that already as Mark has told you, and its voltage drop, from your own admission you know the current in the collector resistor so you should then know it's voltage drop, one further simple calculation should then give you the answer we are all waiting for.

[Terry]

He's already given the answer of 4.3 volts, Trevor, but perhaps you missed it as he immediately wandered off into the world of mystery tables where he seems to be implying, if I've understood him correctly, that the collector voltage should have been shown on the schematic in the first place!

I must say I really admire Mark's patience in this uphill struggle - it has really been painful to watch at times.

[Crackle]

4.3 would be 9 volts minus the 4.7 dropped by the 4700 load. There's another observation. Ok I got it. Let's assume we know the "tables" gave us 4.3 collector volts and supply is 9 volts.

NO! forget the "tables" for now, the answer was not in them.
You have said the correct answer.
4.7v is the voltage dropped across the 4.7k resistor because it has 1mA flowing through it. V=IR   (I is the current)
So the voltage at the collector, (with reference to 0 volts, which is maybe what is confusing you), is 9v - 4.7v = 4.3 volts
The answer Mark was looking for was in your first sentence of your last post.

So to recap the complete answer to Marks original question is b = 1.6v, e = 1v, and finally c = 4.3v

Please, NWM can you go over this again to check how each of the 3 voltages were established, and ask any relevant questions you want if you have any problems.

Mike

[Crackle]

Let's assume we know the "tables" gave us 4.3 collector volts and supply is 9 volts. So we need to drop the difference between 9 and 4.3 which is 4.7. Divide the known 1 milliamp into the 4.7 to drop and you get 4700 R.

Yes you have got it.
If you do not know what the value of the collector resistor should be, then you can work it backwards as you have done above.
But remember don't take the transistor or valve Manufacturer's "table" as gospel.
Use the data available in the service manual for the radio you are working on, these should give you the working voltages of the valves or the transistors. (but not always)
Mike