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

[Mark Hennessy]

4.3 would be 9 volts minus the 4.7 dropped by the 4700 load.

Excellent!

The "trap" is that people who are new to this sometimes forget that they must subtract the 4.7V from the 9V supply rail. Even when folk have had a bit of practice, they still sometimes make that mistake. Most often when they sit the exam, where the pressure gets to them. It might seem unlikely to some of us, but trust me, it happens.

So, to quickly summarise:

1. Find the base voltage.
2. This allows us to see the emitter voltage.
3. With the emitter voltage known, we know the emitter (and hence collector) current.
4. Determine the voltage dropped across the load resistor, and hence determine the collector voltage.

Job done!

Incidentally, there are many ways to bias a transistor - this is a good one, but you'll see others. One notable fact with this method is that we haven't had to use hfe or Beta in any of the calculations. Most other methods do, but because hfe is a highly untrustworthy parameter of a transistor, designs should not depend on a certain value.

This is just the DC conditions. Earlier I said:

DC conditions first. Then add signal.

In other words, you must always bias your valve or transistor before it can amplifier the way you want. The AC signal basically "wobbles" the circuit either side of the quiescent DC condition.

We can take care of the AC stuff whenever you'd like. It's pretty simple. But for now, it's worth trying to be sure you're happy with establishing DC conditions. After all, when fault-finding, this is where we start - when someone reports a faulty amplifier, the first question I have is essentially "what's the collector voltage". So, to that end, I've attached another exercise for you. Exactly the same circuit, but different resistors, and a different supply rail. Have a look at it. It won't take more than 5 minutes to work out, providing you've understood everything we've done so far.

   

Well done for sticking with it. This approach - logically following a path, taking small and simple steps as we go - is the right approach for most engineering challenges, and learning this philosophy is as important as the electronics itself.

Mark

[Nowhere-Man]

The wording of a question for me is crucial. I did old exam questions on this stuff in the past and never got one wrong. I'm also choosy about textbooks and avoid classroom systems. I would start first on PNP germanium because these were the first used and current doesn't leave the N type base. It sinks current at the base. Bear in mind transistors are new to me (about 7 weeks). So, start with the basics is best, PNP first. Also with valves phase is very important and I have to apply this to semiconductors too. What is the phase of cathode anode voltage in relation to the anode load of a triode? That is with signals on the grid. What is the cathode grid phase? I need the full picture to get the details.

[Refugee]

We did get very close back at post #81.
We got a bit of a distraction from the Russian transistor sellers and it has taken us a little time to get back to it again.

I tried to recreate the collector currant and HFE curves for some transistors out of those bags of cheap bits that were on sale in the 1970s.
I basically built a small part of a curve tracer that required 25 measurements to get a low def display on a bit of graph paper.
In those days electronic calculators were expensive and in most places were worshiped and guarded. Pens used to run out fast.
I still have the gadget I built at the time and it still works.
I could offer to do a full service on it and ultimately add an internal DVM to replace the external meter it was originally made for use with.
It is still used to test large audio transistors for matching. and selection.

[Mark Hennessy]

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.

What you're saying here is that we assume no current flows out of the junction of a potential divider. Yes, this is true. If it does, we might need to account for it in some way - although sometimes we can ignore it. If we need to do anything, it might be really simple, or it might be rather more involved. The general term for this is "loading effect".

Let's give a relevant example. Think back to the potential divider at the start of the exercise - we said that the base current was so small that we can ignore it. But perhaps, as you've raised the issue, we should explore it.

Have you come across Thévenin's theorem? This is an incredibly useful tool that I use nearly as much as Ohm's law.

It tells us that the potential divider in my original problem (47k and 10k) could be replaced by a "perfect" voltage source of 1.58V in series with 8k2 (10k in parallel with 47k):

   

Let's assume the hfe of the transistor is 200, meaning the base current will be 5uA (1mA/200). Is this significant? Well, 5uA flowing in an 8k2 resistor gives us 41mV, meaning the voltage will drop slightly from the expected value by about 2.6%. Hmm.... Given that we're most likely using 10% resistors in a vintage radio, I'm not going to lose sleep over that! And bear in mind the 9V comes from a battery, which will measure more than 9.5V when new, and the radio will be expected to work down to about 6V...

Indeed, our assumed 0.6V across the base-emitter junction was just that - an assumption. But absolutely fine.

With experience, you learn when assumptions like this are appropriate, and when more precision is needed. By leaving out certain details and over-simplifying the maths, I haven't told you any lies. An important part of teaching it to give the right level of detail at all times. We build a picture, starting with broad strokes and filling in the fine detail when and where it's needed.

[Nowhere-Man]

The Russians can be very closed. I made good friends there and lots of friendly people. However, they need to mix more. I was not welcome on the forum although the moderators were OK and sent a pm with the info I need. I suspect the Cold War climate is going to get worse and the new thing is Russia will ditch Microsoft with China installing an alternative network. Next time I'll pass myself off as Hungarian or something.

[Mark Hennessy]

I would start first on PNP germanium because these were the first used and current doesn't leave the N type base. It sinks current at the base. Bear in mind transistors are new to me (about 7 weeks). So, start with the basics is best, PNP first.

Actually, I strongly disagree with this.

In a PNP transistor, current is sourced from the base. An NPN transistor sinks current at the base.

If you're familiar with valve circuits, NPN is best.

When it comes to understanding them, there's actually no difference - they both behave the same. If people find a difference between them, they always prefer NPN and struggle (only slightly) with PNP. By sticking to this mantra of "PNP came first", you're just making life difficult for yourself.

However, what it boils down to is just the difference between positive and negative numbers.

Also with valves phase is very important and I have to apply this to semiconductors too. What is the phase of cathode anode voltage in relation to the anode load of a triode? That is with signals on the grid. What is the cathode grid phase? I need the full picture to get the details.

You're asking about AC here. Are you sure you're ready to get onto this?

Here goes:

The collector current is, very roughly, proportional to the voltage at the base-emitter junction.

To be more precise, there is an exponential relationship between the two, as described here: https://en.wikipedia.org/wiki/Bipolar_ju...Moll_model

So, if the base-emitter voltage increases, the collector current also increases. So what happens to the voltage at the collector - up or down? Over to you - everything you need to know to answer this question is contained in this post.

[pwdrive]

The wording of a question for me is crucial. I did old exam questions on this stuff in the past and never got one wrong. I'm also choosy about textbooks and avoid classroom systems. I would start first on PNP germanium because these were the first used and current doesn't leave the N type base. It sinks current at the base. Bear in mind transistors are new to me (about 7 weeks). So, start with the basics is best, PNP first. Also with valves phase is very important and I have to apply this to semiconductors too. What is the phase of cathode anode voltage in relation to the anode load of a triode? That is with signals on the grid. What is the cathode grid phase? I need the full picture to get the details.

It's easy to figure out the phase relationship between a signal on the grid and the signal across a cathode resistor, as the grid voltages increases more current flows through the cathode resistor therefore the voltage across it increases, therefore the grid voltage and cathode voltage increase at the same time, therefore they are in phase with each other, that same increase in current causes a voltage drop across the anode load resistor which means that the voltage on the anode will reduce at the same time, in other words the increase of grid voltage results in a decrease of anode voltage, therefore the grid and anode voltages are not in phase with each other, there is now a phase difference of 180 degrees between the grid signal and the signal between the anode and ground, you now have the basics of a phase splitter because the anode and cathode signals are out of phase by 180 degrees, take that one step further and shift the anode signal by a further 180 degrees and feed it back to the grid in sufficient quantity and you now have the basics for an oscillator.

Same thinking for transistors.

EDIT: Post crossed.

Lawrence.

[Nowhere-Man]

That's interesting, Mark, no I'd not heard of the said theorem. I'll go through it tonight. My situation here is strange. I once got hold of an Owen Bishop classroom book so I guess this is the way you teach. That is, active work-sheets and class tests with group activities. It brings back memories of my old biology classes with Dr Houle. My first introduction was Chas Miller's Valve Radio and Audio book so there were never any classes or tests to do. Chas devoted a few short pages to ohms law and that was it. The rest was diagram understanding and diagnostics. I moved on to American Ham written in 1960 and this is very deep but not active as in a class. Plus the terminology is American. You don't sit exams with these books and most of it is explanation, then example and then one question. There are limits. Being self-taught I sometimes find little gaps because I didn't go fom A to B to C. With high voltages that can spell dangers - I had a couple of zaps when starting out. So this is something to ponder. Some of us out there are self-taught and we sometimes get heat. I think maybe Rick McWhorter was self taught and he got some real criticism on Amazon over Americanism and a maths blunder in his book.

[Refugee]

What we have to remember is that many of the authors of the books learned there trade by working with scientific instrument manufacturers.
The circuits in those books will cover things like small variations in things like Vbe caused by temperature and so on. This makes them into a very big distraction from learning about basic circuits unless you want to make a thermometer with a transistor as a sensor.

[Nowhere-Man]

The AC output (anode load) voltage will be opposite phase to signal voltage. Cathode anode voltage will fall as anode load voltage rises but now my question is, what fundamental capacitor prevents negative feedback to the signal input voltage? I know the answer but am throwing it out to all.
I will take a good look later at Mark's helpful comments. Why doesn't he make a tutorial on the forum? I mean, the other big site does have reference reads.