25-02-2017, 04:22 PM
OK, that's an excellent start 
First of all, let's not say "approximately", and then go on to give an answer with 9 numbers after the decimal point. We need to be realistic here: we are engineers who deal with real-world components and real-world measurements. The resistors might only be within 10% of their stated values, and a basic DVM might only be 1% accurate. I know from experience that some people are uncomfortable initially when adjusting to "engineering" - especially those who have done a lot pure maths or physics. However, they usually get the idea, and find that the world is still turning.
My approach is to always guess at what the answer should be, then reach for the calculator to give me the precision. The advantage of that is simple: it's easy to make a "typo" with the calculator. I spend most of my working life trying to convince graduates and school leavers to do this - after a life of learning by rote to pass exams, it doesn't come easily. However, it does make their lives better...
So here, I'd say that if the 47k was actually 50k, then I'd have a sixth of 9V across the 10k resistor (1.5V). But as the upper resistor is a little lower, that voltage will be a little higher in practice.
Let's take your results and record them with our "engineering" hats on: approximately, we have 7.4V and 1.6V
You've discovered that there is 1.6V across the 10k resistor. Another way of putting this: the base voltage is 1.6V with respect to 0V (or ground). That's the first step done.
We're left with the emitter voltage and the collector voltage. I'll give you this one for free: we can't do the collector just yet. So we must do the emitter first.
Now, if you've read up a bit on transistors already, you might know that it's actually really easy to work out the emitter voltage, given that we now know the base voltage. The sum is literally one you can do in your head. However, you might not have gleaned the vital bit of information yet. If not, just ask
Finally, just an aside: please, please don't worry about NPN vs PNP transistors. They work in exactly the same way. The difference is the polarity of the battery (and any polarised components like electrolytic capacitors). And as for silicon vs germanium, again, they work in exactly the same way. Yes, there are some detail differences - the most obvious one has a bearing on the above question so I won't say more (but that's perhaps a useful hint?) - but let's come back to that later.
So really, transistors are easier than valves. I think so, at least. Ignoring JFETs and MOS-FETs (which you won't find in a 1960s radio) there's only 1 basic type to learn. Compare that to valves, where you have triodes, tetrodes, pentode, et al.
Yes, the BC109 is NPN. You probably know that the direction of the arrow tells you that, as will the datasheet. I picked it because it's used by Hacker in their 1960s radios (it's one of my favourites)
First of all, let's not say "approximately", and then go on to give an answer with 9 numbers after the decimal point. We need to be realistic here: we are engineers who deal with real-world components and real-world measurements. The resistors might only be within 10% of their stated values, and a basic DVM might only be 1% accurate. I know from experience that some people are uncomfortable initially when adjusting to "engineering" - especially those who have done a lot pure maths or physics. However, they usually get the idea, and find that the world is still turning.
My approach is to always guess at what the answer should be, then reach for the calculator to give me the precision. The advantage of that is simple: it's easy to make a "typo" with the calculator. I spend most of my working life trying to convince graduates and school leavers to do this - after a life of learning by rote to pass exams, it doesn't come easily. However, it does make their lives better...
So here, I'd say that if the 47k was actually 50k, then I'd have a sixth of 9V across the 10k resistor (1.5V). But as the upper resistor is a little lower, that voltage will be a little higher in practice.
Let's take your results and record them with our "engineering" hats on: approximately, we have 7.4V and 1.6V
You've discovered that there is 1.6V across the 10k resistor. Another way of putting this: the base voltage is 1.6V with respect to 0V (or ground). That's the first step done.
We're left with the emitter voltage and the collector voltage. I'll give you this one for free: we can't do the collector just yet. So we must do the emitter first.
Now, if you've read up a bit on transistors already, you might know that it's actually really easy to work out the emitter voltage, given that we now know the base voltage. The sum is literally one you can do in your head. However, you might not have gleaned the vital bit of information yet. If not, just ask
Finally, just an aside: please, please don't worry about NPN vs PNP transistors. They work in exactly the same way. The difference is the polarity of the battery (and any polarised components like electrolytic capacitors). And as for silicon vs germanium, again, they work in exactly the same way. Yes, there are some detail differences - the most obvious one has a bearing on the above question so I won't say more (but that's perhaps a useful hint?) - but let's come back to that later.
So really, transistors are easier than valves. I think so, at least. Ignoring JFETs and MOS-FETs (which you won't find in a 1960s radio) there's only 1 basic type to learn. Compare that to valves, where you have triodes, tetrodes, pentode, et al.
Yes, the BC109 is NPN. You probably know that the direction of the arrow tells you that, as will the datasheet. I picked it because it's used by Hacker in their 1960s radios (it's one of my favourites
)
