03-03-2017, 12:18 PM
(03-03-2017, 01:42 AM)Nowhere-Man Wrote: 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.
