how can i calculate Base currents/voltages?
As a rule of thumb, the voltage between base and emitter is about 0.6V.
In theory base current = collector current / hFE.
In practice you can forget about doing any accurate calculation of base current, because you don't know what the hFE is. For example the BC546 datasheet says it's hFE could be anywhere between 110 and 800 at 25 degrees centigrade with collector current = 2mA. Just to make things more interesting, hFE changes with temperature and collector current as well, as the graph below shows.
Btw, if you look at a datasheet from a different manufacturer, you may see a totally different looking graph. That's because the "same" part from different manufactirers aren't actually the same.
Anyway, the point is that "
assume beta = 100" only happens in classroom exercises, not real life.
Having said all of that though, the situation's not completely hopeless since we normally don't care what the base current is anyway, so long as it's low enough to not cause a problem.
What we can do is a sort of "worst case" analysis, to calculate the highest value the base current might be. For example a BC546 with Ic = 1mA should have an hFE of at least 100, so base current should be less than about 10uA. We dont know whether it's 10uA or 2uA, but we
do know it's not more than about 10uA.
First thing is to do a DC analysis.....
Good thinking. Let's start with the voltage gain stage and Vbe multiplier, referring to the schematic below (which I copied from your link, so I hope it's the right one).
For the given supply voltage, the current through R9 and R10 is about 5mA. This is also Q4's collector current, and the current flowing through the Vbe multiplier.
Looking at the Vbe multiplier: To bias the output stage correctly, Q9 must have about 1.2V between collector and emitter, i.e. about 0.6V across each of R16 and VR1. So VR1 will be set to about 1K and there will be about 0.6mA flowing through R16 and VR1.
Since we started with 5mA, the other 4.4mA must flow through Q9. Since Q9's hFE is at least about 110, the base current must be less than about 0.04mA. Since Q9's base current is at least 10 times smaller than the current flowing through R16 and VR1, it doesn't really effect the calculation - we can forget about it.
Moving on to Q4....
Collector current is about 5mA and the hFE is lousy - the datasheet says hFE = 40 to 160 at Ic = 150mA (IIRC), but at 5mA it may be worse. So Q4's base current may be as high as 0.1mA or even a bit more. At the same time Vbe = about 0.6V, so there's about 1mA flowing through R6.
Added together, that means Q1's collector current is probably a bit more than 1mA.
Which brings us (at last) to the input stage and current source.
I don't know exactly what the forward voltage of the green LED is, but it will be high enough to ensure Q3's collector current is more than 2mA. How
much more is a bothersome question, but let's assume Rod's done his homework and designed the circuit so so that the collector currents of Q1 and Q2 are roughly equal i.e. about 1 to 1.5mA each.
This time the base currents actually are important, and cause a slight problem.
Q1's base current flows through R2 and R3 causing a small voltage drop across them, while Q2's base current flows through R5 causing a small voltage drop across it.
If those two voltage drops are equal
and the Vbe of Q1 and Q2 are the same, then there will be zero DC voltage between the amplifiers output and ground. That's highly unlikely though, so there's going to be some (unwanted) DC voltage across the loudspeaker. Let's try figure out how bad it is:
To start with, according to the datasheet the difference between the Vbe's could be as much as 150mV, even if the currents through Q1 and Q2 are the same. Hopefully it won't be that bad though and there's not much you can do about it anyway unless you want to take time out to measure individual transistors and select matched pairs.
Then there's the voltage drop across the resistors. Since Ic is probably a bit over 1mA each for Q1 and Q2, and hFE may not be much over 100, the base currents could be as high as about 10uA, but may also be much lower. The worst case here is if the base currents are very different.
e.g. If the base current is 10uA for one transistor and 2uA for the other, this will cause a DC offset across the speaker of about 8uA * 22K = 176mV.
That's not bad, but there is something we can do to improve it. Without using matched transistors, the next best thing is to use transistors with higher hFE, i.e. choose BC546B rather than BC546A. The "A" suffix indicates hFE = 110 to 220, while the "B" suffix indicates hFE = 200 to 450. Even better, use BC547C or BC550C, which have hFE = 420 to 800. (BC546 isn't available in the "C" grade).