Just did a very quick sum. Dissipation in each emitter resistor is about 220mW for 100mW output. 0.5W or above resistor should suffice. You will soon know if they are under rated cos they will get hot.
Have to disagree with post #18. You have to do the basics first. If you are not prepared to go to classes to learn then buy the Horrowitz book. You do not need to read the entire publication before you start playing but, you must learn the basics first.
The reason that adding the capacitors have a large effect is because they performing what is commonly known as decoupling. Wire has resistance and, importantly inductance. As we should all know (simplifying it a bit) inductance resists the flow of current when the demand is a simple pulse or a...
Have a look at the attachment. It has 2 variations on a simple constant current generator.
They are both essentially the same with Example A using a PNP Darlington whilst Example B uses a P Mosfet.
The current is determined by the Vbe of Q1 and Q3 in conjunction with R1 and R4. R3 and R6...
One point that is often forgotten in these items when discussing the charging of NiCad and NiMh batteries is that the charge process is only 60% efficient and must be considered when calculating charge currents. Example, if charging a 1Ah cell on the 14hour cycle you would set the charge current...
Don't forget that you will have to rectify, smooth and regulate the result otherwise you will blow the nuts out of it if you will pardon the expression.
If you need any more info, ask.
I'm with Kellys-eye. Post a schematic without all extra ephemera as the moving dots make it very difficult to read. Also, give us some idea of what you are trying to achieve, just understand the circuit or, make a proper SLA battery charger!
Further to my earlier comments. There is no quick and easy path to learning about electronics. I suggest that you buy yourself a copy of Paul Horowitz's book "The art of electronics". It is available on Amazon's site and will tell you everything that you need to know. Over the years it has...
Sorry but you simply don't need to calculate the voltages to get the correct answer. I think the method used is the quickest and least confusing. If you wish to know the voltages it may be done afterwards with no confusion. That is my last word on the subject.
Yes but the whole point of my reply was that you don't need to calculate any voltages at all. The original question had no requirement to do so. Your bit of maths is correct, I just couched it such that is was easier to understand.