That's right, you did spec a diode drop of 0 V. My bad. My default is to always start with a drop of 0.7 V unless I'm going for a real quick estimate. In this case, the diode drop of 1.4 V has a noticeable impact on the answer because it lowers the peak output voltage by nearly 10%, so in my calculations I used a sine function with the corresponding peak value to determine the conduction angle (in order to find the current when recharging the cap) and I had that stuck in my head.Hi,
Not sure what you mean because i specified a diode drop of 0v meaning that even 2*Vd=0 volts, or did i not make that clear enough maybe?
To switch to the diode voltage drop version with Vd=0.7v the total drop would be 1.4v and that would be dealt with in a slightly different manner. It would at least have to subtract from the input sine voltage by the amount of 2 diode drops.
I always start with a diode drop of 0v because that establishes a 'base' version of the calculation after which we can start to impose other deviations and reformulate somewhat. The 0v version is like a 'reference' version that can show the basic idea about how the calculations are developed without being too complicated.
Is that what you meant? If not, please elaborate if i overlooked something i'd like to know what it is.
Oh no problem at all, i use 0.7 a lot too when explaining things because often people expect to see some kind of drop at least.That's right, you did spec a diode drop of 0 V. My bad. My default is to always start with a drop of 0.7 V unless I'm going for a real quick estimate. In this case, the diode drop of 1.4 V has a noticeable impact on the answer because it lowers the peak output voltage by nearly 10%, so in my calculations I used a sine function with the corresponding peak value to determine the conduction angle (in order to find the current when recharging the cap) and I had that stuck in my head.
Still waiting to see your best attempt at a solution. We really can't do much more for you until we see that.
I also tried 2.4/220 x 1000 x 10^-6. Which also gave me 10.90 as an answerView attachment 276915View attachment 276914
I tried two approaches but I got wrong answers. I don't really know which formulas to use. Also, were studying a book by Boylestad and Nashelsky called "Electronic devices and circuit theory" if that information helps I guess.
The problem is that you are just wanting to through numbers at equations instead of analyzing the circuit. As a result, you are using (or rather misusing) a slew of equations with little rhyme or reason.I tried two approaches but I got wrong answers. I don't really know which formulas to use. Also, were studying a book by Boylestad and Nashelsky called "Electronic devices and circuit theory" if that information helps I guess.