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I don't understand, how did you calculate the 75.32 ohm
29V/75.32ohm = 385mA
but why 385mA
If the load seek max 350mA and the zener needs min 5mA,
you don't want 29V/355mA = 81.69ohm instead of the 75.32ohm series resistance
I mean where did you find the 385mA, did you allow more current...
Does this schematic is right?
If yes tell me how in that schematic: the C.T. can be use as a 0V reference voltage.
It's really confusing me, there's not suppose to be voltage on the C.T. if my first draw is good?
thanks
Gauthier
Yeah Retrotechie your right on that it's 15Vrms ,but the 1000uF cap will charge at Vpeak, so 15Vrms/.707 = 21.2 Vpeak
21.2V-.7V = 20.5V D.C.
What you mean?
Thanks for your explanation
Gauthier
Does someone would be able to explain me how the current and voltage are distributing in the schematic ....
I tryed by myself for 2 hours, but I'm clueless it gives me impossible results. The center tap is confusing me and R1,R2 too...
It's bugging me because it's my final project at school...
We used standard electrolytic capacitors, , those that explode in reverse polarity!
Maybe those one can tolerate a short and small voltage reverse polarity, unlike tantalum electrolytic one???
Exactly, that's the job of D1,D2,D3,D4.
Yes it is.
It's a drawing error I made the correction in...
At school I had to build this circuit on a protobaord, I started from the left to the right testing it between each layer. Everything was fine until I plugeg the 10uF capacitor on the negative side of the power supply.
Both 10uF capacitors are there to filter the noize that the potentiometers...
But IanP, me too it gives me .0010211897 to be exact , but anyway we have both the right voltage divider formula. Mine is good i'm sure, I used it like 50 times in the past month...
We argue on impossible values... but I think it's important to understand why these values are impossible ...
This is the voltage divider formula: Vx = (Rx/RT)Vs
Vx= the voltage drop at the motor
Rt= Total resistance
Vs= Voltage source
but just like srizbf told you I think that with this resistance your values are not possible.
Imagine that you have the biggest load ! The biggest load would be no...
yeah the thing about the cancellation law was for the string above: loge(Vx/Vs) = loge(e^(-t/RC))
si I puted exposents every where it's my bad lol, I didn't find the real base e on my keyboard..
but tanks for giving you time Len, I learned coples of very interesting things.
OK I watched a coples vids about logarithms on youtube...
so,
Vx= Vs(e(-t/RC))
Vx/Vs = e(-t/RC)
ln(Vx/Vs) = ln (e(-t/RC))
ln = log base e of x
loge^(Vx/Vs) = loge^(e^(-t/RC))
Using the cancellation laws I can keep only the last exposent at the right side.
loge^(Vx/Vs) = -t/RC
RC ln...
Hi,
Let me give you the problem that bring me to that question,
First, I have 24 Vcc source that aliments R1=62KΩ in series with C1= 1000µF, very basic circuit...
The first question asks me to calculate the voltage drop at C1 and curent in the circuit at intervals of 30 secondes and to trace...
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