Ward 6 - page 31

New comment
 
This is how I propose to forget about the indicator. And consider the problem itself. We want to arrange separately set time constants for charging and discharging the capacitor. it's time to write some equations. Which will explicitly connect the input and output potentials. And then code. Where are the equations?
 

Forgive me for interrupting, I think the following formula works for you

Ut =Uo EXP [-t/(R*C)


where t is time in seconds

C is the capacitance of the capacitor

Uo - voltage at the coils in the initial moment

R - resistance

Ut - voltage on the terminals at the final moment

EXP-exponent of 2.712

 
M_Dimens:
A second resistance must also be added to this somehow.
 
M_Dimens:

EXP - exponent of 2.712

I don't know what you mean by "exponent at 2,712", but I would start by saying that we need volt-ampere characteristics of our diodes :-)))

That's not to mention the fact that you seem to intend to charge a capacitor, I gather, and the kind of function you propose is somehow inconsistent...

 
DmitriyN:
A second resistance must also be added to this somehow.
Forget about the second resistance for now. Write as if the VD2-R2 branch were not present. Just at the input potential Phin(t), then through V1-R1-C to the conductor with zero potential, and the question, what is the potential difference at the capacitor coils, Phoet(t). The parameters of the problem are: R1, C, and the WAV of the diode (what did you want?). In our trader-unphysical model we can make diode's WAV even linear. Or parabolic. Or as we want. :-))) I propose to think not in "pips" and "bars", but in volts and seconds. Let's say 1.55 Volts instead of $1.55 per pound :-)))
 
Dr.Drain:

I don't know what you mean by "exponent at 2,712", but I would start by saying that we need volt-ampere characteristics of our diodes :-)))

That's not to say that you seem to be going to charge a capacitor, I gather, and the kind of function you propose is somehow inconsistent...


diodes are not right, they have non-linearities, it's a threshold element

(or if you take a perfect rectifier then it's different)

imho it's better to charge the capacitor via a controlled switch (opens by trigger condition)

 
M_Dimens:


But it's a mistake with diodes because they have non-linearities.

We'll have the diodes as we set them. At least these diodes will be:

We can read it from a separate file or calculate it using a formula. If we want non-linear I-V curve - you're welcome. Our formulas do not depend on it. This is just a task parameter. Our filter will work differently with different V waveforms.

 

Maybe someone likes it better that way, how do I know.

But there has to be an exponent, not the point.

 

Or even so.

in any case, our filter should require this characteristic explicitly.

 
Maybe don't talk about diodes, they don't seem to have anything to do with it. Let them be perfect - only one way current. It's all about the difference in resistance of the resistors.
1...242526272829303132333435363738...78
New comment
Reason: