Simple RC Hardware Circuit Understanding -- Questions

In summary, this circuit has problems with drawing the correct schematic and the incorrect placement of voltage sources.
  • #36
PhysicsTest said:
But it will always be 2.5V since the resistors are equal and by voltage divider rule, it shall be 2.5V.
The Thevenin resistance, Rth, of the 2.5 volt reference is 4k7 / 2 = 2k35 . That Rth is in series with the 1k0 of the LPF, making 3k35 total. If you ignore Rth, the transfer function will be incorrect.

You must close the circuit from ground of the LPF, to the ground of the bias network.

I still do not know what you are doing, nor why.
 
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  • #37
Baluncore said:
I still do not know what you are doing, nor why.
The main reason why i am trying to do is, to derive an expression for Icts as a function of resistors and Vout. Develop firmware to use this formula calculate and display the Icts (additionally convert this into phase current) on lcd. Is my approach correct?
 
  • #38
Hi @PhysicsTest...
Now that you have defined the components and have an easy-to-read schematic, etc, let's try breaking the circuit down to the DC component and then add in the AC component.

We will be making a few assumption along the way, and later add in corrections to cover the assumptions.

For the first part let's try to find the DC voltage at the A/D converter input. We can ignore the CT and the capacitors for this, just assume they do not exist or are open circuits.

What IS needed is the input impedance of the A/D converter. We may decide later that this can be ignored if it is much higher that the circuit resistance values, but we don't know that yet!

Next would be to draw a schematic with the Voltage Source, the Resistors, and the Input Impedance of the A/D converter. From that you can calculate the 'Zero' reading at A/D input.

Now you get to make a decision. Is the calculated DC voltage at the A/D input enough different from the 2.5V 'Ideal' voltage to really matter? If you want to end up with the Complete design equations, this step is complete and you can ignore the next paragraph

For instance, if the voltage difference between using the A/D input, and assuming it is an open circuit is less than the A/D resolution, then consider the A/D as an open circuit and use the 2.5V divider voltage. (In the real world, there may be a variable resistor in the R2 - R3 divider network to trim the Zero voltage.)

Now consider the situation with the CT connected. The CT will have some internal resistance which is in parallel with the 10 Ohm resistor. The equivalent parallel resistance can be substituted in the equation derived above for the 'Zero' DC input to the A/D converter...
Or, again, you may want to substitute the equation for the parallel resistances into the DC equation derived above.

That completes the basic DC bias point calculation. If this was a design for producing many of these circuits, things like the resistors tolerances and maybe their resistance variation with temperature would also be accounted for.

Sorry for hijacking this thread, after you get the DC bias point completed, I will bow out and let the others here carry on their fine support.

Cheers,
Tom

p.s. Remember details count, and the more of them you supply the easier it is for everyone. (except maybe you when you have to type so much :wink:)
 

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