Simple RC Hardware Circuit Understanding -- Questions

[DaveE]

OK, I didn't want to keep beating this horse but I guess I'll have to say it again. Define your terms. Be consistent.

I don't know what $V_{R2}$ is, so I don't know why you said $V_{C1}=V_{R2}$. In fact I don't even know what $R2$ is because you posted two schematics with different names for the resistors.

It would be conventional for $V_{R2}$ to be the voltage across $R2$ with a polarity clearly defined in a schematic (which isn't there, yet). That would be either with a current direction clearly defined or the voltage polarity indicated. Look into "passive sign convention" for more about that. So, for example, you did a good job of telling us that $i_2$ flows through $R1$, but you didn't tell us the direction, so I can't say if your eq 1.1 is correct or not.

Same comments for $V_{C1}$ , $i_1$ , everything, really.

In this case I can't even really guess what you actually meant, because I don't think $V_{C1}$ is equal to any other branch voltage. So there might be a problem there. However, there are also problems with your other equations, but I don't think it's really worth commenting on those until we have a bit more structure in your approach, and communication.

The thing about circuit analysis (using KVL, KCL, etc.) is that the solution is relatively easy, even sort of automatic when you go through all of the steps. The hard part, the part the messes everyone up, is that they don't define all of the voltages and currents well. They try to skip steps, which is often quite tempting. This an analysis technique where how you set up and define the problem is by far the most important part. Khan Academy has some really good tutorials about circuit analysis. I would suggest watching some of those next.

If you want a STEM career, like in physics or EE, you need to learn to define and communicate the details clearly. Slow down, be more careful.

 

[Baluncore]

It also helps if currents have a direction marked with an arrow.

 

[PhysicsTest]

Thank you for all the help, i know i am confusing all, circuit analysis is always tough for me. I am doing another attempt

Loop1 : Consisting of elements C1 and R1
$\frac{-\int I_1dt} {C1} + I_2R1=0$ -> eq1
Loop2: consisting of 2.5V, R1 and R4
$2.5 - I_2R_1-(I_1+I_2+I_{cts})R_4 = V_{out}$ -> eq2
$I_{cts} = I_0\sin(\omega t)$
I need one more equation, Please help me if the equations are correct and where do i apply the next equation?

 

[DaveE]

Sorry, too busy to write a real reply. But that node labeled 2.5V is only 2.5V if there's no current being drawn through those 47K resistors. You can't leave R2 & R3 out, also C2 needs to be included. Those parts are in loops you haven't included yet.

But, good work on defining things, now we can all speak the same language.

 

[PhysicsTest]

But that node labeled 2.5V is only 2.5V if there's no current being drawn through those 47K resistors. You can't leave R2 & R3 out,

But it will always be 2.5V since the resistors are equal and by voltage divider rule, it shall be 2.5V.

also C2 needs to be included. Those parts are in loops you haven't included yet

Loop1 : Consisting of elements C1 and R1
$\frac{-\int I_1dt} {C1} + I_2R1=0$ -> eq1
Loop2: consisting of 2.5V, R1 and R4
$2.5 - I_2R_1-(I_1+I_2+I_{cts})R_4 - \frac{\int (I1+I2+I_{cts})dt} {C_2} = 0$ -> eq2
$I_{cts} = I_0\sin(\omega t)$ ->eq3

The current through $R_2 = \frac{2.5} {4.7k} = 0.53mA$ ->eq4
Similarly current through $R_3 = \frac{2.5}{4.7k} = 0.53mA$ ->eq5
$I_1 + I_2 = 1.06mA$ ->eq6
Can i use the above equations to solve for Vout?

 

[Baluncore]

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.

 

[PhysicsTest]

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?

 

[Tom.G]

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 )