Capacitor equation to be used in electrical analysis

[PhysicsTest]

TL;DR

I am confused with the capacitor equation to be used.

I am confused with the equation to be used for capacitor in electrical analysis
The standard equation we have is Q=CV -> 1
the other equation is is V = Z*I ohms law Z is the impedance of the capacitor. Both are giving me voltage, which one to use ?

 

[Baluncore]

Both are giving me voltage, which one to use ?

That depends on how you got there, and if you know Z or Q.
It also depends on the type of analysis, as an AC circuit or as a charge pump.

 

[Svein]

In an electronic system you mostly use Z_{C}=\frac{1}{\omega C}.

 

[Baluncore]

In an electronic system you mostly use ...

But capacitive reactance has a negative sign.

 

[Svein]

But capacitive reactance has a negative sign.

Yes... actually the full expression is Z_{C}=\frac{-i}{\omega C} to indicate the full phase angle. I just gave the amplitude in my answer.

 

[berkeman]

I am confused with the equation to be used for capacitor in electrical analysis
The standard equation we have is Q=CV -> 1
the other equation is is V = Z*I ohms law Z is the impedance of the capacitor. Both are giving me voltage, which one to use ?

You forgot one...
$$v(t) = \frac{1}{C} \int i(t) \,dt $$

 

[DaveE]

Have you studied calculus yet?

 

[PhysicsTest]

Have you studied calculus yet?

Yes

 

[DaveE]

These are all just variations on one basic equation, depending on how you want to approach your problem.
$$ q(t) = C⋅v(t) \Leftrightarrow \int i(t) \, dt +Q_o = C⋅v(t) \Leftrightarrow \int i(t) \, dt = C⋅(v(t) - V_o) \Leftrightarrow i(t) = C⋅\frac{dv(t)}{dt} $$

For AC analysis where $i(t)$ and $v(t)$ are limited to sinusoids, or for more sophisticated analysis with Laplace transforms, we can use the concept of complex impedance, where $v = Z⋅i$, with $Z( \omega )= \frac{1}{j \omega C}$, or $Z( s )= \frac{1}{sC}$. In these cases $\omega$ must be well defined.

edit: rearranged things to deal with the integration constant.

 

[DaveE]

BTW, it's the same for inductors. If you just swap $i(t) \leftrightarrow v(t)$ and rename the constants, you get this:

$$ \int v(t) \, dt = L⋅(i(t) - I_o) \Leftrightarrow v(t) = L⋅\frac{di(t)}{dt} $$

and $Z( \omega )= j \omega L$, or $Z( s )= sL$.

There is no common name for the equivalent to charge. But if you talk with people that do a lot of magnetics design, they'll mention volt⋅seconds often.

 

[LvW]

But capacitive reactance has a negative sign.

I think we should be more exact (to avoid confusion at the beginners side) : The capacitive reactance is not negative. However, it causes a phase shift of 90 deg between voltage and current.
Therefore, we use the imaginary notation and write Zc=1/jwC=-j/wC.

 

[PhysicsTest]

I keep trying to solve the following circuit with equations but i always fail and never able find a final solution, i am trying again now let me see if i can finish

Now i plan to solve for I1, I2, I3 check the frequency response, am i in correct direction?
$Z_c = -\frac{1}{j\omega c}$

 

[Baluncore]

The capacitive reactance is not negative.

X_L = w·L
X_C = -1 / w·C
X = X_L + X_C
Z = R + j X
Resonance occurs when X_L + X_C = 0; Not when; X_L = X_C .

 

[LvW]

X_L = w·L
X_C = -1 / w·C

Sorry, but I cannot agree. Can you explain WHY you wrote "X_C = -1 / w·C" ?

I think, we should write
Z_L = jw·L and |Z_L|=X_L=wL
Z_C = -j / w·C and |Z_C|=X_C=1/wC

That means "1/j=-j" indicats just a phase shift according to exp(-jPi/2)=-j.
In contrast, a minus sign is equivalent to a phase shift of 180deg.

 

[Baluncore]

https://en.wikipedia.org/wiki/Electrical_reactance#Capacitive_reactance
"There are two choices in the literature for defining reactance for a capacitor."

 

[DaveE]

Sorry, but I cannot agree.

Maybe because you are speaking of impedance, he is speaking of reactance?

 

[DaveE]

I keep trying to solve the following circuit with equations but i always fail and never able find a final solution, i am trying again now let me see if i can finish
View attachment 329618View attachment 329622

View attachment 329623

Now i plan to solve for I1, I2, I3 check the frequency response, am i in correct direction?
$Z_c = -\frac{1}{j\omega c}$

You have some sign errors in eq3.

 

[PhysicsTest]

Thank you i have updated to
$I_1e^3 + I_c(Z_c+31e^3) -31e^3I_3 + (V_{ref} -V_1) = 0$

I will solve for equations and frequency response, i am not sure if i can solve manually or i need to take help of tool.

 

[berkeman]

Thank you i have updated to
$I_1e^3 + I_c(Z_c+31e^3) -31e^3I_3 + (V_{ref} -V_1) = 0$

I will solve for equations and frequency response, i am not sure if i can solve manually or i need to take help of tool.

Sorry to be dense, but what is $e^3$ -- what are you cubing?

 

[PhysicsTest]

Sorry to be dense, but what is $e^3$ -- what are you cubing?

I am trying to represent K in e3 instead of 1000.

 

[berkeman]

I am trying to represent K in e3 instead of 1000.

Yikes. It's best not to make up your own notation, especially when asking for help from others. It is also better not to substitute numerical values into equations until you have arrived at your solution with variable names. That's when you substitute in the numbers (with "k" = 1000) and generate your numerical answer.

 

[DaveE]

Pro tip - This is really important IMO!
Don't formulate your equations with numerical values for the constants like resistors or voltage references. Everything should be either a variable or a constant initially. Then do as much algebra as possible to find a simplified form BEFORE you put the numbers in.

So, for your first equation, I would have written $I_1R1+I_cZ_c+I_3R_4+V_2-V_1=0$.

It may seem like more work at first, but there are many advantages:

1) Everything has a unique name. No one will be confuse as when you say 'what if we change the e3 resistor value?' It is common for circuits, like yours, to have several components with the same value.

2) If someone gives you the same circuit but with different values, you will likely have to start the analysis again at the beginning instead of simply recalculating the final result.

3) Algebraic simplifications with numerical values can obscure the more general result. Very often there are relationships between the circuit components that are hard to see if everything is shown as a specific number. For example, if I say the gain of an amplifier is $Av=(1+\frac{R1}{R2})$ you can immediately see that it is the ratio $\frac{R1}{R2}$ that matters more than just the value of $R1$. In the other scenario the result of your calculation may be $Av=11$ which tells you nothing about that type of circuit, it's just your specific case.

4) This process will help you over the long run to recognize typical forms, common circuit elements, and help you partition circuits into simpler groups for analysis. You can often reuse your analytical work, perhaps by memorization, when you see a common circuit. For example, if you removed R1, C1, and R4 from your circuit I could simply write down the formula for the output voltage from memory because that is a very common application. I don't think I would have learned that if every time I solved it it was a mass of numbers in the equations.

5) Circuit analysis can often be simplified by approximations. For example $1M\Omega + 4\Omega \approx 1M\Omega$. This looks simple with the numerical values, but it doesn't show you when you can and can't simplify things and doesn't allow you to relate those simplifications back to the schematic that generated them.

6) Putting in numbers early destroys your ability to check your work with dimensional analysis. For example, if you tell me that your circuits time constant is $\tau = \frac{C1C2}{C1+C2}⋅\frac{R1+R2}{R4}$ I can immediately see that you are wrong, I don't even need to see the circuit or the equations. This has units of capacitance, not seconds as it should. [note that this assumes we are using the naming convention for circuit elements that includes the basic units, i.e. you never name a resistor $C6$, or a voltage source $I3$].

7) Putting the numbers in early is usually an irreversible process. It is hard to see the reasoning in a derivation or to go back to find and fix errors sometimes. Of course if you do need a numerical answer at any point you can just put the values in and work your calculator. You often can't do that in reverse.

8) In the real world, engineers have to check their own work to see if they made a mistake. Because everyone makes mistakes! There is no TA with a red pen. Your boss doesn't want to, and probably can't do it. If your boss does have to check everything you do, what value have you added to that company? Who gets laid off first? Several of the things I've previously mentioned make it hard to verify BY YOURSELF that you have done the analysis correctly. Always check your own work, this is a skill that isn't taught or practiced enough in schools IMO.

 

[PhysicsTest]

Very challenging equation to solve but arrived at the following equation for I1 current

I will try to plot frequency and see if i am in correct direction.

 

[PhysicsTest]

With LTspice i tried to simulate only one portion of the schematic

The filtering happens only at $f=\frac{1}{2\pi rc}$ I substituted r = 2K, C = 56PF
f = 1.4MHz. Why do we need to attenuate at this frequency?

 

[Baluncore]

Why do we need to attenuate at this frequency?

The circuit is a low-pass filter. It attenuates higher frequencies. That keeps RF and EMI out of the op-amp. What is the gain⋅bandwidth product of the op-amp used?

 

[PhysicsTest]

The opamp is MCP6021 opampdatasheet
The data sheet shows 10MHz. Is it related with 1.4MHz?

 

[Baluncore]

GBWP = 10 MHz;
Circuit gain = 15 ;
BW = 10 MHz / 15 = 667 kHz.

 

[berkeman]

With LTspice i tried to simulate only one portion of the schematic

Which analysis type did you use this time?

What variables did you plot?