First order networks are passive networks which includes a single element capable to store the energy (C or L). Let me try to derive the equations which will help us to understand the I/V characteristics with time.

**RC network, C is connected to the ground (low-pass RC filter)**

Assume that input voltage instantly has changed from V0 to Vin, we can write:

Which can be rewritten as:

This is an usual first order differential equation, solving now!

I should mention, I am not sure in the proper naming – not my native language, the university math lectures and practice are almost gone from my head, so I might make mistakes here and there. But I, at least, always trying to make things simple.

So back to the math, I remind that the solution of the differential equation looks like:

where xp is particular solution and xh is a general solution.

solving for xp, we can rewrite the equation to:

We can try to “guess” the solution:

dVin/dt is 0 for the initial moment of time, so looks like my guess is worked out.

So we have xp, now we need to figure out how to solve for xh:

the general solution should looks like:

If we substitute such view to the equation we will have:

From where:

Then the total solution will look like:

Thought we still have here unknown A, how to find it? Actually, kind of easy, just remember that in the initial moment of time t=0, the voltage is equal to whatever settled voltage V0, then:

Now we can insert everything to the main equation:

Thus, we got the voltage equation for the low pass RC filter, but we still need to derive the current. Just refresh memory about the current equation which is CdV/dT, so:

Now we are ready to plot some waveforms, which are depends on the initial voltage value.

1. Assume that V0 = 0, VIN = 5, C = 1uF, R = 1 kOhm

The current has a maximum value from very beginning and decreasing while the voltage is increasing and becomes 0 when the Vin value is reached.

If one would substitute t in the voltage equation by RC, then he could make sure that voltage level at this time will reach 63.2% from its high value, after this the charging will be slowed down and to get to the 99% we would need 5RC intervals.

2. Now high to low transition, V0 = 5, VIN = 0, C = 1uF, R = 1 kOhm

The current behaves exactly like in the previous case, just has a different sign. If we would again put t=RC then we could see that voltage has decreased to 36.8% from its initial value, i.e. the discharge happening faster than the charge.

**RL network, the inductance is series connected (low pass filter)**

Refreshing memory for the inductor:

and gaining the understanding that we should use not a KCL, but the KVL:

Solving:

P:

H:

But i0*R = V0, then the solution will look like:

1. if V0 = 0, VIN = 5, L = 1 mH, R = 1

2. if V0 = 5, VIN = 0, L = 1 mH, R = 1

The waveforms really similar to previous one, although current now follows the voltage, and this is kind of expected since we are measuring it on the resistor and Ohm’s law is still working, yup.

**RC network, ac coupling (high pass filter)**

This one somehow was quite tough for me, let me try to elaborate. Firs of all, the trick here is to examine changes of voltages rather than try to solve for dc values, it wont be helpful at all.

Now let me write the KCL:

Where:

Rewriting this in a common for differential equation view:

and lets solve it for this delta V

In the initial time t=0, and delta is 0, so A = Vin, then:

BUT, this equation gives the voltage across the capacitor, while we want the output voltage:

Again, we don’t have dc Vin now, we rather have the change of the voltage.

1. dVin = +5 В, R = KOhm, C = 1 uF:

Again, the current is repeating the voltage on a resistor. But the voltage reaching the max value immediately.

2. dVin = -5 В, R = KOhm, C = 1 uF:

**RL network, the inductance in the load (high pass filter)**

Following the KVL :

rewriting it a little:

P:

H:

T:

rewriting it once more:

but i*r is V, so we have V on the left side and for t= 0:

And the final equations:

1. L = 1 mH, R = 1 Ohm, V0 = 0, Vin = 5

1. L = 1 mH, R = 1 Ohm, V0 = 5, Vin = 0

OK, I also made a summary picture for convenience:

I’d like to add a couple interesting cases with the current source at the input:

**RC network, the capacitance in the load**

So the current is I = CdV/dT

But we have a source, so current is predefined, plus the KCL says that it will be equal to the one passing thru a resistor and the same limited current will be used to charge a capacitor. Lets pull the voltage from the equation:

Which is solved by It/C and which is just literally straight line with a slope regulated by I/C, just giving the simulation result.

Looks correct.

**RL network, the inductor in the load, the current source is a driver**

The output voltage is coming straight out from the equation:

I.e. the output voltage is proportional to the rate of the current change and will be equal to 0 when current is constant.