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Hello. And welcome to the TI Precision Lab series on passive components and circuits. Today we'll be discussing RC circuits. RC circuits are composed of resistive and capacitive elements that can be formed in one of the two ways shown here.

RC circuits can be both intentionally designed and occur parasitically. Because they are so commonly seen with op-amps, understanding their function is important for analog designers. In this presentation, we will observe both the transient and AC response of RC circuits.

Recall that a capacitor stores energy in the forming of an electric field created by the charge on two metallic plates. When a change in voltage is applied to a capacitor, current flows through the capacitor, and the stored charge is changed.

The same is true for an RC circuit. When a step voltage is suddenly applied to a discharged RC circuit, current through the circuit causes the capacitor to charge at an exponential rate. This is referred to as the step response of an RC circuit.

Similarly, when an RC circuit with stored charge is suddenly tied to ground, a current is induced in the capacitor, which causes it to discharge. This is referred to as the natural response of an RC circuit. For these equations, V sub c refers to the voltage across the capacitor at a given moment in time. V sub s refers to the supply voltage.

V sub i refers to the initial voltage across the capacitor. And t refers to time. Note that both responses are exponential and depend on the product of the resistance and capacitance. We refer to this important product as the RC time constant of the circuit and represent it with the Greek letter tau.

To gain a better understanding of these equations, consider these plots for a charging RC circuit. These plots were obtained through a TINA-TI transient analysis. Note that at time 100 microseconds, a sudden step voltage is applied. This causes the voltage across the capacitor to rise exponentially from its initial voltage of 0 volts and asymptotically approach the final voltage of 1 volt, as predicted by the step response equation.

The current through the capacitor at any given point in time is proportional to the derivative of the voltage across the capacitor. A similar SPICE analysis can be run for a discharging RC circuit. Here, an initial voltage of 1 volt is applied to the capacitor.

Then at time 0, the RC circuit is suddenly switched to ground. The voltage across the capacitor decays exponentially as expected. And the current through the capacitor corresponds to the change in voltage. Note that, in this case, the sign of the current direction is flipped because the current is set to flow in the opposite direction.

Regardless of the values of the resistor and capacitor in an RC circuit, the response will have the same exponential form. However, the resistor and the capacitor will determine the time it takes the capacitor to approach its final voltage level. This time is often measured in RC time constants, which are represented by the letter tau.

For example, a charging RC circuit will take one RC time constant to reach 63% of its final charge value, three time constants to reach 95% of its final charge value, and five time constants to reach 99% of its final charge value. Time constants can similarly be used to determine the time an RC circuit will take to discharge.

Let's now look at some examples of this in TINA-TI. Let's first simulate a charging circuit. The TINA circuit is embedded in the presentation for your convenience.

First, open the TINA circuit. Note that the supply voltage is set to emulate a step function with an amplitude of 1 volt and a 100-microsecond delay. Next, click Analysis and Transient to set up the simulation. We will run the sim for 1 millisecond.

Press OK to run the sim. We can now see the transient response of the circuit for a 1-volt step function. For this simulation, we used a resistor of 10 kiloohms and a capacitor of 10 nanofarads. Multiplying these two together, we get a time constant of 102 microseconds.

From before, we know that a charging RC circuit should reach 95% of its final output voltage after three time constants. To find this point, we take the beginning of the step function at time 100 microseconds and add 300 microseconds. So we expect to see 95% of our final output level at 400 microseconds, and similarly, 99% of the final output at 600 microseconds.

Considering that our initial voltage was 0 volts and our final voltage is 1 volt, we expect to see 950 millivolts across the capacitor and 990 millivolts across the capacitor at these points in time. This can be verified with a cursor in TINA-TI.

The transient response of a discharging RC circuit can be found in the same manner. This time, the voltage supply is disconnected, and an initial condition of 1 volt is placed across the capacitor. The circuit is now tied to ground.

The TINA file for the circuit used is embedded on this slide. The natural response of the simulated RC circuit shows an exponential decay asymptotically approaching 0, as expected. After three time constants, or 300 microseconds, 95% of the original charge is lost, and the voltage level has fallen to 5% of the initial voltage. In this case, that would be 50 millivolts. Similarly, allowing five time constants or 500 microseconds to pass will leave only 1% of the original voltage across the capacitor.

To conclude our discussion on the transient response of an RC circuit, we provide a real world example. In general, RC filters are placed at the output of op-amps and before certain types of ADCs, notably SAR ADCs. The RC circuit acts as a charge bucket, storing charge in an electrical field to provide current to the ADC during the acquisition phase.

This is critical to ensuring an accurate and fast conversion of an analog signal to a digital signal. And more accurate ADCs will require more time constants for the capacitor to settle. This topic is covered in much more detail in the TI Precision Lab series on ADCs.

Having covered the time domain, let's now move on to the frequency domain. Recall that a resistor has an impedance that is independent of frequency, while a capacitor has an impedance that is inversely proportional to frequency. Analog engineers make use of these properties to design frequency filters with RC circuits.

Let's begin by discussing the low pass filter. A low pass RC filter is created by placing a resistor followed by a capacitor down to ground. The output is taken from between the two components.

An intuitive understanding of this circuit's AC behavior can be obtained by considering the behavior of the capacitor at low and high frequencies. From the previous presentation on passive components, we know that at very low frequencies, approaching DC, a capacitor will act as an open.

If this is done for the low pass filter, the low frequency signal will effectively see an open circuit. Current will stop flowing, and the input voltage will appear at the output. Thus, the gain will be of 0 dB.

On the other hand, a capacitor will act as a short circuit at very high frequencies. In the case of the low pass filter, this will effectively short the output to ground. Thus, the gain will approach negative infinity on a decibel scale.

The behavior of the low pass filter can be understood using the equation of its pole. At low frequencies, where the capacitor acts as an open circuit, the filter has a gain of 0 decibels. But at higher frequencies, the capacitor has a lesser and lesser impedance, causing the gain to drop by 20 decibels per decade.

The key transition point between these two regions is referred to as the pole. At the pole frequency, the gain of the circuit reaches minus 3 dB and begins to fall rapidly. Being able to determine this frequency is key to a successful filter design.

For this RC circuit, the pole can be found using the equation f sub p equals 1 over 2 pi RC, where f sub p is the pole frequency. For a low pass filter of 10 kiloohms and 10 nanofarads, we expect to see a pole frequency of 1.59 kilohertz. Poles and their plots are covered in much more detail in the TI Precision Lab series on op-amps.

Let's now validate our analysis by running a TINA-TI simulation. The TINA file used is embedded in this presentation for your convenience. After opening the file, we can now run an AC sim. This can be done by clicking Analysis, then AC Analysis, and finally, AC Transfer Characteristic.

The simulation settings box will then pop up. We will sweep our input signal from 10 hertz to 1 megahertz. When you're ready, hit OK to run the sim. On the right, we can see our simulation results.

Indeed, the overall response of the circuit matches our expectations. There is a low frequency segment of 0 dB of gain, followed by a pole frequency and a region of negative 20 decibels per decade of gain. True to its name, the low pass filter passes signals of lower frequency and attenuates those at higher frequency. Using TINA's cursor function, we can place a cursor on the minus 3 dB point to double check our calculation and indeed find that our pole comes in at 1.59 kilohertz.

The complement of the low pass filter is the high pass filter. This can be created by simply flipping the position of the capacitor and resistor. Given that they look so similar, it can be hard to remember which is a low pass filter and which is a high pass filter.

To aid your memory, simply look at the location of the capacitor. If the capacitor comes first in the signal path, it's a high pass filter. If it comes second, then it's a low pass filter.

The high pass filter has the opposite behavior of the low pass filter. At low frequencies, the capacitor acts as an open, such that the output is disconnected from the input and pulled down to ground. The gain approaches negative infinity on the decibel scale.

At high frequencies, the capacitor acts as a short circuit and shorts the input to the output. In this scenario, the gain approaches 0 dB. Just as a low pass filter has a critical frequency called the pole, the high pass filter has a critical point called the zero.

Prior to this frequency, the circuit attenuates the input. As the zero frequency's approached, the attenuation is lessened by 20 dB per decade. The zero frequency has a gain of negative 3 decibels, and the frequencies above this point show a gain of about 0 dB.

Note that the zero frequency equation is the same as the pole frequency equation. However, we now use f sub z to denote the zero frequency. For this high pass filter, we use the same component values as before. However, we now expect to see a 0 at 1.59 kilohertz rather than a pole. 0's and their plots are covered in much more detail in the TI Precision Lab series on op-amps.

Let's again validate our analysis by running a TINA-TI simulation. Download the embedded TINA file and open it. Run an AC sim as before by clicking Analysis, then AC Analysis, and AC Transfer Characteristic.

The simulation settings box will pop up. Sweep the input signal from 10 hertz to 1 megahertz, and hit OK to run the sim. Again, the overall response of the circuit matches our expectation. The high pass filter attenuates signals at low frequencies and allows the higher frequency signals to pass.

Using TINA's cursor function, we can place a cursor on the minus 3 dB point to double check our calculation and indeed find that our 0 comes in at 1.59 kilohertz. That concludes this lesson. Thanks for your time, and please try the quiz.

Question 1, fill in the blank-- the natural response of an RC circuit involves a blank capacitor and is described by the equation blank. The correct answer is b. The natural response of an RC circuit refers to its ability to discharge and is described by the equation v sub c equals v sub i times e to the minus t over RC.

Question 2, assume a capacitor in an RC charge bucket has no charge stored and begins at 0 volts across its nodes. Suddenly, a step voltage is applied. How many time constants will it take to reach 95% and 99% of its final voltage level, respectively? The correct answer is d. It will take three and five constants for the capacitor to reach 95% and 99% of its final voltage level, respectively.

Question 3, true or false-- the circuit on the left is a high pass RC filter, and the circuit on the right is a low pass RC filter. The correct answer is true. Remember that for a high pass filter, the current first passes through the capacitor and then the resistor, while the opposite is true for the low pass filter.

That's all for now. Thanks again for your time.