Feb 13: Capacitors (High Pass/Low Pass Filters)

Reactance of a Capacitor
-Consider a circuit with a time varying source and a capacitor (without a resistor).
-Using the definition of capacitance and differentiating, we obtain I = C*dV/dt, which we can then show that there is a 90 deg phase shift between V(t) and I(t), with the ratio of voltage amplitude to current amplitude of 1/ωC. 
-Recall that for resistors driven by a time varying source, using Ohm's Law, we found that V(t) and I(t) were in phase (0 deg phase shift) with the ratio of voltage amplitude to current amplitude of R. Thus, 1/ωC can be considered as being analogous to the resistance. This quantity 1/ωC is the capacitive reactance.
-Note that ω = 2*pi*f
-In many ways, capacitors behaves as a resistor (a freq-dependent resistance of 1/ωC). Thus, at high frequencies (which is like having a "resistance" of 0), capacitors start to look like a "short circuit," whereas at low frequencies (which is like having an infinite "resistance"), capacitors start to look like an "open" circuit.

An RC differentiator, experimentally

-Only works at low frequency regime ... after that range, the circuit behaves as just a regular RC circuit. (voltage across resistor << voltage across capacitor)

-We were able to see this as we manually varied the frequency of a square wave from 500 Hz to 100 kHz where we noted the behavior of the "differentiator" circuit for cases where ω << 1/RC was no longer satisfied. (1/RC = 1/(100k Ohms*100pF) = 100 kHz). The frequency we measure and can adjust is f, which is ω/2pi, so ω was still relatively large (not << 1/RC). 

(square wave output voltage at 100 kHz)

-After constructing the RC differentiator, we drove it with a square wave with a frequency of 100 kHz and 1 Vpp amplitude. We then tried the triangle wave and sine wave.

-We expected this output voltage (vs. time) with peaks having a width ~RC. The circuit is unable to differentiate the high frequency components of the square wave and as a consequence of the time lag by the RC circuit, we see these peaks with widths. 

As the name suggests, we saw that the input voltage function was differentiated:

 


*Side Note: Because we saw some significant oscillations on the output voltage on the oscilloscope, we used scope probes instead of regular coaxial cable. After the replacement, we saw significant decrease in these oscillations: (before and after)

Input Impedance

• Input Impedance at DC (impedance to get to ground) = infinity (f = 0 --> ω = 0 --> z_c = 1/0 = infinity)

• Input Impedance at infinite frequency = resistance of resistor (100 kOhms) because impedance of capacitor = 1/infinity = 0 (looks like a wire)

Filters

Low Pass Filter

-Circuits containing capacitors behave differently as frequency changes (we can think of capacitors as frequency-dependent resistors).

-Then, we can see this circuit as being a type of divider.


-Similar to a voltage divider with two resistors:

• If 1/ωC >> R, then Vout is approx equal to Vin. 

This is because if R is very small compared to 1/ωC , the voltage drop across R is negligible (compared to the Vin or Vp value), making Vout close to Vin. 

• If 1/ωC << R, then Vout << Vin 

This is because if 1/ωC is very small compared to R, the voltage drop across R is significant, making the voltage drop across 1/ωC negligible (or much less than Vin or Vp).

Conclusion: Thus, for low frequencies (low ω), Vout is approx equal to Vin, and for high frequencies, Vout << Vin, making this circuit a LOW PASS FILTER. It lets through low frequencies while high frequencies are attenuated. 

Complex Impedance (Still about Low Pass Filter)

-Using complex numbers to characterize impedance (because it has both a magnitude and a phase), the reactance of a capacitor can be written as -i/ωC. Then, just as before (using resistance), Vout = (X/(R+X))*Vin, where X is the reactance. 

-Using the relative amplitudes of the signals, |Vout|/|Vin|, we see that if ω << 1/RC, then |Vout|/|Vin| approx = 1, whereas if ω >> 1/RC, |Vout| << |Vin|.

-We see that the cutoff frequency (threshold) of 1/RC roughly shows the boundary between frequencies that pass or don't pass.

-Also, using Vout = (X/(R+X))*Vin again, but now looking at the imaginary portions, we can simplify to 1/(1+iωRC). We can see that at low frequencies, the imaginary term doesn't contribute significantly, leading to input and output voltages being nearly in phase. Here, the capacitor's impedance is more significant vs. impedance of the resistor. On the other hand, at high frequencies, the impedance of the capacitor is less significant, leading to a more significant imaginary term. Thus, the voltage across the capacitor will be nearly 90 degrees out of phase with the input voltage. 

Decibels

-ratio of two voltages usually given in decibels

-In low pass filter circuit, at cutoff freq ω0, the ratio of signal (Vout/Vin) attenuated by a factor of 1/sqrt(2), which is equal to -3.01 dB.

Q: Beyond the cutoff, the response of a low pass filter drops at a rate of 6dB per octave. Explain.

-Beyond the cutoff, meaning at high frequencies, Vout is attenuated at a rate of 6 dB/octave. This is 2*3 dB. And frequency doubles per octave. This means that the amplitude of Vout drops by a factor of 2 per octave at these high frequencies as frequency is doubled. 

In other words:

Low Pass Filters, Experimentally

-Integrator vs. Low Pass Filter 

Integrator is a special case of a low pass filter. This circuit is a decent integrator only at high frequencies. 

What do you calculate to be the filter's -3 dB frequency? ω = 1/(RC) = 1/(1.5 kOhms * 0.01 uF) = 6.7 kHz or f = 1.1 kHz.

-We drove the circuit with a sine wave, sweeping over a large frequency range. 

Observation: We noted that as frequency increased (to a relatively large magnitude), Vout decreased, which was what we expected from theory discussed above.

-We then found the f-3dB experimentally by measuring the frequency at which the filter attenuates by 3 dB (or Vout cut down to 70.7% of full amplitude or Vout = 0.707*Vin) and got a value of 1.2 kHz (which was relatively close to the calculated value, considering uncertainties in the resistor and capacitor). 

What is the limiting phase shift, both at very low frequencies and at very high frequencies?

DC or high freq?

Low freq (0.5 Hz) --> no phase shift.

High freq (25 kHz) --> 90 deg phase shift. 

We determined this phase shift by measuring the delta t using the cursor from peak to peak then dividing by the period. Here, we got (10 us / 40 us)*360 deg = 90 deg

Check to see if Low Pass Filter attenuates 6 dB/octave for frequencies well above -3 dB point (i.e. check if output voltage drops by a factor of 2 as frequency is doubled. Frequencies we choose have to be well above ~1 kHz).

We chose to take voltage measurements at 10 (10 times f-3dB), 20 (20 times f-3dB), and 40 kHz .

Data:

f= 10 kHz --> Vout = 348 mV

f= 20 kHz --> Vout = 178 mV

f= 40 kHz --> Vout = 88 mV

 

(left to right: 10, 20, 40 kHz)
Observation: We do see that at frequencies well above the -3 dB point, the filter attenuates 6 dB/octave.

Phase shifts:
f << f-3dB --> no phase shift (tested above)
f = f-3dB --> (108 us/ 833.33 us)*360 deg = 46.7 deg (close to 45 deg)
f >> f-3dB --> 90 degrees (tested above)

Finally, measure attenuation at f = 2f-3dB and write down the attenuation figures at f  =2f-3dB, 4f-3dB, and 10f-3dB. 

Data:
f = 2 * f-3dB --> 1.32 V/3.00 V
f = 4 * f-3dB --> 740 mV/3.00 V
f = 10 * f-3dB --> 296 mV/3.00 V

Observation: We noted a steep "rolloff" in the attenuation figure (Vout/Vin) as freq increased.


High Pass Filter

-Similar to a voltage divider with two resistors (opposite of low pass filter)

• If 1/ωC << R, then Vout is approx equal to Vin. 

This is because if 1/ωC  is very small compared to R , the voltage drop across 1/ωC  is negligible (compared to the Vin or Vp value), making Vout close to Vin. 

• If 1/ωC >> R, then Vout << Vin 

This is because if R is very small compared to 1/ωC , the voltage drop across 1/ωC is significant, making the voltage drop across R negligible (or much less than Vin or Vp).

Thus, for high frequencies, Vout is approx equal to Vin, while for low frequencies, Vout << Vin, making this circuit a HIGH PASS FILTER. It lets through high frequencies (unattenuated) while low ones are. 

Again, the cutoff freq ω0 = 1/RC is the boundary. 


High Pass Filter, Experimentally

Where is this circuit's 3 dB point?

Because the resistor and capacitor is the same as the one we used for the low pass filter, the 3dB point should be the same (~1.1 kHz). Our measured value was 1.04 kHz, which was relatively close, considering our resistors have 10% and capacitors have 20% uncertainties.

Check out how the circuit treats sine waves.

(out of phase at first glance; haven't played around with frequencies yet)

Check to see if the output amplitude at low frequencies (well below 3dB point) is proportional to frequency.

Yes, we saw the "linear" portion of the curve and checked for the proportionality by plotting several points and noting a high R^2 value of the regression line and a slope of 0.9208 (almost output voltage equaling frequency?)

Frequency (Hz)	Vout (mV)
500	440
250	256
125	136
62.5	76

 

Limiting Phase Shifts

Low (62.5 Hz) --> (4ms/16ms)*360 = 90 deg

High (25 kHz) --> 0 deg

(Left: low freq, Right: high freq)

Garbage Detector

-Used a "transformer" to make a low voltage (thus "safe") replica of 110-volt power line signal and then use the high pass filter to attenuate the 60 Hz portion of the signal so we can clearly see the high freq "garbage" (unwanted signal/noise) that "sits on top of it." 

-"Transformer" --> reduces the 110 V to 6.3 V and "isolates" circuit we're using from the potentially lethal power line voltage

Looking at A, the output signal looked like a classical sine wave. 

However, looking at B, we saw the glitches and wiggles (the output of the filter). 

What is the filter's 3dB frequency?

Again, it should be around 1.1 kHz, and experimentally, we measured 1.06 kHz.

What is the filter's attenuation at 60 Hz? 

We know that amplitude grows linearly with frequency at frequencies well below f-3dB, so we measured the "slope" or rate at a low frequency. We got 55.2 mV / 6.3 V = 0.009. BUT high pass filters don't (significantly) attenuate high frequency signals (although I thought these frequencies had to be well over f-3dB.) I was expecting a higher attenuation value at 60 Hz, which is a relatively low frequency. I may want to redo this exercise to check. The value we got seemed small. Here, 60 Hz would be considered as "noise."

Blocking Capacitor

-"tug of war" between two power supplies --> Think of the circuit as Thevenin equivalent circuits for each power supply. Voltage at point A = 15 V - I(Rth1) using Kirchoff's Law(?) which equals 15 V - (10V/(Rth1 + Rth2))*Rth1.

From here, we can see that if Rth1 << Rth2, V at A is close to 15 V and if Rth1 >> Rth2, V at A is close to 5 V (current flowing in the way of least resistance). Thus power supply with smaller output impedance tends to win out. 

-For DC voltages, the function generator is effectively no connected to output (whereas for AC, it is).

-Thus, the AC voltage supplied will "ride on top of " the 5V DC voltage supplied by the voltage divider. 

-This is what we've been talking about more broadly about AC coupling mode. AC coupling "blocks" any DC signals (with a blocking capacitor in series with a 1 MOhms input impedance of the scope). In other words, this RC circuit is a high pass filter with a very small cutoff freq that lets "high" (low cutoff so doesn't take much to be considered high) signals through and we are able to view a small AC signal that rides on top of a large DC offset. 

Blocking Capacitor, Experimentally

-Instead of attenuating noise, blocking unwanted DC signals

-Wiring up circuitry A

Observations:

This circuit let the AC signal "ride on" 5 V, but it in fact "rides on" 1.6 V (1/3 of 5) offset. This portion seemed to have "blocked" about 2/3 of the DC offset.

-Wiring up B

Observations:

By adding another blocking capacitor, this effectively blocked the DC signal.

Note the 1.60 V offset.

f-3db = 1/RC = 1/0.47 = 2.13 Hz --> low freq limit for this blocking circuit

At low freq:

We observed a phase of 90 degrees, with the circuit acting as a high pass filter.

-This circuit (full circuit) acted as a high pass filter, but as long as f > f-3dB, all it did is block DC signal without attenuating AC signal. If f is low, it would attenuate AC signal as well, which leads to distortion in signal and phase shift between Vin and Vout.