Day 6: Diodes and More Capacitors

In order to add a DC offset to any Vin, we can simply use a blocking capacitor and build this circuit

Second picture by Hannah

This particular circuit causes a 4 V offset since we are using a 12 V power supply. 

Picture by Hannah

However, at low frequencies, the offset becomes much smaller. In the picture above, the frequency is very high -- 100 kHz. At a lower frequency, like 1 Hz, we merely observe

Picture by Hannah

The offset is still 4V, but because of the similarities of the circuit to a high pass filter, the low frequency wave has been attenuated!

Second picture by Hannah

We can also bring the offset back to 0 V with the circuit above. At 100 kHz, the oscilloscope reads a Vout that looks exactly like the Vin coming out of the function generator. At lower frequencies like 1 Hz, the magnitude of Vout is decreased. So... basically yet another iteration of the high pass filter!

We can figure out the low frequency limit by measuring 3dB. This is, if you recall, the point at which Vout is 70.7% of Vin. So since Vout goes from 1500 mV at 1 MHz to 1050 mV at 11 Hz, the low frequency limit is somewhere in the 11 Hz range.

The next section is on diodes! 
An ohmmeter measures 0.605 V as the forward bias of the diode, and ~5 Mohm as the resistance (though the latter is largely not useful unless we measure a lot of diodes with the same ohmmeter). And of course, we get an infinite voltage if we try to measure the reverse bias.

Now we can build a half wave rectifier!

Second picture by Hannah

We used a variable transformer set to 6 Vac. 
Recall that Vpp = 2 sqrt(2) * Vrms
So, to get a Vrms of 6 V, the power supply should be adjusted until Vpp = 16.8 V

Alternatively, our oscilloscopes will automatically calculate the Vrms and Vpp of a wave.

Picture by Hannah

Yellow is Vin; Blue is Vout

And it lets only positive voltage through, as expected! Of course, we get a Vp > 6 V, since Vp > Vrms

The next task is to build a ripple. 
A 47 μF capacitor is added in parallel with the 2.2 kΩ resistor in the half wave rectifier circuit, being careful to keep the negative end of the capacitor towards ground.

Picture by Hannah

Yellow is Vin; Blue is Vout

As seen above, the measured ΔV = 900 mV

We can calculate ΔV as follows:
ΔV = Vmax (discharge time / (Rload C))
       = 8.4 V (14.8E-3 s / (2.2E3Ω 47E-6 F))
       = 1.2 V
We had to measure discharge time of the capacitor, since it is obviously not discharging for the entire period. Another adjustment could have been measuring the actual Vmax, since Vmax of the output is not the same as that of the input.

Recall that a capacitor has 20% error, so our result has an error of approximately the same order.

Next, we can build a signal diode (aka rectified differentiator)

Second picture by Hannah

By driving the system with a 10 kHz square wave at 20 Vpp (the max our function generator can generate), we ge the following:

Picture by Hannah

Blue shows Vin; Yellow shows Vout

Only the positive derivative is produced!

Without the 2.2 kΩ resistor, the circuit is a mere RC circuit. With the oscilloscope's resistance being 1 MΩ, the time constant is extremely large, so the discharge time is much slower. The 2.2 kΩ resistor therefore draws current through itself, allowing the derivative to correctly be a spike with a short discharge. 

Next, a diode limiter!

Second picture by Hannah

A diode limiter is a wonderful creation that cuts off high voltage. For the following images, we are using a Vin of 100 Hz at the max amplitude of 20 V. 

Pictures by Hannah

Blue is Vin; Yellow is Vout

Since 20 V is much larger than the cut off, we can see clearly how the diode limiter cuts off high voltages.

Picture by Hannah

Blue is Vin; Yellow is Vout

However, the above image has a 100 Hz frequency but only a 1 V amplitude. It does not get cut off very much! This would be very useful if we had an instrument that was damaged by high voltage.

Now we will consider the impedances of test instruments! While we'd like our measurements to not affect the system at all, of course our instruments are not perfect. 

By driving our oscilloscope with a 100 Hz sine wave and the following circuit

We can see that at the attenuation is approximately 1/2.
At the higher frequency of 10 kHz, the attenuation is 1/4.
So, as frequency increases, Vout decreases.

The oscilloscope behaves as a low pass filter!

So, at low frequencies, we should observe the circuit behaving as a simple voltage divider. Since the attenuation was 1/2, we know the resistance of the oscilloscope must be approximately 1 MΩ. This also means the capacitance will be small, somewhere in the 10s of pF.

Recall that the magnitude of the impedance of the capacitor will be 1/ωC. 
At a higher frequency of 10 kHz, the attenuation is 1/4
We can draw the two resistors as one resistor with Rth = 0.5 MΩ.
Now, we can say that for an RC circuit,
Vout = Vin /(1+iωRC)
attenuation = 1/(1+ωRC)
C = (4 - 1) / ωR = 95 pF, using a ω of 2 Pi 10 kHz and R of 0.5 MΩ.

The coaxial cable also has capacitance which makes this capacitance higher than the capacitance of the oscilloscope.

To make the circuit a divide by two attenuator, we must decrease the capacitance of the oscilloscope. This will make the resistance of the scope will dominate even at higher frequencies and still behaves as a voltage divider.

Probes are particularly useful because they increase the impedance.
With a 10x probe where Vout is above, Vout ~ Vin.