Feb 19: Diodes

Diode
-"one-way streets" of electronic circuits; allow current to flow through them in one direction only (in direction of the arrow)
-Ideal diode (I-V curve):

Note the instantaneous increase in current towards infinity as soon as the voltage changes sign. In reality, the sharp increase isn't as instantaneous. There is a bit of a gradual rise (in the beginning) and a nonzero change in voltage.

-Effective resistance = V/I

so for a forward-biased diode, R about 0 (current flows without opposition)

and for a reversed-biased diode, R approaches infinity (current will not flow)

-The forward-biased diode we used that carried an appreciable current has a fixed voltage drop across the diode of about 0.6 V (our "regular" silicon diode).

Measuring the Resistance of a Diode:

-Changed settings on the meter

• At the 20 MOhm scale, we read resistance of 4.84 MOhms.
• At the 2000 MOhm scale, we read resistance of 11 MOhms. 

The 20 MOhm (smaller resistance) scale produced larger current (the beginning of the steeply rising portion of the I-V plot). 

To summarize, the magnitude of the current depends on the scale selected. Smaller resistance scale users larger currents.

For a forward-biased diode, the resistance that the meter measures will be smaller as scale becomes more sensitive (smaller scale). 

Half Wave Rectifier

-Rectifier = converts AC signal into DC signal (unchanging polarity or sign)

-Use: to provide a constant, stable voltage supply... but readily available power supply usually wall socket which provides AC voltage

Explanation: Until the voltage hits 0.6 V, the diode doesn't conduct (by the nature of diodes described previously). And when the sign of voltage is negative, there is no flow (there would be a negative flow but the diode prevents that. So in these regions, V = 0). So, at every positive V point, we expect to see it be 0.6 V smaller than what we saw in the input.

Half Wave Rectifier, Experimentally (Lab 3-2)

circuit used

We observed what we expected (only positive voltages. Negative voltages gone because no current flow in that (opposite) direction).

We then compared the input (blue) and output (yellow) voltages and saw that the amplitudes of the bumps of the output voltage has also decreased by 0.6 V, which was what we expected.

Closer Look:

We measured the voltage difference between Vin and Vout at the time when the input and output voltages peak, which was 0.640 V (relatively close to 0.6 V). When we repeated for one other part of the waveform when Vout > 0, we measured 0.480 V, so the voltage difference at other points of the wave seemed to be smaller than that at the peak. Also, Vpeak > 6.3 V because the 6.3 V on the transformer referred to its rms, which is different from the amplitude of the wave. The amplitude is greater than the rms by a factor of sqrt(2). We measured the rms value of the transformer (using the voltmeter), which was 7.5 V, and the input voltage difference of the circuit was 10.8 V. By looking at this ratio, amplitude of input voltage/rms, we got 10.8 V/7.5 V = 1.22 = sqrt(2), which matched the theory. Thus, Vpeak > 6.3 V as amplitude (or Vpeak)  > rms by a factor of sqrt(2). 

Ripple

-Now adding a capacitor in parallel with the load resistor...

-What does this do? AND SIGNIFICANCE?

This adds a RC time lag so that the output voltage will be effectively "bridged" between the peaks we observed above. In other words, before the output voltage just falls as the falling portion of the sine wave, the addition of the capacitor causes the charged capacitor to discharge/drain more slowly (so output voltage doesn't go down to 0 like last time). And in addition, if R-load and C of the circuit are large enough such that R-load*C >> 1/60 (time between two positive portions of the cycle observed in the experiment above), then the capacitor will not discharge very much before it starts charging again, so this "ripple" will be small. Ideally, if it is small enough, it can almost be like having a constant or nearly constant voltage. But even with large R and C, input voltage may fluctuate (need a voltage regulator to stabilize voltage).

Voltage much more "constant" than when just using the half wave rectifier circuit

Ripple Amplitude

-Added a 47 uF capacitor in parallel with the 2.2 kOhm load resistor in the half wave rectifier circuit.

Calculate ripple amplitude:

delta V approx = (1/60 s)/(R-load*C) = 10.8(1/60 s)/(2.2 kOhms * 47 uF) = 1.6 V

Measure ripple amplitude:

delta V = 1.4 V

The values are relatively similar, taking into account the uncertainties of both the resistor and capacitor.

Signal Diodes, Experimentally (Lab 3-5)

Square wave, 10 kHz, 5 V peak to peak

Observation: What we saw on the oscilloscope was just like what we saw last week with the differentiator. The square wave was differentiated, in which the voltage became a straight line (constant) except the relatively sharp peaks at where the square wave began to rise or fall very quickly. 

Removed the 2.2 kOhm resistor and saw:

Long decay time (even after a cycle of square wave ends)

Looks familiar... but we never used a capacitor in the circuit...

We see this on the oscilloscope due the "parasitic capacitors" in the circuit effectively in series from the capacitance of the scope, cable, breadboard, and etc (not actual capacitors but the effect of having capacitors in series).

RC discharge with 2.2 kOhms:

Time constant = RC = 96 ns --> C = 44 pF (R = 2.2 kOhms)

RC discharge without 2.2 kOhm resistor:

Time constant = RC = 31 microseconds --> C = 31 pF where R = 1 MOhms from the scope

SIGNIFICANCE: So, the output voltage appears to be differentiated and rectified in that  only the positive portion of the square wave is getting through as the output (like the exercise before).

Diode Clamp

Outputs:

Observations:

-Voltage never got greater than 6.6 V (expected this value to be 5.6 V because we used a 5 V input (Arduino)... so on the other side of the diode, we expected 5 + 0.6 = 5.6 V). This may have been due to use of AC coupling on the scope (vs. DC). We will fix this soon!

-Also, though the voltage output "plateaus" at 6.6 V, Vout here is not flat, showing that resistance of the diode is not static (or "dynamic resistance") and how it depends on the voltage across the diode. More quantitatively, we'd have to look at dV/dI, the slope of the V vs. I curve. We could use the triangle wave to do this, since portions of wave are linear, where slopes are easier to determine. 

Update: We tried DC coupling.

Output voltage = 6 V, which is closer to 5.6 V.

Diode Limiter

Why might we care?

-Some instruments can be damaged if applied voltage is too large. So, we need a way to limit the input voltage to < 0.6 V. So, we can keep the shape of the applied signal while limiting the voltage to less than 0.6 V.

-works for both positive and negative input voltages

How does this work?

As mentioned before, this limiter can work for both positive and negative input voltages.

1) For positive input voltage:

-Since the voltage drop across a diode is up to 0.6 V, most of the voltage drop in this circuit is across the 1 kOhm resistor. Then, note that no current goes through the diode on the left. So the voltage across the right diode (0.6 V) is the output voltage. Note that in the photos above, even after changing the amplitude of the applied voltage, the output is limited up to 0.6 V.

2) For negative input voltage:

-Again, most of the voltage drop in this circuit is across the 1 kOhm resistor. However, this time, the current goes through the diode on the left because the negative input voltage is less than the voltage of the ground. (No current flows to the diode on the right.) Therefore, the output is now limited down to -0.6 V.

*These can be used as "protection circuits" to keep instruments from frying from high applied voltage.

Impedances of Test Instruments

-Good for voltage measuring device to have as large input impedance as possible (to prevent perturbing circuit we're measuring as little as possible)

-For instruments to measure high frequency signals (e.g. oscilloscope), input impedance (Zin) is due to 1 MOhm resistor in parallel with ~30 pF of "stray capacitance." (No actual capacitor component). This stray capacitance limits scope's ability to observe high frequency signals. So, we want to make this capacitance as small as possible. 

Magnitude of capacitor's impedance = -i / omega*c

At low frequencies, input impedance close to 1 MOhm (close to the resistance of resistor). In a parallel circuit with resistor and capacitor, smaller resistance dominates (low frequencies imply large impedance for capacitor). Vice versa.

So input impedance of oscilloscope frequency dependent. 

Impedances of Test Instruments, Experimentally 

(Lab 3-8)

1) Measuring Internal Resistance of the DVM

DC Voltage Source: +5V pin on Arduino

"DVM good at what it does" --> high input resistance so we chose to use a 1 MOhm resistor

Created this voltage divider:

where R1 = 1 MOhm, Vin = +5 V

Measured Vout = 4.57 V

So, 4.57 V = (Rint / (Rint + 1MOhm))*5 V 

---> R-int = 11 MOhms

2) Measure the Input Resistance of the Scope

100 Hz sine wave

Similar voltage divider as before:

where R1 = 1 MOhm (because again, a good voltage measuring instrument expected to have high resistance), Vin = 5V

Vout = 24.6 V

So, 24.6 V = (R2/(R2 + 1MOhm))*5V

---> R2 = input resistance of the scope = 1.3 MOhms

3) Measure scope's input impedance

*Note: We were having a hard time getting the expected output signal for this (we expected Vout's amplitude to be half of that of Vin). Then, we realized that we were using the scope probes for Vout, and these scope probes are used to decrease attenuation, which was something we wanted to see in this case.

-We know that the output impedance of the function generator is << 1 MOhm because for a power supply, good to have 0 or very small resistance so voltage is delivered to output instead of dropping significantly across supplier.

What is the low frequency (f < 1 kHz) attenuation?

At low freq --> C high impedance so in parallel with resistor, smaller impedance dominates --> resistance of resistor --> 1 MOhm 

-In this case, this circuit is like a divider (both resistances around 1 MOhm... 1 MOhm resistor and we measured about 1.3 MOhms for the input resistance of the scope). So attenuates by 2 (decreased by a factor of 2... "cut" in half). 

At high frequency? (5 kHz)

-C low impedance so now impedance of C dominates --> Zc = -i/omega*C. 

omega = 2*pi*5 kHz

C = ~ 30 pF (our oscilloscope has ~30 pF of stray capacitance)

(R = 1.3 MOhms (solved in #2))

so Zc = impedance of capacitor = 1.1 MOhms

Check:

Req = (1.3 MOhm*1.1MOhm)/(1.3 MOhm + 1.1 MOhm) --> Req  = 6E5 Ohms

Then voltage divider.

V-out = (6E5/(6E5+1E6))*2V = 0.75 V = 750 mV

Our V-out values were not very close, so we'll try it again next time!

4) Looking at Oscilloscope Probes

-Since Z-in (input impedance) is frequency dependent, different frequency components of a signal can be attenuated by different amounts --> distortion. To minimize this, we use an oscilloscope probe.

-suppresses or exaggerates high frequencies by changing capacitance (not actual capacitors but behave/result in having effects of capacitors). 

-The x10 probe raises input impedance of oscilloscope by a factor of 10 for all frequencies--> attenuates signal

-but decreases sensitivity by a factor of 10 (bad for looking at very weak signals)

-variable capacitance 

suppressing vs. exaggerating

When using these probes, want to adjust the "compensation" screw to have "perfect" rectangle/square waves:

5) Measure the internal resistance of the function generator. 

-Knew this value would be relatively low because again, for a power supply, good to have 0 resistance so voltage is delivered to output instead of dropping significantly across supplier. 

-So used R = 47 Ohms

Vin = 1V using the function generator

First circuit to look at Vin. Then, second to look at Vout.

Vin = 1 V

Vout = 0.504 V (which indicated R1 would be close to 47 Ohms)

0.504 V = (47/(R1 + 47))*1 V

--> R1 = 46 Ohms