2/14: More with capacitors; Diodes

in which Julia continues capacitor investigations and explores voltage-limiting diode circuitry

ΔμΩ

H&H Lab 2-6: Filter Application: Garbage Detector

The garbage detector makes the input signal (purple)'s noise visible (yellow).

Here we used a transformer to make a low-voltage signal from the standard 110 V outlet, then added a high-pass filter to attenuate the 60 Hz portion of the signal.  This circuit clearly shows the high-frequency "garbage" that sits atop the 60 Hz signal.

We expect the voltage to be attenuated by

-24 = 20*log(Af/Ai)
Af/Ai = 10^(-1.2) = 6.3%

We measured Vout = 2.88 V and Vin = 187 mV.  Our actual attenuation value was in good agreement with the expected value:

Vin/Vout = 0.187/2.88 = 6.3%

Lab 2-8: Blocking capacitor

(orange: red of FG. yellow: black of FG.  Brown: Red of scope.)

The circuit diagram in H&H.

First, we built only circuit "A", which added +5 V to the AC signal from the function generator.
For DC voltages, the capacitor lets no current pass; as a result, this circuit lets AC voltage from the function generator sum with DC voltage from the voltage divider.  

DC offset adds to AC signal...at a scale of 2.5 ms?!

Unfortunately, what we assumed was a constant DC voltage produced by the transformer was actually (although always positive) highly variable.

The transformer's "DC" voltage is always positive, but not constant.

We decided to use a battery pack as the voltage source instead (although this gave us only about +4.2 V, this was a constant source).

Constant voltage source shows DC offset to the AC signal.

Adding the circuit part "B" caused the circuit to do nothing (as confirmed by oscilloscope readings and noting no change when removing the battery voltage source from the circuit).

Here the output signal matches the input: the circuit has no net effect.

Diodes!

Next we investigated the circuit element known as a "diode".  This element limits current flow to only one direction; its J-shaped IV curve causes high current flow (low resistance) at high voltages (at about +0.6 V for our 1N914 diodes), and essentially zero current flow (very high resistance) at lesser voltages.  When we measured the 1N914 diode with an ohmmeter, we measured lower resistance at higher voltage scales (because the voltage applied during testing was higher).

Lab 3-2: Half-wave rectifier

Only the positive portion of a half-wave-rectified signal is visible (shifted down by 0.6 V by the diode).

The output (shown above) had the expected flatlines between positive peaks, which shows that above a certain voltage (0.6 V) all of the current flowed through the diode, through the resistor, to ground; but below that voltage, no current flowed through the diode (and so no voltage drop was seen over the resistor).
Although we used a variable transformer with dial set to approximately 6 Vac, our output signal had an amplitude of 1.04 V, suggesting that the transformer's scale is not particularly accurate for AC voltages.

Pre-Lab 3-5:

Add a 47 μF capacitor (we used a 4.7 μF capacitor, as 47 μF were unavailable) in parallel with the 2.2 kΩ resistor in the half wave rectifier circuit.  Explain the output; calculate the "ripple" amplitude, compare to measured peak height.

The resulting "ripple" had an amplitude of 0.640 V.

We measured the fall to be 640 mV (0.640 V) in this circuit; the maximum voltage (without capacitor) was 960 mV.

We calculated the "ripple" amplitude using the equations in Electronics.pdf:

ΔV ≈ (0.960 V)/[(60 s)*(2200 Ω)*(4.7 x 10^-6 F)] = 1.55 V

However, our measured ΔV  was 680 mV, which is far from this calculated value.  We thought perhaps the approximation was invalid for small capacitance, and tried the exact formula:

ΔV = (0.960 V) - (0.960 V)*exp(-1/[(60s)*(2200 Ω)*(4.7 x10^-6 F)]

ΔV = 768 mV

Although this is certainly closer than the approximate formula's result, the calculated value still did not agree with our observations.

Concerned, we tried a capacitor with a higher capacitance: 470 μF.

As expected, the ripple circuit with a 470 μF capacitor showed smaller peaks.

We then measured the peak height to be 16 mV (0.016 V), which agreed with our approximate calculation:

ΔV ≈ (0.960 V)/[(60 s)*(2200 Ω)*(470 x 10^-6 F)]

ΔV ≈ 155 mV

The exact calculation gave a similar value:

ΔV = (0.960 V) - (0.960 V)*exp(-1/[(60s)*(2200 Ω)*(470 x10^-6 F)]

ΔV = 153 mV

Still, we were confused why our calculations were so inaccurate for smaller capacitance.  We tried a 0.22 μF capacitor, which caused no noticeable change in the capacitor-free output.  We also investigated the effects of a 10 μF capacitor, which did cause a noticeable change: a measured ΔV of 420 mV.

The ripple circuit with a 10 μF capacitor showed an intermediate ripple height.

The approximate value was much higher than our measured ΔV of 420 mV:

ΔV ≈ (0.960 V)/[(60 s)*(2200 Ω)*(10 x 10^-6 F)]

ΔV ≈ 727 mV

While the "exact" calculated value was closer to our measured value, it was still higher than that found in the experiment:

ΔV = (0.960 V) - (0.960 V)*exp(-1/[(60s)*(2200 Ω)*(10 x10^-6 F)]

ΔV = 510 mV

In general, it seems that the approximation is less valid for smaller capacitance values--as expected, because the Taylor series approximation of exp(-x) = 1 - x is less valid when x is inversely proportional to a small capacitance value.  However, the "exact" formula also does not yield a ΔV in agreement with experiment for small capacitance values, which suggests that this formula neglects some part of the circuit.  For these relatively small voltage values, perhaps the resistance of the wires and breadboard is significant?

Lab 3-5: Signal diodes

Here we used a diode to make a rectified differentiator, as shown above (though we used a 470 pF cap, as 560 pF was unavailable).  When driven with a 10 kHz wave, the output undergoes rapid RC decay.

At the 5 μs scale, the circuit appears to generate a sharp peak.

RC decay is visible at the 250 ns scale (0.25 μs).

The 2.2kΩ resistor amplifies output signal by providing resistance to current flow so it slows current flow to ground.  Without this resistor, all of the current flowing through the diode goes to the output, which makes the ratio of RC decay to total signal smaller, so it is more difficult to see the RC decay.  The first resistor (1 kΩ) and the capacitor (470 pF) determine the RC decay in this circuit.

Lab 3-6: Diode clamp

H&H's diode clamp circuit.

The diode clamp begins to limit voltage at about 10.6 V; it only limits the positive voltage(top peaks) because this circuit uses a diode, so current cannot flow through the diode when voltage is negative.  10.6 V is a reasonable value: half of this (because the diode only works in positive voltages) is 5.3 V; subtracting 0.6 V (to get into the diode's low-resistance IV curve region) yields 4.7 V.  While we expected a final value of about 4.2 V, because that is the voltage supplied by our batteries, 4.7 V is fairly close; the small difference may be due to internal resistance in the batteries.

The diode limiter clearly limits voltage with a 13.7 V input signal.

The diode limiter flattens the signal's top peaks.

While the diode clamp flattens the peak, there is still some curvature in the output signal (this is more clearly evident in the zoomed-out image above), which reflects the diode's non-zero impedance.

Next we added a voltage divider to the circuit:

H&H's diode clamp + voltage divider circuit.

This caused a more rounded voltage-limited output peak:

Adding a voltage divider made the output peaks rounder.

Voltage flows through the second 1kΩ resistor only if the applied voltage (Vin) is greater than 4.7 volts.  If Vin > 4.7 V, all of the "extra" voltage goes through the diode, through the second 1kΩ resistor, to ground.  Increasing the extra voltage increases the voltage drop through the resistor, which makes this output curve rounded.

Driving the circuit with a triangle wave caused a similar output voltage shape: a triangle beneath a triangle.  In the upward-sloping region, the capacitor is charging, which limits the amount of extra current supplied through the load circuit; in the downward-sloping region, the capacitor is discharging, and thus limiting the "negative" current provided to the load circuit.

Triangle wave input (yellow) through the diode limiter yields a symmetrical output peak of lesser slope (purple).

Adding a second capacitor over the final resistor in the circuit flattened this peak.

Triangle wave input (yellow) through the diode limiter with a 15 μF secondary capacitor yields a symmetrical, flattened output (purple).

Here, the extra current (that would have increased the voltage across the second 1 kΩ resistor) instead goes to charging the capacitor.  Decreasing the applied voltage slowly decreases the capacitor's charge; even though voltage is decreasing through the resistor, capacitor provides extra voltage to keep the voltage drop over the resistor steady.
We were curious about how the size of the capacitor affects this flattening, so we tried a tiny capacitor (0.01 μF) for comparison.  This charges very quickly (so the first half of the peak is approximately triangular), and discharges slowly (the second half of the peak is curved).

Triangle wave input (yellow) through the diode limiter with a 0.01 μF secondary capacitor yields an asymmetrical output (purple).

3.7: Diode Limiter

If the diode receives over +0.6 V, current flows through 1; if the diode receives less than -0.6 V, current flows through 2 to ground.  A diode limiter such as this would be useful to protect a sensitive instrument from high voltage.

The diode limiter (yellow) limits an input sine wave (purple) to +/- 0.6 V.

Next lab: Instrument Impedances and Op Amps!