Lab 5 -- Circuit Deux

To:Students of EE 222


From:Owen Grace, Esquire and Bob Chandler XVII


Subject:Laboratory #5 -- Nifty Diode Circuits


Greetings fellow EE's. We are going to present to you a brief run-through of Nifty Diode Circuit #2, what it does, how it does it, and some groovy applications for you guys to try at home with your common, every-day, household wall socket. Check out our circuit in Figure 1.

Figure 1 - Schematic of Nifty Diode Circuit #2
Schematic goes here

We were asked to analyze the effect that varying the load resistor, R2, had upon Vout, and to determine the minimum load resistance necessary to maintain a useful output.

Beginning with Vin equal to 4V (peak) AC, a 0.56 nF capacitor for C1, a 1N4148 diode and an open circuit (infinite resistance) as R2, we observed how Vout varied with Vin. A plot of both Vin and Vout vs. time may be seen as Figure 2, below.

Figure 2 - Vout and Vin vs. time of circuit with infinite load. Vout is above Vin.
Plot goes here

C1 = 0.56 nF
Diode = 1N4148
R2 = Open circuit


As seen in the oscilloscope plot, Vout was the same sinusoidal wave as Vin, yet Vout was shifted upwards by Vin (peak), or 4V. Because of this, this circuit can be used as a positive voltage clamp.

Now you're probably wondering how this circuit works, eh? So are we. Just kidding. Here's what we came up with: Assuming the circuit has reached steady state, we see that when Vin is at it's negative peak, Vout is approximately 0 V, because the voltage drop across a diode in forward bias is zero. During this stage, C1 was positively charged to Vin (peak). As Vin rises toward 0 V, C1 begins to dominate, creating a reverse bias potential difference across the diode. This continues while Vin ri ses to it's positive peak of 4 V, making Vout equal to Vin + V(C1), or approximately 8 V. Because there is no load to allow the capacitor's voltage to discharge, Vout will always remain V(C1) above Vin. Ideally, this would clamp Vout's minimum voltage a t exactly 0 V, however, due to our diode's turn-on voltage of roughly 0.7 V, Vout (min) will be -0.7 V. This phenomenon can be seen in Figure 2 as well.

There is a direct, linear relationship between Vin and Vout, as expected, which may be seen in Figure 3.

Figure 3 - Vout vs. Vin of circuit with infinite load.
Plot goes here

C1 = 0.56 nF
Diode = 1N4148
R2 = Open circuit


Once we add a load, R2, within the circuit, the story may change. With extremely high loads, little current is pulled from the circuit, thus barely affecting the previous analysis of Vout. However, if R2 draws a significant amount of current, the ca pacitor is able to discharge when Vin is positive. Check out Figure 4.

Figure 4 - Vout and Vin vs. time with load resistance of 10 kohms.
Plot goes here

C1 = 0.56 nF
Diode = 1N4148
R2 = 10 kohms


Notice the drop in the voltage across the diode as it's in reverse bias. Instead of having nowhere to go, energy from the capacitor dissipates across the load resistor! Check out the Vout vs. Vin characteristics as seen in Figure 5.

Figure 5 - Vout vs. Vin of circuit with load resistance of 10 kohms.
Plot goes here

C1 = 0.56 nF
Diode = 1N4148
R2 = 10 kohms


By trial and error, we determined the minimum load resistance which still allowed the circuit to properly act as a positive voltage clamp to be approximately 510 kohms.

We were curious as to what effects switching the polarity of the diode would have upon the output, Vout. After building this circuit we found that it only made the circuit a negative voltage clamp. That is, the output never "ideally" is greater t han zero volts. Also, ideally, the frequency does not have an effect on the output.

One general use for this circuit could be to power a system requiring positive voltage only. This also could be used for any application needing an oscillating DC source. Party on!!
This page written by JUICE (grace@bucknell.edu) and LBC (chandler@bucknell.edu)


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