"load " Negative rail of battery (-) terminal Positive rail of battery (+) terminal. Diode inserted here for experiment #3 (+)

Transcription

1 A Four Cell Lime Battery, in Series, After Freezing the Limes SECTION #1 & #2 - Experiment #1 & #2 1. The following experiment uses 4 limes, as the cells, connected in series. 2. The negative electrode is a Zn plated iron nail. 3. The limes were frozen over night and then allowed to thaw for several hours before the experiment. 4. In the first experiment (labeled 1st), I used 4 Canadian Pennies as the positive electrode. 5. All other experiments (2 to 5), 4 Canadian Silver Dollars as the positive electrode: 80%Ag/20% Cu. 6. Experiments were run continuously, i.e., once connected to the battery, the experiment was allowed to run for the indicated time period before being disconnected from the battery. Each cell = 1 lime. 7. The experiment used the following components: - four limes, previously frozen, then thawed (termed a freeze/thaw cycle) - 4 galvanized (Zn plated ) nails - four Canadian coins: 4 x CDN 1994 Penny; 4 x CDN$ (80% Ag/20% Cu) - five 470 μf electrolytic capacitors (rated at 16 volts) - resistors = 10,000Ω (circuit load); 22,000Ω (transistor biasing resistor) - one diode (1N4148) - one transistor (2N4401) 8. Set up for Experiment #1 (Cu electrodes) and #2 (Ag electrodes): "load " Resistor 10,000Ω A A = Ammeter, to measure the current - + five 470 μf capacitors in parallel Negative rail of battery (-) terminal Positive rail of battery (+) terminal (-) Diode inserted here for experiment #3 (+) Zn {Cu 1st / Ag 2nd } Zn {Cu 1st / Ag 2nd } Zn {Cu 1st / Ag 2nd } Zn {Cu 1st / Ag 2nd } Each line represents one of the four cells connected in series

2 SECTION #1 & #2 - The Data for Experiment #1 and #2 Table 1 - Experiment #1: Values of Circuit Current for a Circuit Load of 10,000 Ohm Resistance 4 galvanized nails, Zn, (negative) / 4 Pennies, Cu (positive); 10,000 Ω Resistor Load Resistance Current Measured with a load resistance of 10,000Ω highest value obtained in 1st minute 1.200V ma 1 min 1.070V ma 5 min 0.928V ma 10 min 0.951V ma 15 min 0.970V ma 20 min 0.942V ma 36 min 0.903V ma Battery Resistance measured = 3000 Ω = 750 Ω/cell Table 2 - Experiment #2: Values of Circuit Current for a Circuit Load of 10,000 Ohm Resistance 4 galvanized nails, Zn, (negative) / 4 CDN$, Ag (positive); 10,000Ω Resistor Load Resistance Current Measured with a load resistance of 10,000Ω highest value obtained in 1st minute 1.65V ma 1 min 1.63V ma 5 min 1.47V ma 10 min 1.41V ma 20 min 1.34V ma 30 min 1.35V ma Battery Resistance measured = 1600Ω/4 = 400 Ω/cell SECTION #1 & #2 - Discussion for Experiment #1 and #2 1. From Table 1, experiment #1, it can be seen that the four cell battery, with four Cu electrodes, is producing an average voltage of 1.05V ( V) and an average current of 106μA ( mA). 2. This is approximately 1.05/4 = 0.263V per cell. 3. The measured resistance through the "Cu" battery is 3000Ω, which is 750 Ω/cell. A good result. 4. From Table 2, experiment #2, it can be seen that the four cell battery, with four Ag electrodes, is producing an average voltage of 1.5V ( V) and an average current of 124μA ( mA). 5. This is approximately 1.5/4 = 0.375V per cell. 6. The measured resistance through the "Ag" battery is 1600Ω, which is 400Ω/cell. A very good result. 7. The voltages are less than I hoped for, but that is what I measured. 8. The best result is the internal resistance of the battery and in the "Ag" battery, it is about 400Ω/cell and when compared to the very first cells that I obtained in experiment #1 (which had a resistance of 4000 Ω/cell), this current design (no pun intended) is great news! 9. I attribute this result to the fact that I froze (and thawed) the limes prior to use. As expected, the cell contents were released into the extracellular spaces and created a "more fluid" environment, thus, less resistance.

3 SECTION #1 & #2- Discussion 10. Even though I am not getting voltages as high as I was hoping for, I am still getting a voltage of 1.35V and a current of 137μA at 30 minutes into the experiment, for a 10,000 ohm load, which is a good result. 11. It seems like I am going to have to continue to use the Ag electrodes for my battery design, if I am going to get an adequate voltage to run my project. 12. As a final comment on these two experiments, the batteries seeped fluid continuously from the insertion points of the electrodes (further evidence of the freeze/thaw event) and this meant more mopping up around the battery during the experiment. This result is an obvious lead into a whole new set of experiments, but as per my original goal, I will stay with the lemons (or limes) as a unit cell. SECTION #3 - Experiment #3 (Zn/Ag electrodes) 1. For experiment #3, I added the silicon diode in series with the battery and the load resistance (as in previous experiments, see the experiment entitled "Storing Charge - Capacitors, Section #2). As well, in Section #1, #8 above, I have indicated where I had inserted the diode. Zn/Ag electrodes were used. 2. The following is a partial description of the setup, i.e., addition of the diode and, moving the ammeter to a point just after the diode, as opposed to the positive side of the load resistor (as in expts #1 & #2). A Additional notes: - a digital volt meter was connected in parallel with the load resistor for the entire length of the experiment - a second digital ammeter was connected in series as noted to the left and remained in the circuit for the entire length of the experiment - a third, older analog volt meter, was used intermittently to measure either the voltage across the battery terminals or across the diode, during the progress of the experiment Zn (negative electrode) SECTION #3 -Discussion of Data for Experiment #3 (Table #3, below) 1. From Table 3 below, experiment #3, it can be seen that the four cell battery, with four Ag electrodes, the voltage across the load resistor went from 1.189V to 0.949V over a period of 60 minutes. Thus, the battery is still producing 0.949V/1.189V = 80% of the starting voltage. 2. From Table 3, experiment #3, it can be seen that the four cell battery, with four Ag electrodes, the current measured at a point just after the diode went from 0.120mA (120μA) to 0.103mA (103μA) over a period of 60 minutes. Thus, the battery is still producing103μa/120μa = 86% of the starting current. 3. But, the battery is still producing about 100μA after 60 minutes of operation, not a bad result at all. The total time that battery was in operation to this point, experiment #1, #2 & #3 was 126 minutes! 4. At the independent time points, i.e., 26, 31, 41 & 58 minutes, the voltage of the battery was 1.50V. 5. Consistent with this, was the summed voltage (the voltage across the load resistor and the diode) at the independent time points, i.e., 7, 14, 19, 29, 39 and 58 minutes of 1.574V, 1.552V, 1.531V, 1.480V, 1.433V and 1.393Vor an average of 1.49V. 6. As with a previous observation, the voltage across the diode remained constant ( V). 7. Finally, the current after the diode drops to 0.000mA instantaneously after the battery is disconnected; whereas, the voltage across the load resistor drops off slowly. This is further evidence that, after the positive lead is disconnected, electrons move from the negative plate of the capacitor to the positive plate of the capacitor, i.e., electrons are not moving from the negative plate of the capacitor to the negative terminal of the battery.

4 SECTION #3 -Discussion of Data for Experiment #3 (Table #3, below) 8. Lastly, the diode is dropping a large percentage of the battery voltage, i.e., 0.45V/1.5V = 30%. Thus, the need for a diode with better characteristics, e.g., a germanium diode (or equivalent), which should drop only about 0.3V or 20% of the battery voltage. This would allow for more voltage to be supplied to the remainder of the circuitry. Table 3: Experiment #3 - Values of Circuit Current for a Circuit Load of 10,000 Ohm Resistance (5 x 470 μf capacitors in parallel with battery; 1N4148 diode in series with battery) (ammeter inserted just after the diode); standard time points in bold; shaded cells - data not recorded 4 Zn plated iron nails (negative) / 4 coins containing 80% Ag (positive) / 10,000Ω Load Resistance, digital meter Current Measured with digital ammeter inserted after diode highest value obtained in 1st minute 1.189V ma 1 min 1.184V ma the diode, 5 min 1.137V ma 7 min, 1.104V 0.47V 10 min 1.131V ma 14 min 1.082V 0.47V 15 min 1.075V ma 19 min 1.061V 0.47V the battery, 20 min 1.054V ma 26 min 1.50V 29 min 1.030V 0.45V 30 min 1.047V ma 31 min 1.50V 39 min 0.983V 0.45V 40 min 0.981V ma 41 min 1.50V 58 min 0.943V 0.45V 1.50V 60 min 0.949V ma Current through diode drops to ma at the instant that the positive lead is disconnected. The voltage across the load resistor drops off slowly once the positive lead is disconnected. SECTION #4 - Experiment #4 (Zn/Ag electrodes) 1. For experiment #4, I added a transistor so that the collector current will be in series with the load resistance (10,000 Ω). I introduced a biasing resistor, 22,000Ω, at the base of the transistor. 2. Additionally, I connected the biasing resistor to the positive terminal of the third cell of the battery. 3. Additional voltages measured were the Base-Emitter (V BE ) and Base-Collector (V BC ) voltages. 4. Total time the electrodes were in place, including expt #1 (30 min), expt #2 (36 min), expt #3 (60 min) & expt #4 (min) = 186 minutes (3.1 hours!). 5. The ammeter was placed at the positive side of the load resistor.

5 SECTION #4 - Experiment #4 "load " Resistor: 10,000Ω A A = Ammeter, to measure the current - + Base Collector 22,000 Ω resistor Emitter diode negative rail of battery positive rail of battery positive terminal of 3rd cell Zn {Cu 1st / Ag 2 to 4} Zn {Cu 1st / Ag 2 to 4} Zn {Cu 1st / Ag 2 to 4} Zn {Cu 1st / Ag 2 to 4} negative terminal, cell #1; cell #2; cell #3; positive terminal, cell #4

6 Table 4a: Experiment #4 - Values of Circuit Current for a Circuit Load of 10,000 Ohm Resistance (5 x 470 μf capacitors in parallel with battery; 1N4148 diode in series with battery) (transistor 2N4401); standard time points in bold; shaded cells - data not recorded 4 Zn plated iron nails (negative) / 4 coins containing 80% Ag (positive) / 10,000Ω Load Resistance, digital meter Current Measured with digital ammeter inserted after load resistor the diode, the battery, highest value obtained 0.920V ma in 1st minute 4 min 0.45V 1.45V 5 min 0.862V ma 9 min, 0.45V 10 min 0.813V ma 1.45V 13 min 0.45V 1.45V 19 min 0.45V 20 min 0.816V ma 1.43V 25 min 0.45V 1.45V 30 min 0.730V ma 1.45V 59 min 0.45V 1.40V 60 min 0.713V ma Table 4b: Experiment #4 - Base-Emitter (V BE ) and Base-Collector (V BC ) Voltages dropped by the Transistor V BE Voltages V BC 8 min 0.47V 0.23V 15 min 0.50V 0.25V 21 min 0.45V 0.23V 29 min 0.50V 0.17V 60 min 0.47V 0.18V Average 0.48V 0.21V SECTION #4 -Discussion of Data for Experiment #4 (Table #4a & 4b, above) 1. From Table 4a, experiment #4, it can be seen that the four cell battery, with four Ag electrodes, the voltage across the load resistor went from 0.920V to 0.713V over a period of 60 minutes. Thus, the battery is still producing 0.713V/0.920V = 78% of the starting voltage. 2. From Table 4a, experiment #4, it can be seen that the four cell battery, with four Ag electrodes, the current went from 0.093mA (93μA) to 0.072mA (72μA) over a period of 60 minutes. Thus, the battery is still producing72μa/93μa = 78% of the starting current.

7 SECTION #4 -Discussion of Data for Experiment #4 (Table #4a & 4b, above) 3. As noted in the description of the set up for experiment #4 (point #4), the total time that the electrodes were in place up to and including the end of experiment #4 was 3.1 hours. Thus, these voltages and currents are truly amazing. 4. The voltage measured across the battery ranged from V, over a period of 60 minutes. 5. From Table 4a, it can be seen that the voltage dropped the diode is 0.45V. A voltage drop across a diode of 0.6V would indicate that the diode is conducting a significant current (i.e, the voltage drop is greater than the forward voltage). Since the measured voltage is 0.45V, the diode is below the forward voltage, but is still allowing electrons to flow through it, as indicated by a current of 93μA-72μA being measured by the ammeter. 6. Considering the data obtained at the 59 & 60 minutes time points, the sum of the voltages across the load resistor and the diode = 0.713V V = 1.163V. Note that this is not equal to the voltage across the battery (1.40V). 7. Thus, the remaining voltage is: 1.40V V = 0.237V and must dropped by the transistor/biasing resistor and the internal resistance of the battery. 8. As found in experiment #2, Table #2, the measured resistance per cell was 1600Ω and using the current from Table 60 minutes (72 μa) and using E=IR: E = ( A)(1600 Ω) = 0.115V Thus, 0.273V V = 0.122V, which should be the voltage dropped by the rest of the circuit. 9. From Table 4b, it can be seen that the average voltage of the Base-Emitter (V BE ) is 0.48V. A transistor is similar in construction to the 1N4148 diode in the circuit, i.e., the base-emitter is essentially the same as the 1N4148 diode, in a forward biased mode. 10. A transistor is considered to be "turned ON" at a voltage of 0.6V or greater, which then allows for a large collector current. Thus, at 0.48V, the transistor is not "turned ON", but this voltage is allowing electrons to flow through the transistor. 11. The data from Table #4a, shows a current range of 93μA to 72μA; thus, makes sense based on the 10,000 Ω resistor, i.e., E = IR = x 10,000 = 0.93V (0.92V measured). The battery to this point has been operating for 3.1 hours and can still produce 82.5 μa, (average of 93 &72). 12. From Table 4b, it can be seen that the average voltage of the Base-Collector (V BC ) is 0.21V. I was expecting the voltage to be similar to V BE (i.e., 0.48V). SECTION #5 - Experiment #5 (Zn/Ag electrodes) 1. Experiment #5 is the same as experiment #4, with the exception that I changed the load resistor from10,000ω to 15Ω. Table 5: Experiment #5 - Values of Circuit Current for a Circuit Load of 15 Ohm Resistance (5 x 470 μf capacitors in parallel with battery; 1N4148 diode in series with battery) (transistor 2N4401); standard time points in bold; shaded cells - data not recorded 4 Zn plated iron nails (negative) / 4 coins containing 80% Ag (positive) / 15Ω Load Resistance, digital meter Current Measured with digital ammeter inserted after load resistor the diode, the battery, highest value obtained 0.001V ma 1.4V in 1st minute 5 min 0.001V ma 0.43V 10 min 0.001V ma 1.35V 20 min 0.001V ma 0.45V 1.27V Battery 20 minutes = 3000Ω/4 or 750Ω/cell

8 SECTION #5 - Experiment #5 Discussion 1. From Table 5, experiment #5, it can be seen that the four cell battery, with four Ag electrodes, the voltage across the battery went from 1.4V to 1.27V over a period of 20 minutes. Thus, the battery is still producing 1.27V/1.4V = 91% of the starting voltage. 2. From Table 5, experiment #5, it can be seen that the four cell battery, with four Ag electrodes, the current went from 0.076mA (76μA) to 0.064mA (64μA) over a period of 20 minutes. Thus, the battery is still producing 64μA/76μA = 84% of the starting current. 3. A load of 15Ω (resistor) is a very high load circuit, so it is amazing that the battery is still producing 84% of the current after 20 minutes. 4. It should be noted that the electrodes of the battery have been in place for a total of minutes or 3.4 hours! 5. As noted before, the diode is still dropping about 0.45V. Using E = IR: 0.001V = I (15Ω) I = A or 66μA, which is very close to the 64μA measured in the experiment. 6. As well, the internal resistance of the battery has gone up from 1600Ω (see Table #2) to 3000Ω. This may look like a large increase, but one has to consider the fact that the battery has been in operation for about 3 hours and with a very high load (15Ω); thus, not a bad result at all. SECTION #6 - Lessons Learned 1. Subjecting my lime (or lemon) cells to a freeze/thaw cycle has greatly improved the battery performance, i.e., the internal resistance of the battery is about one order of magnitude lower (400Ω versus 4000Ω). 2. I definitely need a germanium diode versus the silicon diode, which should drop only 0.3V from the battery, leaving more voltage for remaining circuitry. 3. I must retain the Ag electrodes as the positive terminal of the battery. 4. I will have to use a set of earphones that are at least 2000Ω or higher resistance (preferably piezoelectric which are on the order of MΩs) so that the earphones do not download the entire circuit. 5. I do not know if I can get a germanium transistor, but it would be best because it would have a lower base-emitter voltage (thus easier to turn the transistor into the "ON mode"). This would provide a better collector current, which could be used to drive the earphones.