I had to import some libraries to get the parts I need to use, and then it was simply place parts in a sensible pattern and connecting them with wire.
Below is my PSpice for working out the last problem of the 8th WebAssign assignment.
Wednesday, March 23, 2011
PSpice Fun
I had to import some libraries to get the parts I need to use, and then it was simply place parts in a sensible pattern and connecting them with wire.
Below is my PSpice for working out the last problem of the 8th WebAssign assignment.
Tuesday, March 22, 2011
More pictures from Node Analysis
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| This shows a reading for our V1 during the last problem when we were trying to make it and V2 equal to 9V. |
Friday, March 18, 2011
Nodal Analysis is for Knowing Nerd Noggins
We (James Dunn and I) used nodal analysis to determine the theoretical voltage between two points and a reference node AKA "ground" and the currents coming from two voltage sources.
Our theorectical values for V2 and V3 were 10.26V and 8.67V respectively.
We determined that IBat 1 and IBat 2 were 17.4mA and 1.48mA, which would give power supplies of 208.8mW and 13.35mW respectively.
In the last part, for extra fun, we figured out what voltages we would need to load across the circuit to make both V2 and V3 equal 9V. We worked out that VL1 would be 9.9V and VL2 would be 10.98V. To get these particular load voltages, we would need to use a voltage divider to bring 12V down to these voltages.
We did this with a resistor box of 213Ω in series with Vbat1 and 103Ω in series with Vbat2. We got as close as 9.09V for V2 and 8.99V for V3, after fiddling with the resistances a bit. :) Our currents became 9.90mA and 8.61mA for Ibat1 and Ibat2 respectively.
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| Circuit Diagram |
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| I'll get a better picture from James soon. |
Our theorectical values for V2 and V3 were 10.26V and 8.67V respectively.
We determined that IBat 1 and IBat 2 were 17.4mA and 1.48mA, which would give power supplies of 208.8mW and 13.35mW respectively.
In the last part, for extra fun, we figured out what voltages we would need to load across the circuit to make both V2 and V3 equal 9V. We worked out that VL1 would be 9.9V and VL2 would be 10.98V. To get these particular load voltages, we would need to use a voltage divider to bring 12V down to these voltages.
We did this with a resistor box of 213Ω in series with Vbat1 and 103Ω in series with Vbat2. We got as close as 9.09V for V2 and 8.99V for V3, after fiddling with the resistances a bit. :) Our currents became 9.90mA and 8.61mA for Ibat1 and Ibat2 respectively.
Tuesday, March 15, 2011
Voltage Dividers
Given 1 to 3 loads of parallel 1kΩ resistors, we calculated the Vsource and Rsource of an unregulated power supply to make the voltage variations on the loads + or - 5% of 5V.
We recognized that the REQ,max and the REQ,min were 333Ω and 1kΩ, respectively, and that the larger resistance corresponded with the larger VBUS. I worked out then the following:
We made sure nothing would be broken by the circuit we were planning to make, so we put it together. Here's a picture with all 3 loads connected.
We recognized that the REQ,max and the REQ,min were 333Ω and 1kΩ, respectively, and that the larger resistance corresponded with the larger VBUS. I worked out then the following:
- VS = 5.54V
- RS = 55.5Ω
- IBUS,max = 14.26mA
- IBUS,min = 5.25 mA
Below is collected data. (I'll fix the format later.)
| Color Code | Nominal Value | Measured Value | Wattage (W) | |
| Brwn,Blck,Red,Ag | 1 kΩ | 0.995 | kΩ | 0.25 |
| Brwn,Blck,Red,Ag | 1 kΩ | 1.001 | kΩ | 0.25 |
| Brwn,Blck,Red,Ag | 1 kΩ | 0.989 | kΩ | 0.25 |
| Resistor Box | 55 Ω | 56.4 | Ω | 0.3 |
| Power Supply | 6 V | 6.06 | V | NA |
We made sure nothing would be broken by the circuit we were planning to make, so we put it together. Here's a picture with all 3 loads connected.
These are the results:
| Configuration | REQ (Ω) | VBUS (V) | I BUS(mA) | Pload(mW) | |
| 1 Load | 1001 | 5.75 | 5.74 | 33.0 | |
| 2 Loads | 499 | 6.74 | 10.9 | 66.3 | |
| 3 Loads | 332 | 5.20 | 15.6 | 81.2 |
The power of 2 loads was found by P = V^2/R = (5.46 V)^2 / 499Ω = 66.3 mW
The actual % variations were quite different from the theoretical because we used 6.06 V instead of 5.54V.
There would be theoretically a -9.4% variation from 5V if we were using a 5.54V power supply and added a fourth 1kΩ load.
If we redesign it for a ±1% variation of VBUS from 5V, then we would have a RS = 10.2Ω and VS= 5.10V.
Saturday, March 12, 2011
LED circuit
On Monday we built a circuit to light two LEDs (light emitting diodes), a red (LED1) and a yellow (LED2), maximally without burning them out with too much wattage.
The red LED's given voltage and current ratings were 5V and 22.75mA respectively, yellow's were 2V and 20mA.
The LEDs were placed in a circuit in parallel with a resistor (R1 and R2) in series on each. A 9V supply was given by a power supply.
Ohm's Law was used to calculate the equivalent resistances of the LEDs.
RLED1 = 219.8Ω, RLED2 = 100Ω
Using the current and voltage ratings, theoretical voltages and resistances values were found for R1 and R2.
V1 = 4V, V2 = 7V
R1 = 176Ω, R2 = 350Ω
We only had access to 100, 150, 220, 360, and 470-Ohm resistors, so we approximated R1 with a 150Ω and R2 with a 360Ω. Their measured values were 148.2Ω and 360Ω respectively. Both were quarter-watt resistors.
Three configurations were used:
The red LED's given voltage and current ratings were 5V and 22.75mA respectively, yellow's were 2V and 20mA.
The LEDs were placed in a circuit in parallel with a resistor (R1 and R2) in series on each. A 9V supply was given by a power supply.
Ohm's Law was used to calculate the equivalent resistances of the LEDs.
RLED1 = 219.8Ω, RLED2 = 100Ω
Using the current and voltage ratings, theoretical voltages and resistances values were found for R1 and R2.
V1 = 4V, V2 = 7V
R1 = 176Ω, R2 = 350Ω
Three configurations were used:
- Both LEDs in the circuit
- Without LED2
- Without LED1
These are the results:
Using this data, I calculated a few things.
a) If the supply was just a 9V battery with a 0.2 A-hr capacity of useful voltage, the circuit in config1 would last 4.68 hours.
b) The % error between experimental and desired values of LED current were 36.3% for LED1 and 1.45% for LED2. This was caused by resistor restrictions.
c) The efficiency of the circuit in configuration 1 was found to be 40.8%, which is pretty poor.
d) Assuming current is the same, if a 6V battery was used instead of a 9V battery, ... I really don't understand this problem.
Oh, yeah, and we blew up a red LED by overloading it with voltage.
| Configuration | I of LED1 (mA) | V of LED1 (V) | I of LED2 (mA) | V of LED2 (V) | I of supply (mA) |
| 1 | 14.5 | 6.68 | 19.71 | 19.71 | 35.1 |
| 2 | 14.8 | 6.74 | NA | NA | 15.1 |
| 3 | NA | NA | 20.2 | 1.65 | 20.3 |
Using this data, I calculated a few things.
a) If the supply was just a 9V battery with a 0.2 A-hr capacity of useful voltage, the circuit in config1 would last 4.68 hours.
b) The % error between experimental and desired values of LED current were 36.3% for LED1 and 1.45% for LED2. This was caused by resistor restrictions.
c) The efficiency of the circuit in configuration 1 was found to be 40.8%, which is pretty poor.
d) Assuming current is the same, if a 6V battery was used instead of a 9V battery, ... I really don't understand this problem.
Oh, yeah, and we blew up a red LED by overloading it with voltage.
Sunday, March 6, 2011
Modelling Power lines
In our first official lab, we worked out the maximum length of cable between a load and its power source using a resistor box to model the length of cable.
The ideal power source of 12V was modeled with a power supply. The load required 11 V across it and it was worked to have a resistance of 1 kΩ so it was modeled with the appropriate resistor (brown black red).
This is the setup of another group. Gerrek (I have no idea about the spelling), my lab partner, took pictures of our set up, but I haven't received a message from him (I gave him my card with my contact info) and didn't get his contact info, so I may have to upload the picture he took after class on Monday.
We used two multimeters to simultaneously measure the current in the circuit and the voltage across the load resistor. We varied the resistance of the resistor box until the voltage across the resistor was 11.00V.
At this point: current = 10.93mA and R of total cable = 97Ω.
I worked out the time the hypothetical battery rated at 0.8Ahr would discharge in 73.2 hours, that the power to the load was 0.1195W and the lost in the cable was 0.01159W, and thus the efficiency was 91.2%.
The resistor box's rating of 0.3W was not exceeded in this experiment. Finally, I worked out that if the cable were AWG #30 wire, the load could be as far as 142m away from the power source.
The ideal power source of 12V was modeled with a power supply. The load required 11 V across it and it was worked to have a resistance of 1 kΩ so it was modeled with the appropriate resistor (brown black red).
We used two multimeters to simultaneously measure the current in the circuit and the voltage across the load resistor. We varied the resistance of the resistor box until the voltage across the resistor was 11.00V.
At this point: current = 10.93mA and R of total cable = 97Ω.
I worked out the time the hypothetical battery rated at 0.8Ahr would discharge in 73.2 hours, that the power to the load was 0.1195W and the lost in the cable was 0.01159W, and thus the efficiency was 91.2%.
The resistor box's rating of 0.3W was not exceeded in this experiment. Finally, I worked out that if the cable were AWG #30 wire, the load could be as far as 142m away from the power source.
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