To start off a DC analysis, you must first launch the Capture Student application. Then click File>New>Project from the menu.Click on Analog or Mixed. The next page will ask if you want to open an existing or a blank project, click blank project. To start on the circuit, all wires, resistors, current sources, and voltage sources will be found on Place on the menu bar. Once the circuit is finished go to PSpice>new simulation profile. Type Bias in the name field and click create. Select Bias Point under the Analysis type and click OK. Finally, run the simulation by selecting PSpice from the menu and selecting run.
This an example done for one of the Homework problems from assignment 7
Friday, March 25, 2011
Friday, March 18, 2011
Nodal Analysis
For this lab, we were asked analyze a "reliable" circuit that contains multiple loads and two power supplies. We were given the values for all resistors and both power supplies. With these we calculated what our voltages were supposed to be using nodal analysis. When we finished, we needed to then make a circuit to be exactly like the one we used to calculated our voltages. With the multi-meters, we measured the voltage and current values and compared them with our calculated values.
Variable Theoretical value Measured Value Percent Error
I_Batt1 17.45 mA 17.43 mA 0.11%
I_Batt2 14.82 mA 15.24 mA 2.79%
V_1 10.255 V 10.17 V 0.88%
V_2 8.674 V 8.58 V 1.09%
Monday, March 14, 2011
Voltage Dividers
In a house where there are multiple loads in a circuit, there is going to be some kind of variation in the voltage source. We are going to create a circuit that will help us understand what our voltage source should be. With the given values for the resistance loads and voltage bus max and min, we can calculate what our voltage source and our bus resistance should be.
V_source= 5.54 volts, R_s=55.56 ohms
Voltage bus max = 5.25 volts, Voltage bus min = 4.75 volts. (A 5 percent variation about 5 volts).
This is what our circuit was like when all resistance loads where connected to each other.
Calculated:
Voltage bus min= 5.18 volts, Voltage bus max= 5.72 volts
Because our Voltage source could not have been set to 5.54 volts, we were strictly forced to make it 6 volts. This new voltage changed our voltage variation by 9.91 percent.
V_source= 5.54 volts, R_s=55.56 ohms
Voltage bus max = 5.25 volts, Voltage bus min = 4.75 volts. (A 5 percent variation about 5 volts).
This is what our circuit was like when all resistance loads where connected to each other.
Calculated:
Voltage bus min= 5.18 volts, Voltage bus max= 5.72 volts
Because our Voltage source could not have been set to 5.54 volts, we were strictly forced to make it 6 volts. This new voltage changed our voltage variation by 9.91 percent.
Friday, March 11, 2011
Introduction to Biasing
In this lab we are supposed to light up two LEDs connected in parallel with a 9V power supply. However these LEDs are rated for 2V and 5V. Which means in parallel configuration, both are receiving 9V of the power supply and will burn out almost instantly. To prevent this from happening we have to add a resistor to accompany each of the LEDs. Given the rated voltage of the LEDs and the current flowing throw them, we calculated our resistors to be:
R_1= 175.82 ohms. To be connected in series with LED rated at 5 Volts
R_2= 350 ohms. To be connected in series with LED rated at 2 Volts.
The closest available resistors we can get for this set up is
R_1= 150 ohms
R_2= 360 ohms
Our data for this configuration:
Voltage for LED_1 = 6.68
Voltage for LED_2 = 1.64
R_1= 175.82 ohms. To be connected in series with LED rated at 5 Volts
R_2= 350 ohms. To be connected in series with LED rated at 2 Volts.
The closest available resistors we can get for this set up is
R_1= 150 ohms
R_2= 360 ohms
Our data for this configuration:
Voltage for LED_1 = 6.68
Voltage for LED_2 = 1.64
Sunday, March 6, 2011
Introduction To DC Circuits Lab
In this lab, we were to set up a circuit that consists of a power supply, a voltmeter, an ammeter, and two resistors. Our goal in this lab was to find the distribution efficiency of power in this circuit. We found the efficiency by making the voltage of the load to be 11.00 Volts. From here we take the value of current from the ammeter and the resistance of the resistor. In the end we found that our calculated efficiency was 91.86 percent.
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