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### Re: Circuit Lab B/C

Posted: September 23rd, 2018, 9:13 am
All right!

Describe the behavior of an RC circuit which is a) charging and b) discharging.
a) The voltage through the resistor starts high and decays exponentially as the capacitor becomes saturated.  The charged capacitor has an extremely high voltage drop.
b) The voltage through the resistor starts high and decays exponentially as the capacitor discharges.  The capacitor acts as a voltage source in the absence of other sources.

### Re: Circuit Lab B/C

Posted: September 26th, 2018, 2:53 pm
Do current and voltage sources exist in real circuits?

### Re: Circuit Lab B/C

Posted: September 26th, 2018, 3:34 pm
Do current and voltage sources exist in real circuits?
Not ideal ones but they can be approximated; however, real-life sources have a limited amount of charge (which can be measured in ampere-hours) and have internal resistance that sometimes needs to be accounted for.

### Re: Circuit Lab B/C

Posted: September 26th, 2018, 4:47 pm
Do current and voltage sources exist in real circuits?
Not ideal ones but they can be approximated; however, real-life sources have a limited amount of charge (which can be measured in ampere-hours) and have internal resistance that sometimes needs to be accounted for.

### Re: Circuit Lab B/C

Posted: September 26th, 2018, 4:55 pm
What is doping in the context of PN junctions and why is it done?

### Re: Circuit Lab B/C

Posted: September 26th, 2018, 8:13 pm
What is doping in the context of PN junctions and why is it done?
Pure silicon has four valence electrons that are able to conduct electricity. Doping adjusts the concentrations of the charge carriers within the silicon crystals to either increase the number of electrons or increase the number of "holes" (aka lack of electrons). Doping with elements that have 5 valence electrons such as arsenic increases the number of electrons in the silicon crystal. This makes it "negative" or n-type. Doping with elements that have 3 valence electrons such as boron or indium creates silicon crystals that have a high concentration of holes, making it "positive" or p-type. This causes electrons to move only in one direction: from n-type to p-type. This is essentially a silicon diode. The property of allowing electrons to only flow in one direction forms the basis for more advanced transistors such as BJTs and MOSFETs.

### Re: Circuit Lab B/C

Posted: September 27th, 2018, 1:19 pm
What is doping in the context of PN junctions and why is it done?
Pure silicon has four valence electrons that are able to conduct electricity. Doping adjusts the concentrations of the charge carriers within the silicon crystals to either increase the number of electrons or increase the number of "holes" (aka lack of electrons). Doping with elements that have 5 valence electrons such as arsenic increases the number of electrons in the silicon crystal. This makes it "negative" or n-type. Doping with elements that have 3 valence electrons such as boron or indium creates silicon crystals that have a high concentration of holes, making it "positive" or p-type. This causes electrons to move only in one direction: from n-type to p-type. This is essentially a silicon diode. The property of allowing electrons to only flow in one direction forms the basis for more advanced transistors such as BJTs and MOSFETs.

### Re: Circuit Lab B/C

Posted: September 27th, 2018, 2:13 pm
Calculate the RMS voltage of a 12V peak-to-peak sinusoidal AC signal and a 0-6V square wave with a time-on of 8ms and a period of 10 ms.

### Re: Circuit Lab B/C

Posted: September 27th, 2018, 3:31 pm
Calculate the RMS voltage of a 12V peak-to-peak sinusoidal AC signal and a 0-6V square wave with a time-on of 8ms and a period of 10 ms.
$12\ V \cdot \frac{\sqrt2}{4} = 3\sqrt2\ V$

$\sqrt{\frac{8\ ms}{10\ ms} * (6\ V)^2} = 5.367\ V$

### Re: Circuit Lab B/C

Posted: September 27th, 2018, 7:45 pm
Calculate the RMS voltage of a 12V peak-to-peak sinusoidal AC signal and a 0-6V square wave with a time-on of 8ms and a period of 10 ms.
$12\ V \cdot \frac{\sqrt2}{4} = 3\sqrt2\ V$

$\sqrt{\frac{8\ ms}{10\ ms} * (6\ V)^2} = 5.367\ V$
Looks good!