## Circuit Lab B/C

mdv2o5
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### Re: Circuit Lab B/C

UTF-8 U+6211 U+662F wrote: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.

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### Re: Circuit Lab B/C

mdv2o5 wrote:
UTF-8 U+6211 U+662F wrote: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.

mdv2o5
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### Re: Circuit Lab B/C

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.

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### Re: Circuit Lab B/C

mdv2o5 wrote: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.
$12\ V \cdot \frac{\sqrt2}{4} = 3\sqrt2\ V$

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

mdv2o5
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### Re: Circuit Lab B/C

UTF-8 U+6211 U+662F wrote:
mdv2o5 wrote: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!

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### Re: Circuit Lab B/C

All right! Derive an equation for the power given off by a resistor from Ohm's law.

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### Re: Circuit Lab B/C

UTF-8 U+6211 U+662F wrote:All right! Derive an equation for the power given off by a resistor from Ohm's law.
By Ohm's law, $V = IR$.  By the definition of power, $P = VI$.  Thus, $P = (IR)I = I^2R$.

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### Re: Circuit Lab B/C

Jacobi wrote:
UTF-8 U+6211 U+662F wrote:All right! Derive an equation for the power given off by a resistor from Ohm's law.
By Ohm's law, $V = IR$.  By the definition of power, $P = VI$.  Thus, $P = (IR)I = I^2R$.
Yep, your turn, although I was thinking of a derivation using
$P = \frac{dW}{dt}$
But that's a little more complicated (and equivalent).

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### Re: Circuit Lab B/C

Two resistors are combined in a circuit. First, they are combined in series. Then, they are combined in parallel. What is the equivalent single resistance of each setup?

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### Re: Circuit Lab B/C

Jacobi wrote:Two resistors are combined in a circuit. First, they are combined in series. Then, they are combined in parallel. What is the equivalent single resistance of each setup?
$R_{series} = R_1 + R_2$
$R_{parallel} = \frac1{\frac1{R_1} + \frac1{R_2}}$

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