The new technology is expected to come out soon, or so they say.
http://www2.tek.com/cmswpt/prdetails.lotr?ct=PR&cs=News+Release&ci=17452&lc=EN
Sunday, October 17, 2010
200 GHz processers?
Tektronix, Inc plans on taking some IBM chips well beyond 30 GHz. They're aiming for 200!
The new technology is expected to come out soon, or so they say.
http://www2.tek.com/cmswpt/prdetails.lotr?ct=PR&cs=News+Release&ci=17452&lc=EN
The new technology is expected to come out soon, or so they say.
http://www2.tek.com/cmswpt/prdetails.lotr?ct=PR&cs=News+Release&ci=17452&lc=EN
Saturday, October 16, 2010
Bush Gardens
Hey guys,
I went to the Bush Gardens amusement park nearby today. Just got back, so no neat post for today, obviously. Sorry about that.
I planning a few things for tomorrow's, and I think I've got something together. Lot's of homework coming up next week as well, so plenty of material I need to contemplate. See you 'round!
I went to the Bush Gardens amusement park nearby today. Just got back, so no neat post for today, obviously. Sorry about that.
I planning a few things for tomorrow's, and I think I've got something together. Lot's of homework coming up next week as well, so plenty of material I need to contemplate. See you 'round!
Friday, October 15, 2010
Physical constants
Hey guys,
In order to get the A's that the girl in my class is getting, I have to memorize some stuff. Here's some of the physical constants I'll be using in my equations. Time to type some stuff up in excel!
I might do some derivations for the equations I have to memorize later today. Either that, or I'll post the properties of intrinsic Silicon and intrinsic Gallium Arsenide I have to memorize. Thanks for sticking around, and ask any questions you might have about any posts.
In order to get the A's that the girl in my class is getting, I have to memorize some stuff. Here's some of the physical constants I'll be using in my equations. Time to type some stuff up in excel!
| Physical Constants | ||
| Angstrom | Å | 0.1 nm |
| Avagadro constant | Nav | 6.022x10^23 |
| Boltzmann constant | k | 1.38066x10^-23 J/K |
| Elementary charge | q | 1.602x10^-19 C |
| Electron rest mass | m0 | 9.1094x10^-31 kg |
| Electron Volt | eV | 1.602x10^-19 J |
| Gas constant | R | 1.98719 cal/mol*K |
| Permeability in vacuum | μ0 | 1.25664x10^-8 H/cm |
| Permittivity in vacuum | ε0 | 8.85418x10^-12 F/cm |
| Planck's constant | h | 6.62607x10^-34 J*s |
| Reduced Planck's constant | ℏ | 1.05457x10^-34 J*s (or h/2π) |
| Speed of light in a vacuum | c | 2.99792x10^10 cm/s |
| Thermal voltage at 300K | kT/q | 0.025852 V |
| Wavelength of 1 eV quantum | λ | 1.23984μm |
I might do some derivations for the equations I have to memorize later today. Either that, or I'll post the properties of intrinsic Silicon and intrinsic Gallium Arsenide I have to memorize. Thanks for sticking around, and ask any questions you might have about any posts.
Thursday, October 14, 2010
Failed Tests and The Hall Effect
Ok guys,
Just got back my Semiconductor materials test, very low grade (like most of the class, actually). :(
Talked with some of the students from last year, and they said it was a similar situation then. The tests this teacher give out are monsterously difficult, nothing like the homework or examples, and hardly any information is given about each problem. One person scored 100, and she actually memorized dozens of physical constants as well as nearly all the parameters of intrinsic Silicon and intrinsic Gallium Arsenide semiconductors at 300 kelvin...so yeah. She did a great job (and she's rather cute, too). We'll have to put in as many hours as she does if we're going to pass this class. I'm still upset the teacher refused to give us most of the physical constants and none of the equations we needed, but it really made me think during the test.
My burning academic career aside, let's talk about the Hall Effect. Here's a nifty picture taken from my professor's lecture (forgive my larcenous ways!)
Let's say we have a chunk of semiconductor material that's p-type (doped so that the carrier concentration favors holes left by absent electrons).
Let's say we jam some wires on the left and right side and run a voltage through that thing positive side on the left, negative on the right. What will happen? The voltage will push the holes through the semiconductor torwards the right (x-axis).
Now let's say we get an electromagnet and place a pole on the long, close side and another pole on the far side(z-axis). Now what happens? The Hall Effect.
A magnetic field perpendicular to the hole's path forms on the z-axis, inducing an electric field downwards through the semiconductor material. This electric field applies an upwards force on the hole. Like in the picture, the hole curves up under the effect of two forces (one right, one up). This leads to a collection of negative charge on the bottom of the material, and a collection of positive charge on the top of the material.
Now if you were to break out a meter and read the top and bottom of the material, you'd get a voltage reading! Magnet off, nothing. Magnet back on, Voltage!
Just by applying a voltage on either end and a magnetic field, a new voltage can be created. This new voltage can even help you calculate how many atoms of 'doping' material are in the semiconductor, a difficult feat if you don't know it and it isn't labeled.
There are several pages of equations (I have to memorize like everything else -_-) and derivations to actually prove all this. I could touch on it later if anyone wants me to go through that torture.
Later guys
Just got back my Semiconductor materials test, very low grade (like most of the class, actually). :(
Talked with some of the students from last year, and they said it was a similar situation then. The tests this teacher give out are monsterously difficult, nothing like the homework or examples, and hardly any information is given about each problem. One person scored 100, and she actually memorized dozens of physical constants as well as nearly all the parameters of intrinsic Silicon and intrinsic Gallium Arsenide semiconductors at 300 kelvin...so yeah. She did a great job (and she's rather cute, too). We'll have to put in as many hours as she does if we're going to pass this class. I'm still upset the teacher refused to give us most of the physical constants and none of the equations we needed, but it really made me think during the test.
My burning academic career aside, let's talk about the Hall Effect. Here's a nifty picture taken from my professor's lecture (forgive my larcenous ways!)
Let's say we have a chunk of semiconductor material that's p-type (doped so that the carrier concentration favors holes left by absent electrons).
Let's say we jam some wires on the left and right side and run a voltage through that thing positive side on the left, negative on the right. What will happen? The voltage will push the holes through the semiconductor torwards the right (x-axis).
Now let's say we get an electromagnet and place a pole on the long, close side and another pole on the far side(z-axis). Now what happens? The Hall Effect.
A magnetic field perpendicular to the hole's path forms on the z-axis, inducing an electric field downwards through the semiconductor material. This electric field applies an upwards force on the hole. Like in the picture, the hole curves up under the effect of two forces (one right, one up). This leads to a collection of negative charge on the bottom of the material, and a collection of positive charge on the top of the material.
Now if you were to break out a meter and read the top and bottom of the material, you'd get a voltage reading! Magnet off, nothing. Magnet back on, Voltage!
Just by applying a voltage on either end and a magnetic field, a new voltage can be created. This new voltage can even help you calculate how many atoms of 'doping' material are in the semiconductor, a difficult feat if you don't know it and it isn't labeled.
There are several pages of equations (I have to memorize like everything else -_-) and derivations to actually prove all this. I could touch on it later if anyone wants me to go through that torture.
Later guys
Today's Post
Today's post is going to be later in the day than usual. I'm in class all day, so I don't think I'll get around to it until later. See you then, guys!
Wednesday, October 13, 2010
Power's relationship with Ohm's Law
Hey guys,
Today we're going to quickly review power's relationship with Ohm's Law (what we covered last time, V=I*R). Power is what your energy company charges you for, so you might want to pay attention!
We'll add on P = I * V
Where P = Watts (w) of power. One watt is equal to one Joule of energy expended in one second.
Well, what if we only have Voltage and Resistance? What if we only have Current and resistance? How can we solve for the power in a circuit? Ohm's Law (you'll see Ohm's Law a lot)
Since we have V = I*R, then we also have I = V/R (from dividing both sides by R)
Now substitute IR for V, or V/R for I (depending on what we're given to solve for P, and we have two more equations for power:
P = (V^2)/R
P = (I^2)*R
Now it's easy to solve for P if you're only given two of the variables in a circuit.
A great picture can be found at the12volt.com that neatly illustrates the relationship between power, voltage, current, and resistance. Check it out.
http://www.the12volt.com/ohm/ohmslaw.asp
Today we're going to quickly review power's relationship with Ohm's Law (what we covered last time, V=I*R). Power is what your energy company charges you for, so you might want to pay attention!
We'll add on P = I * V
Where P = Watts (w) of power. One watt is equal to one Joule of energy expended in one second.
Well, what if we only have Voltage and Resistance? What if we only have Current and resistance? How can we solve for the power in a circuit? Ohm's Law (you'll see Ohm's Law a lot)
Since we have V = I*R, then we also have I = V/R (from dividing both sides by R)
Now substitute IR for V, or V/R for I (depending on what we're given to solve for P, and we have two more equations for power:
P = (V^2)/R
P = (I^2)*R
Now it's easy to solve for P if you're only given two of the variables in a circuit.
A great picture can be found at the12volt.com that neatly illustrates the relationship between power, voltage, current, and resistance. Check it out.
http://www.the12volt.com/ohm/ohmslaw.asp
Tuesday, October 12, 2010
Understanding Ohm's Law (and other equations)
Ohm's Law, I believe, is the most used equation in electronics analysis. It is the fundamental building block for knowing what's going on in any particular circuit.
V = I * R
Where:
V is the potential in Volts (V)
I is the current flow in Amperes (A)
R is the resistance in Ohms (Ω)
So, one Volt of potential is equal to one Ampere of current multiplied by one Ohm of resistance. Using this equation, if you have two know variables you can easily solve for the third.
Let me explain what these variables mean.
Voltage (V)
Voltage is the electric potential of a circuit. Electricity flows like water. If you have two bodies of water, one higher than the other and both connected with a sloping pipe, water would flow from the higher potential body of water to the lower potential body of water. In the diagram below, the water from Tank A should flow towards Tank B because of the potential difference they have.
Current (A)
The electric current in a circuit is the flow of electrons through it. A single electron has a charge of roughly 1.602 x 10 ^-19 Coulombs. One Ampere is equal to one Coulomb divided by one Second.
This is the flow of charge per second. An example would be to measure the amount of water leaving Tank A above and flowing into Tank B per second. That would be the current of water flowing.
Resistance (Ω)
The resistance of a circuit is a measurement of how hard it resists the flow of charge. In our water example above, there are two examples of resistance. One is the friction from the pipe slowing the water down, and the other is the size of the pipe itself. If the pipe is too small, only a limited amount of water can flow through. In a circuit, the resistivity of the materials is like the friction inside the pipe. The length of the wires connecting the devices is kind of like the width of the pipe.
So we have Ohm's Law at
The electrical potential of a circuit is equal to the flow of charge each second multiplied by the resistance to that charge. A concept to be thought about, rather than a bunch of letters.
With all the equations being thrust upon you at school, it's not hard to see them as just letters and numbers to be memorized and later forgotten. If you look at a formula and break it down into its units, and try to voice out what the formula implies, you can really grasp what's going on and learn a lot.
V = I * R
Where:
V is the potential in Volts (V)
I is the current flow in Amperes (A)
R is the resistance in Ohms (Ω)
So, one Volt of potential is equal to one Ampere of current multiplied by one Ohm of resistance. Using this equation, if you have two know variables you can easily solve for the third.
Let me explain what these variables mean.
Voltage (V)
Voltage is the electric potential of a circuit. Electricity flows like water. If you have two bodies of water, one higher than the other and both connected with a sloping pipe, water would flow from the higher potential body of water to the lower potential body of water. In the diagram below, the water from Tank A should flow towards Tank B because of the potential difference they have.
Current (A)
The electric current in a circuit is the flow of electrons through it. A single electron has a charge of roughly 1.602 x 10 ^-19 Coulombs. One Ampere is equal to one Coulomb divided by one Second.
This is the flow of charge per second. An example would be to measure the amount of water leaving Tank A above and flowing into Tank B per second. That would be the current of water flowing.
Resistance (Ω)
The resistance of a circuit is a measurement of how hard it resists the flow of charge. In our water example above, there are two examples of resistance. One is the friction from the pipe slowing the water down, and the other is the size of the pipe itself. If the pipe is too small, only a limited amount of water can flow through. In a circuit, the resistivity of the materials is like the friction inside the pipe. The length of the wires connecting the devices is kind of like the width of the pipe.
So we have Ohm's Law at
The electrical potential of a circuit is equal to the flow of charge each second multiplied by the resistance to that charge. A concept to be thought about, rather than a bunch of letters.
With all the equations being thrust upon you at school, it's not hard to see them as just letters and numbers to be memorized and later forgotten. If you look at a formula and break it down into its units, and try to voice out what the formula implies, you can really grasp what's going on and learn a lot.
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