shrimants
1000+ Head-Fier
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scannon18: I am a computer engineering major who just "teamed up" with an EE major and we are trying to design a few basic amps. Here is part what we fould (or rather, what he wanted me to realize before we started).
Lets say that we are designing an amplifier. We want to be able to power a 600 ohm set of cans and we want it to be able to deliver 2 watts of power through these 600 ohm cans. Look at the bottlehead S.E.X. It is a 2 watt amplifier.
Power (P) is measured in watts. Ohm's Law states that Voltage (V) = Current (I) * Resistance (R). Voltage between A and B is the difference between the number of electrons at A and the number of electrons at B. Current THROUGH A or THROUGH B is the rate of flow of electrons through those points. When you say that "my usb port has +5v", you are saying that "if point B has 0 electrons, point A has 5 volts worth of electrons." Lets look at our equations now because I dont want to delve too deep into something wiki can explain better.
Power = Voltage*Current.
Doing a bit of shuffling around with our ohms law equation and substituing that equation into our power equation gives us 2 more equations for power that are a bit more useful to us.
P=I^2 * R & P=V^2/R
Now lets look at what we know. We want to be able to put 2 watts of power into a pair of 600 ohm cans. Using the above 2 equations, we can figure out how much voltage and current will be present at full power for this amplifier.
2=I^2 * 600 & 2=V^2 / 600
I = approx. 57.74milliAmps = .05774 Amps
V= approx 34.64 volts.
As you can see, we cant possibly make this a portable amplifier because we'd need a battery capable of providing nearly 35 volts constantly. Not a good idea. We can make this a desktop amp though if we design our power source correctly.
Anyways, now we've seen what happens if we plug a large load into this 2 watt amplifier. We have, effectively, the "upper limit" of our amplifier. Now, lets figure out what will happen if we plug in our crappy ipod headphones into it.
2=I^2 * 32 & 2=V^2 / 32
I=250.0 milliAmps = .25Amps
V=8Volts
See how we dropped our voltage by 1/4 but we basically quadrupled our amperage? Since we have a low impedance load in our amplifier and our amplifier still is trying to output 2 watts, we will have a gigantic current (relatively speaking) but merely 8 volts.
Why is this a problem? Because electronic components are not really built or designed to put out .25 amps of current. Thats a heck of a lot, and the components can get pretty hot. furthermore, it requires a pretty thick wire unless you want the friction of electrons to set the thing on fire.
I'm nowhere near knowledgeable enough to really know the specifics of what happens at this point. I have to contact my partner and figure out what the next step is. He has designed a current amplifier based on mosfets but he doesnt know too much about this. I figured I'd ask him what the next step is, now that we know our power requirements. We obviously need to design a 35 volt +/- power supply and some sort of power conversion circuit. We need to figure out "parts jargon" so we know what to search for when picking parts out, we need to make sure we dont exceed the limitations of our parts and such, but we also need to make sure our design can support such massive requirements.
I THINK, but im not SURE, that if we design our amp to output X volts at Y amps, that we cannot change this. So for example, I think we would need to design this amplifier to actually output 35 volts and .25 amps. That is much higher power than we need, but I think adding the headphones fixes it. Either that or we pick a single worst case scenario, ie the 600ohm headphone, and we design it around that, or we go with 35 volts max and .25 amps max and we design it like that (making it an 8 watt amplifier, which is way too much so that cant be right). I'm not entirely sure where to go from here.
Lets say that we are designing an amplifier. We want to be able to power a 600 ohm set of cans and we want it to be able to deliver 2 watts of power through these 600 ohm cans. Look at the bottlehead S.E.X. It is a 2 watt amplifier.
Power (P) is measured in watts. Ohm's Law states that Voltage (V) = Current (I) * Resistance (R). Voltage between A and B is the difference between the number of electrons at A and the number of electrons at B. Current THROUGH A or THROUGH B is the rate of flow of electrons through those points. When you say that "my usb port has +5v", you are saying that "if point B has 0 electrons, point A has 5 volts worth of electrons." Lets look at our equations now because I dont want to delve too deep into something wiki can explain better.
Power = Voltage*Current.
Doing a bit of shuffling around with our ohms law equation and substituing that equation into our power equation gives us 2 more equations for power that are a bit more useful to us.
P=I^2 * R & P=V^2/R
Now lets look at what we know. We want to be able to put 2 watts of power into a pair of 600 ohm cans. Using the above 2 equations, we can figure out how much voltage and current will be present at full power for this amplifier.
2=I^2 * 600 & 2=V^2 / 600
I = approx. 57.74milliAmps = .05774 Amps
V= approx 34.64 volts.
As you can see, we cant possibly make this a portable amplifier because we'd need a battery capable of providing nearly 35 volts constantly. Not a good idea. We can make this a desktop amp though if we design our power source correctly.
Anyways, now we've seen what happens if we plug a large load into this 2 watt amplifier. We have, effectively, the "upper limit" of our amplifier. Now, lets figure out what will happen if we plug in our crappy ipod headphones into it.
2=I^2 * 32 & 2=V^2 / 32
I=250.0 milliAmps = .25Amps
V=8Volts
See how we dropped our voltage by 1/4 but we basically quadrupled our amperage? Since we have a low impedance load in our amplifier and our amplifier still is trying to output 2 watts, we will have a gigantic current (relatively speaking) but merely 8 volts.
Why is this a problem? Because electronic components are not really built or designed to put out .25 amps of current. Thats a heck of a lot, and the components can get pretty hot. furthermore, it requires a pretty thick wire unless you want the friction of electrons to set the thing on fire.
I'm nowhere near knowledgeable enough to really know the specifics of what happens at this point. I have to contact my partner and figure out what the next step is. He has designed a current amplifier based on mosfets but he doesnt know too much about this. I figured I'd ask him what the next step is, now that we know our power requirements. We obviously need to design a 35 volt +/- power supply and some sort of power conversion circuit. We need to figure out "parts jargon" so we know what to search for when picking parts out, we need to make sure we dont exceed the limitations of our parts and such, but we also need to make sure our design can support such massive requirements.
I THINK, but im not SURE, that if we design our amp to output X volts at Y amps, that we cannot change this. So for example, I think we would need to design this amplifier to actually output 35 volts and .25 amps. That is much higher power than we need, but I think adding the headphones fixes it. Either that or we pick a single worst case scenario, ie the 600ohm headphone, and we design it around that, or we go with 35 volts max and .25 amps max and we design it like that (making it an 8 watt amplifier, which is way too much so that cant be right). I'm not entirely sure where to go from here.
