|Feb3-13, 02:01 AM||#18|
electronic devices operating voltage and current
A fact that is technically necessary is that you can't get more power out of a converter ( or anything) than you put into it.
So, if you have 5 volts as your input voltage, as you said you did, then you need at least 2 amps in to get 10 watts out.
And the converters have losses, so more than 10 watts input would be needed.
DC to DC converters are sold on EBay and these sellers give very precise information about what input and output voltages and currents are possible with each converter.
So, if you want to use some other voltages, you would read the specifications to make a decision about each converter.
|Feb3-13, 05:53 PM||#19|
Consider this: if all electrical appliances / devices were designed to work from a particular current and power supplies were required to feed them with that current then HOW would you connect multiple devices to the same power supply? They would need to be connected in SERIES, for a start. You would need to specify that they all take the same amount of current - so the more you connected in series, the greater voltage you would need to supply and they would all have different voltage drop (depending upon the power of each one. Also - if you didn't want any power out of your supply, you would need to SHORT CIRCUIT it or the volts would go up and up and up - as the device 'tried' to deliver its nominal current. Removing an appliance from its socket, you would need to replace it with a shorting link, so the other devices could be powered. A proverbial nightmare and a truly upside down world.
The highest power devices would need to have the highest resistances (having the highest voltage) - compared with voltage driven devices where the highest power device has the lowest resistance (taking the highest current).
There are some components that need to be supplied with a particular current, of course, but they are exceptions and need specially designed circuitry.
|Feb4-13, 02:23 PM||#20|
The reason that power supplies are usually sources of voltage with low source (internal) resistance is because they dissipate very little power internally over a massive range of supply currents.
Ok, I see that: the power sources does not consume too much power if the internal resistance is zero. That does not assure that the most power possible goes to the load (impedance matching theorem, correct?)
In the case of a battery, the voltage is rather constant and the current can change depending on the load resistance. That is why a battery is a constant voltage source and not a constant current source.
Electrical devices need electrical power to function. Power is simply energy per time.
Power is given if both a nonzero current and voltage are applied to the device. The current seems to be determined by powered device while the voltage by the power source....is that a correct general statement?
For instance, in the case of a single solar cell at max illumination and connected to a certain load, both the voltage and current coming out of the solar cell depend on the cell internal resistance and electronic device impedance, correct?
A solar cell, under ideal illumination, reaches a max open load voltage (like a battery has a fixed voltage). The current can increase depending on how much solar light is incident on the cell. The more the light, the higher the max drawable current can be.....correct?
But all depends on the load, correct?
|Feb4-13, 04:55 PM||#21|
Solar cells are interesting because they definitely don't have a low internal resistance and, to get the most energy our of them for any particular light flux, you would really need to be able to vary the load dynamically. I don't know if they use switch mode control, in practice but I'd bet that could be made to give the optimum power output.
|Feb5-13, 08:51 AM||#22|
"you'd be wasting half the power you generated".
I think the correct statement would be that we would be wasting half of the maximum amount of power we can generate.
"Solar cells are interesting because they definitely don't have a low internal resistance and, to get the most energy our of them for any particular light flux, you would really need to be able to vary the load dynamically."
Why do you say that the internal resistance is not low?
Why would we need to change the load dynamically? Because the internal resistance varies continuously?
I have a small device (http://store.solio.com/Solio-Store/S...ger-S620-AH1RW) that can change via usb any sort of device. What do you think is inside it in terms of electronics components and how do manage to change all these difference devices?
|Feb5-13, 12:37 PM||#23|
Yes. It's a non linear device and the optimum value of load will depend upon the flux hitting it. Any device that will not supply much current cannot be said to have a low source resistance. Of course, if you're talking in terms of an array of several msquared, then the resistance is somewhat lower.
Your little solar charger will be based on 5V - the standard USB supply voltage- and will have a regulator to limit the voltage out to 5V. Nothing more, I'd bet. That 5W figure is a bit hopeful for that apparent cell area, I think. Have you ever measured the output?
It will rely on the individual devices being charged to make the best use of what it gives them - they will self protect against any overcharging.
|Feb5-13, 03:04 PM||#24|
Well, the max power transfer theorem implies impedance matching. When we impedance match load and source we are surely splitting the power 50%, but that 50% is a ratio.
Example: no impedance matching, most of the power (99%) is given to the load but that is 99% of little total power. If total power is 10 W, the load gets 9 W.
Impedance matching: 50% to the load and 50% to the source. But now the total power is, maybe, 100 W. So the load gets 50% of 100W, which is 50 W. So in this situation is gets the most power it gets, even if it is a smaller percent of the total power.
No, I have measured the solar cell output.
Regardless of size, a typical silicon PV cell produces about 0.5 – 0.6 volt DC under open-circuit, no-load conditions. If a load with finite impedance is attached, the voltage goes down but the current goes up (current is zero in open-circuit configuration).
So, a specific load, based on its impedance, will receive its own particular voltage and current (i.e. power). The load may have a minimum power requirement to work properly. Hopefully the cell can give the right current and voltage to obtain that power.
5V? I did not know that was the typical USB port voltage. If the device needs more or less,
voltage converters apply I guess.
|Feb5-13, 03:36 PM||#25|
I think we're talking a bit at cross purposes about the implications of the maximum power theorem, Suffice to say that an efficient power supply system will not be aiming at a match. You want more power into your load - you use a bigger generator / battery that will still have a low enough source resistance. There is no advantage in any way to match the load - at the very least, you are doubling the cost of supplying the power to the customer - and you also need a way of getting rid of all that wasted energy at the power station; an additional massive cost.
It is not even common to match a source to a load in RF transmitting installations when high power is involved because this immediately reduces your efficiency and requires higher power devices (valves and transistors - ££££). Matching is more important at the antenna end of a feeder because you do not want reflections, which will cause echos and high standing wave voltages (but this is not relevant to supplying circuit power)
You need to pick a PV array with sufficient current delivering capacity to suit your load at its operating voltage. If you are really smart, you can possibly get around this problem (if the PV provides too many volts, perhaps) by DC-DC transformation with a switch mode circuit. This can avoid just wasting power in a dropper resistor. Usually, the cheaper solution is to use more PV area.
Look up the USB spec. It's all on line. The Phone manufacturer builds to that requirement. 5V was the chosen rail voltage way back when TTL technology was the rage. It has stuck for most standard applications.
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