How Can I Convert Multiple 5-6V 0.35-0.8A Power Sources to 10-15V 10A?

[Massiveoni]

heya

i don't know if this is the right area for this question, but i am going to ask anyway,

i want to turn upto 4 power sources which is 5 - 6V 0.35 - 0.8A into 10 - 15V 10A

this is a link of the item/s i want to use to power it, and a link of the item i want to power

http://cgi.ebay.com.au/ws/eBayISAPI.dll?ViewItem&item=260646396607

http://www.ebay.com.au/itm/Car-Charger-DELL-Inspiron-6400-640M-PA-10-/320655706641?pt=AU_Laptop_Accessories&hash=item4aa891ca11

but pretty much, the items are solar panel with internal battery, which puts out the power of a usb port, and what i want to do is turn that power into the 12v you get from a car cigarette socket

can anyone help me on how to do this, or is it possible?

thanks

massiveoni

 

[DragonPetter]

Well first thing you have to realize, your source has less power capabilities than your load.

5V -.8A = 4 Watts

10V-10A = 100Wattsthis means, even if you built 100% efficient circuitry capable of doing the conversion you want, you'd still only be able to have your device on for 4/100, or 4% of the time. You'd need a voltage converter and some type of energy storage device to accumulate the charge at over 10Volts so you have enough to source 10Amps until your charge drops back down to 10Volts. You will also need a regulator at this point to output 10Volts, which will give you more losses as you're unloading the 10Amps down to 10Volts.

You of course can boost your 5V up to 12V at all times, but you won't have the current behind it to handle the loads unless you collect the charge and unload it at intervals of time.

 

[Massiveoni]

the solar panel has a 2600mah battery internal, and i am willing to use upto 4 of the solar panel things

 

[Mech_Engineer]

The point is your solar panel source only supplies 6V @ 0.8A max (4.8 watts max). So to supply 15V @ 10A (150 watts), you'll need to stack up 32 of the solar planels.

 

[DragonPetter]

the solar panel has a 2600mah battery internal, and i am willing to use upto 4 of the solar panel things

edit:
it says it can only output up to .8A, 2600mAh only tells the energy capacity of the battery, not the output current max.

 

[Massiveoni]

lol, i have no idea about any of this, is it possible, to use 4 of them in series,
cause 1 is 5v 0.5a, but wouldn't 4 of them be 20v 2a?
is there anyway of taking that down to 12v 10a?

 

[DragonPetter]

lol, i have no idea about any of this, is it possible, to use 4 of them in series,
cause 1 is 5v 0.5a, but wouldn't 4 of them be 20v 2a?
is there anyway of taking that down to 12v 10a?

Well, consider kirchhoffs current law, the current into a device is equal to the current out of a device. Even if you put 2amps into the converter, you can only get 2 amps out at the same time, regardless of voltage differences.

 

[Averagesupernova]

Source in series will have the same current. So you would have 20 volts at .5 amps with 4 in series. If you want to even come close you would put 3 in series. Then take 50 of those groups of three and parallel all of them. That would get you 15 volts at 10 amps.

 

[Massiveoni]

lol, in other words, don't bother, it won't work

 

[DaveC426913]

You haven't mentioned what you want to do with it.

Is your load going to use 10 amps? Just because a cigarette lighter can put out 10A doesn't mean you're going to need it.

 

[DragonPetter]

lol, in other words, don't bother, it won't work

like I said, if you only need to use the power in short bursts of time- around 4% of the total time, you can charge up a capacitor bank and unload the 10Amps every so often.

 

[Massiveoni]

if you have a look at the send ebay link in my first post, then youll see what i want to power, its a car charger for a dell laptop

 

[DaveC426913]

if you have a look at the send ebay link in my first post, then youll see what i want to power, its a car charger for a dell laptop

Right. Didn't know if a laptop was the thing you were powering. I don;t think they require 10A.

 

[Dickfore]

When one models a realistic voltage source, one usually models it as an EMF source with an "open circuit voltage" $\mathcal{E}$ connected in series with a resistor of "internal resistance" $r$. When connected to a load with resistance $R$, it supplies current:

$$I = \frac{\mathcal{E}}{r + R}$$

The voltage drop across the load is:

$$V = \frac{\mathcal{E} \, R}{r + R}$$

You can see that the current is a monotonically decreasing function with a maximum value at a short circuit ($R = 0$):

$$I_{\mathrm{max}} = \frac{\mathcal{E}}{r}$$

Instead of denoting the internal resistance of the voltage source, sometimes (as I think it is the case for your device), this max current is given.

The important thing is that the voltage drop across the load is a monotonically increasing function of the load resistance (and it is actually zero for $R = 0$). Thus, the product of the current and voltage drop has a maximum. It is a matter of solving a simple optimization to show that this maximum power is achieved for:

$$R_{0} = r \Rightarrow P_{\mathrm{max}} = \frac{\mathcal{E}^{2}}{4 r} = \frac{\mathcal{E} \, I_{\mathrm{max}}}{4}$$

This is only a quarter of the simple product of the voltage times the max current.

If you make a battery of m such voltage sources connected in series, then the EMF is $m \mathcal{E}$, but the internal resistance is $m \, r$ as well. This means that the short circuit current does not change.

If, on the other hand we make a battery of n voltage sources connected in parallel, the EMF does not change, but the internal resistance decreases as $r/n$. This increases the max current n times.

Finally, if you make a "rectangular" battery of n parallel rows each containing m voltage sources connected in series, then the EMF is $m \, \mathcal{E}$, the internal resistance is $m \, r/n$ and the max current is $n \, \mathcal{E}/r = n \, I^{(0)}_{\mathrm{max}}$.

Using these formulas, you might be able to find the values for m and n that you need.

 

[DragonPetter]

When one models a realistic voltage source, one usually models it as an EMF source with an "open circuit voltage" $\mathcal{E}$ connected in series with a resistor of "internal resistance" $r$. When connected to a load with resistance $R$, it supplies current:

$$I = \frac{\mathcal{E}}{r + R}$$

The voltage drop across the load is:

$$V = \frac{\mathcal{E} \, R}{r + R}$$

You can see that the current is a monotonically decreasing function with a maximum value at a short circuit ($R = 0$):

$$I_{\mathrm{max}} = \frac{\mathcal{E}}{r}$$

Instead of denoting the internal resistance of the voltage source, sometimes (as I think it is the case for your device), this max current is given.

The important thing is that the voltage drop across the load is a monotonically increasing function of the load resistance (and it is actually zero for $R = 0$). Thus, the product of the current and voltage drop has a maximum. It is a matter of solving a simple optimization to show that this maximum power is achieved for:

$$R_{0} = r \Rightarrow P_{\mathrm{max}} = \frac{\mathcal{E}^{2}}{4 r} = \frac{\mathcal{E} \, I_{\mathrm{max}}}{4}$$

This is only a quarter of the simple product of the voltage times the max current.

If you make a battery of m such voltage sources connected in series, then the EMF is $m \mathcal{E}$, but the internal resistance is $m \, r$ as well. This means that the short circuit current does not change.

If, on the other hand we make a battery of n voltage sources connected in parallel, the EMF does not change, but the internal resistance decreases as $r/n$. This increases the max current n times.

Finally, if you make a "rectangular" battery of n parallel rows each containing m voltage sources connected in series, then the EMF is $m \, \mathcal{E}$, the internal resistance is $m \, r/n$ and the max current is $n \, \mathcal{E}/r = n \, I^{(0)}_{\mathrm{max}}$.

Using these formulas, you might be able to find the values for m and n that you need.

I think you might be making things unnecessarily complicated.

First, a maximum current rating is not necessarily just another way of giving the internal source resistance in the specification; it is also dependent on the maximum power dissipation of the components in the device. Two devices can have equal internal source resistances and voltage sources, but different maximum current outputs because of power dissipation. Imagine your internal resistance r being a real component that has to dissipate heat, and this r can have a heat sink on it or be inside an oven so you cannot put the same maximum current through both. Often the power dissipation of a regulator's switching transistors are limiting the maximum output current, and a solar cell device is using a regulator at its output stage. But I agree with you that the internal source resistance is a limiting factor of the output current. Ultimately, you should take the maximum current output spec for what it is, the maximum current you can draw from the device at its specified output voltage, rather than infer an internal source resistance from this spec. Then you just use Ohms law with these two values to determine the maximum load you can have, which is not even a requirement to know with the OP's question, as you can simply use the power conversions to see the bottleneck.

As far as the Pmax equation you give, I'm not exactly sure why this calculation is necessary. The specifications are giving the recommended range of operation of the device, and I don't think you can infer your equations and the internal source resistance and your electrical model from the specifications of this device. The specifications tell you the min and max voltage and current ratings of the device, and you can solve for your best or worst case power output from these specs: Vmin*Imin or Vmax*Imax. . I think anything else is over analysis of the specifications. If I'm wrong on this, and your equation for Pmax is relevant, I'd really like to know because this is really relevant to calculations I have to make and I've been doing it wrong in that case.