DC > DC converter problems please help!!

[Moha99]

What is the efficiency of the boost converter? It is probably in the datasheet for the IC. This will give you a very close approximation to what is lost in the circuit.

You put in 100W of power and get out 95W of power, then the efficiency of the circuit is 95%.

There is not a current loss. The current used on input side times the input voltage must equal the current used on the output times the output voltage, except that is is not 100% efficient. So the equation is:

Vin * In * efficiency = Vout * Iout

Bob

What is your output voltage?
What is your input voltage?
How much current does the converter need to supply on the output?



~42W/13V = 3.23A available on the output. Should probably figure 3A max.

Between 13 to 14V/DC as Output.
Input voltage is 4.8Volts and a current of 10Amps because each battery has about 2500mAh

The converter has a MIN of 1 A and a MAX of 3 A.

 

[jackorocko]

Input voltage is 4.8Volts and a current of 10Amps because each battery has about 2500mAh

You do realize the batteries can not deliver all their power at once right?

 

[Moha99]

You do realize the batteries can not deliver all their power at once right?

Ow now I do! But how can I calculate its expected power at least?

 

[Moha99]

Just read a few things about batteries i think i found the solution and got more of an understanding of my problem and how to solve it.

Im using more than one battery connected in series where voltage will increase and the capacitance will not so the current will be te same so no worries about that.

But I wonder if a battery has 2600mAh how much current could it give out?

 

[jackorocko]

Here is an example of a typical AA battery. Less then 2 hours with a 1A draw. Notice voltage drops off as the battery discharges.

http://208.173.184.182/OEM/Primary/Alkaline/mx1500.asp

 

[Moha99]

Here is an example of a typical AA battery. Less then 2 hours with a 1A draw. Notice voltage drops off as the battery discharges.

http://208.173.184.182/OEM/Primary/Alkaline/mx1500.asp

Pretty good Information thanks!

 

[(*steve*)]

Im just using 4 for now... But later I'll be using more. The D Cell that i bought said it has about 2600mAH x by 4 thatsabout 10K mAH.

No, in series that would still give you 2600mAH.

They're just fancy AA cells in a big housing.

 

[Moha99]

No, in series that would still give you 2600mAH.

They're just fancy AA cells in a big housing.

Perfect just what I needed to hear!

If its 2600mAh its perfect for the converter!

Thanks guys for everything you helped me alot.

 

[(*steve*)]

Just ensuring you realise that 2600mAH is a reasonably normal rating for AA NiMH. There's no point in carrying around D sized cells that are really just AA cells.

 

[Moha99]

Just ensuring you realise that 2600mAH is a reasonably normal rating for AA NiMH. There's no point in carrying around D sized cells that are really just AA cells.

I would've done that.
But I already got those batteries.

 

[BobK]

Max Efficiency: 86%
The batteries im using each have about 2500mAh which is about 2.5 Amps and about 1.2 Volts so I have to supply the converter with about 3.5Volts in so 1.2 x 4 = 4.8V/DC so now we have 2.5x4 = 10Amps
which is 7 amps above the MAX. range...

Also In what way can a DC TO DC converter increase charge? Now I know charge is being lost how can we increase it then...?

lets say I have about 4 volts with 160 mAh if I step it up to about 20 Volt wont the current also increase?

1. You are very confused about current (Amps) and battery capacity (Amp Hours.) They are two different things.

2. When you put batteries in series the current and the capacity do not increase. I.e. 4 batteries with 2500mAH capacity in series still has 2500mAH. 4 batteries that can supply 1A each do not supply 4 amps when in series.

3. When you increase the voltage, the current has to decrease because the product of the two (which is the power) must remain the same -- you cannot create power. And you always lose some power based on the efficiency of the converter.

Bob

 

[Moha99]

1. You are very confused about current (Amps) and battery capacity (Amp Hours.) They are two different things.

2. When you put batteries in series the current and the capacity do not increase. I.e. 4 batteries with 2500mAH capacity in series still has 2500mAH. 4 batteries that can supply 1A each do not supply 4 amps when in series.

3. When you increase the voltage, the current has to decrease because the product of the two (which is the power) must remain the same -- you cannot create power. And you always lose some power based on the efficiency of the converter.

Bob

I finally understood things and how they go around previously but I thank you for this information so far.

The one thing I think about so far is the 3rd point: "When you increase the voltage, the current has to decrease because the product of the two (which is the power) must remain the same -- you cannot create power. And you always lose some power based on the efficiency of the converter."

Could you clarify this more? I know im working on DC circuits and batteries but what I wonder about the most is: I heard in AC circuits and transformers its possible to increase the voltage and the amperage is that correct?

And when you say " the current has to decrease because the product of the two (which is the power) must remain the same"

Im not creating power but rather using power from the batteries that were conserved.
If it is possible to converter batteries from DC > AC then tamper with the power output.

Please clarify that point more and I thank you all so far for the help and support.

 

[davenn]

I finally understood things and how they go around previously but I thank you for this information so far.
The one thing I think about so far is the 3rd point: "When you increase the voltage, the current has to decrease because the product of the two (which is the power) must remain the same -- you cannot create power. And you always lose some power based on the efficiency of the converter."
Could you clarify this more? I know im working on DC circuits and batteries but what I wonder about the most is: I heard in AC circuits and transformers its possible to increase the voltage and the amperage is that correct?

No, its still applies as Bob said.

eg. your transformer has 120VAC(240VAC depending on your country) going in
depending on the ratings of the transformer it can transfer xx Watts of power. If you are stepping the voltage down to say 12VAC on the secondary then you have that XX Watts in power available minus losses in the transformer. Without going into indepth transformer theory, you can google that. lets say you can produce 50W of power on the secondary so 50W / 12V = 4.16Amps. If you want more current out of that transformer you could rewind the secondary with heavier gauge wire, but since there are less turns of wire cuz of its additional thickness there's going to be less voltage induced into those windings. so you may increase to 8 amps but you are not going to have 12VAC any more ....
50W / 8A = 6.25VAC. You can see from the 2 examples, I doubled the current and halved the voltage.

Its still the same pricnipal of you go the other way. You step up from 120V AC input to 1000VAC (1kV) output. The transformer needs much more windings on it to induce that higher voltage, the thickness of the wire is going to be very fine and hence very little current capacity. Lets still work with our 50W capacity of the transformer
50W / 1000V = 0.05Amp

you CANNOT increase current without decreasing voltage
you CANNOT increase voltage without decreasing current
The Volts x Amp must = Watts

And when you say " the current has to decrease because the product of the two (which is the power) must remain the same"
Im not creating power but rather using power from the batteries that were conserved.
If it is possible to converter batteries from DC > AC then tamper with the power output.

Please clarify that point more and I thank you all so far for the help and support.

The batteries are capable of a given current out at a given voltage. That = a given power in Watts available to do work. You CANNOT change it


Dave

 

[Moha99]

ok.

If there was a device that I supplied with enough voltage but not enough current what would happen to that device?

From my understanding both can never work the same. If voltage increases the current decreases and vice versa but what I wonder the most is how do powerplants and solarplants and also wind plants make enough power to power the world? Is there enough current or voltage?


The electric world is intresting but very complicated !!

 

[(*steve*)]

If there was a device that I supplied with enough voltage but not enough current what would happen to that device?

The voltage would fall.

If you assume that the resistance of the load was R (ohms), and the current you could provide was I (amps), then the voltage (volts) would fall to I x R

From my understanding both can never work the same. If voltage increases the current decreases and vice versa but what I wonder the most is how do powerplants and solarplants and also wind plants make enough power to power the world? Is there enough current or voltage?

Ever experienced a brownout? That's what happens when the generation capacity is exceeded. Typically your power provider will shut down sections of the grid (blackouts) to prevent brownouts. In some countries this is standard practice every day.

The generation capacity needs to exceed the peak load in order to prevent this from happening.

Think of your car. The engine is the "generator" and it can generate a certain peak power. The hill you're going up is the load. As the load gets higher and higher (hill gets steeper and steeper), the engine has to work harder and harder for you to maintain the same speed (voltage). At some point, you need to either slow down (brownout), or reduce the load (find a less steep road).

With generation capacity you typically have some amount of spinning reserve (generators running but not actually providing power) that can be called on quickly to provide power. There are also other power generation plants that can be started quickly (gas turbines are an option) that can be brought on-line for short periods of high demand.

The electric world is intresting but very complicated !!

It just requires you to get your head around some stuff. Once you've done that it's easy (until you find the next concept to learn )

 

[Moha99]

Well that's good information right there!

I wonder one thing now if I supply enough voltage but not good current the voltage drops correct? Well that shows the problems of my circuit and also the battery mix.
Now if I supply a unit more current than it needs and the same what would happen?

The thing is people say that if you increase the voltage current will increase as well now I can prove them wrong.

 

[(*steve*)]

You can't (kinda) supply more current than a circuit required, unless you raise the voltage to a point at which more current flows.

OK, that's only true for resistive loads, something employing a switchmode power supply will actually draw more current at lower voltages and reducing the input voltage can result in an increased current demand (I once had a very expensive HP PC that would immediately destroy the power supply if you had a brownout EXACTLY for this reason).

In general though, the current demanded by a device is determined by the voltage supplied, and in most cases the current will rise with increasing voltage.

 

[Moha99]

You can't (kinda) supply more current than a circuit required, unless you raise the voltage to a point at which more current flows.

OK, that's only true for resistive loads, something employing a switchmode power supply will actually draw more current at lower voltages and reducing the input voltage can result in an increased current demand (I once had a very expensive HP PC that would immediately destroy the power supply if you had a brownout EXACTLY for this reason).

In general though, the current demanded by a device is determined by the voltage supplied, and in most cases the current will rise with increasing voltage.

So lets say I have a generator and it generates under perfect resistance if I stepped up the voltage the increase would probably increase current as well.


I started to understand this more and it gave me more confidence to work out my work.

Thank you all again for all that support!

 

[(*steve*)]

So lets say I have a generator

yep

and it generates under perfect resistance

If it had a load that was resistive...

if I stepped up the voltage the increase would probably increase current as well.

Yes, an increase in voltage would produce a proportionate increase in current.

 

[cjdelphi]

Here's an example, take a very dead very flat 9v battery, it says 4volts when you take a reading... you connect the 9v battery and connect it to an LED rated 40ma , forget the resistor for now, it's a dead battery..

The LED is rated 3.2volts 20ma, the battery is supplying 4volts still, you'll find the LED will barely light up and if you read the voltage it will be <1.8volts because there's not enough amps to drive the LED, there's been a few occasions where i've measured the voltage (just fine) and then nothing would work..

That's pretty much the same issue as the OP has, there's simply not enough current to do what you need.