Current/Ampere Booster

Good day. I have only 0.75A but I need 2A. How can I boost my 0.75 to become 2A? Need urgently. Thanks

If your voltage gain will be "1" you can use power bjt like "bd139" with proper resistor selection you can boost your current desired level.

Until we work out how to create energy from thin air, I'm afraid the rules say you can't get more out than you put in.

However, there may be ways to increase the available current if there is sufficient source for it already but you are not using it. You will have to give us far more information though. The first things being why do you need the extra current and what voltage are you expecting it to be at?

Current is drawn from the power source by something, you may need that source to be capable of providing 2 Amps but except in special circumstances you wouldn't boost the demand for current. In human terms it would be like asking "how to I increase my thirst" when you really mean "where can I get more to drink".

Ignore post #2 which is complete nonsense!

Brian.

FvM said:

Ignore post #2 which is complete nonsense!

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Im sorry i just wanted help him. Why it is nonsense if a totem pole is used feeding from his voltage source to boost current ?

Maybe i dont even know how to proper boosting for current but you as a moderator should not in behaviour like "this is nonsense, that is bullsh*t" am i wrong ? I wish you were in the mood that you help anyone even giving wrong answer. Because, we all need just information not anger especially from the ones like you

The question leaves us guessing about voltage, power, existence of an independent power supply. In other words is it about an amplifier, a regular voltage converter or attempting some kind of over unity respectively perpetual motion machine?

Learning to ask clear questions...

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Correct citation would be much appreciated!

I don't think that it's exactly nonsense, but involving assumptions about the unsaid intentions of the original poster. In this case, assuming he wants to make a current amplifier and has an independent power supply available.

Do you need the 2A current has a burst current which lasts for some 2 or 3 seconds or it has to be continuous current..

FvM said:

Correct citation would be much appreciated!

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I'm sorry for that. Also, op did not mention whether it is different source or not so i wanted to share my information about boost current from the same source thanks.

Do you need the 2A current has a burst current which lasts for some 2 or 3 seconds or it has to be continuous current..

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I appreciate your attempt to get a clearer statement from the original poster. I think we should await his clarification.

Harryz said:

Good day. I have only 0.75A but I need 2A. How can I boost my 0.75 to become 2A? Need urgently. Thanks

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A current does not exist but itself. You need to specify other parameters like voltage, load resistance etc.

Also it seems you are talking about DC current, so I will not suggest you simply use a step-down transformer. Unless you want to convert the DC into ac by (say) chopping, and then use a transformer with 3:1 step down ratio,

Harryz said:

Good day. I have only 0.75A but I need 2A. How can I boost my 0.75 to become 2A? Need urgently. Thanks

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When you have a simple power supply or battery , current is limited by the ratio of voltage over resistance.

So if you have insufficient current, the total resistance must be reduced (by proper selection) or the voltage increased. This ratio also applies between two voltages like a charger and battery, or a battery and LED, This is Ohm's Law.

Please lets not argue over what the current is for, wait for a reply from Harryz. Guessing it to be the output of an amplifier, AC or DC or to do with series resistance is futile (with apologies to the Borg!).

Brian.

Sorry to confuse you guys.
My source is a battery, of course DC, that has an output of 19V 3.5A. I have to supply 4 circuits that need 12V 2A, 5V 2A, 12V 1A and 12V 0.9A. I divided 3A into 4 and I got 0.75A. I want that 0.75A boosted to become 2A for those loads.
Thanks for helping.

Now we are getting somewhere!

Firstly, batteries are not rated in Amps, they are rated in AH (Amp hours). Although there are obviously limits to how much current a battery can deliver, the specification is an indication of how long they can sustain different current over a period of time. For example, if the battery is 19V 3.5AH, it suggests it can sustain a current of 3.5A for one hour or 0.35A for 10 hours. Can you confirm where you are seeing the reference to 3.5A or whether it is really 3.5 AH.

Unless you use exotic power circuits, the total current is the sum of all the individual currents, regardless of the voltages so the load you are trying to feed is 5.9A. To some extent you can trade voltage for current but you are looking at fairly sophisticated electronics, are you confident with working with these, I doubt an 'off the shelf' solution exists.

Brian.

Presumed the input power supply is actually 19V 3.5 A, the intended set of output voltages could be achieved by switch-mode (boost) converters with about 85 % efficiency, which is feasible.

Sounds somehow like an artificial homework problem specification.

Because I was curious what kind of battery that would be...
I think the OP has one of those external battery charger bricks they make to recharge your portable electronics. They typically have multi-output voltages with various amounts of maximum current they can supply at those voltages. Most of the ones I've seen have a maximum current rating of 3.5A but that typically isn't on the 19V output. The few I found on the internet being sold on amazon, ebay, and aliexpress had 19V/3A.

Regardless your requirement for 12*2 + 5*2 + 12*1 +12*0.9 = 56.8W isn't going to fly when you can only get 57W max from the battery. Any circuit that can output those various voltage rails will not be 99.7% efficient.

Hmmm, I did find a number of them that are 19V/3.5A so if you can select the proper DC-DC converters you might be able to get 85+% efficiency up to perhaps 92% on some of the rails.

They do specifically say it's a battery and 19V is used in some laptops so perhaps that is what it is. It does sound more like a power supply specification that a battery though.

With respect to Harryz, when they made the simple division by four I assumed they were not tech savvy and custom SMPS design was not a realistic propositon.

Another possibility to be explored: Harryz, are all the currents you mentioned a continuous figure for the equipment or is it likely they are quoted as maximums? For example, most audio amplifiers draw more current at higher volume levels but a power supply to them would be rated for maximum volume, in real life it would be lower most of the time. Possibly, if the currents are not all needed simultaneously, the PSU design can be simplified.

Brian.

Thanks guys.
The battery I am saying is a high power bank 30000mAH I found on the internet.
Yes the load will demand a continuous supply for at least 2 hours. I just made a greater assumption to avoid less supply on my circuit. I admit I am not good at this subject so I need your help really badly.
Thanks again.

30000mAH = 30AH so I would guess we are talking about something the size of a small car battery. It should therefore be able to supply your 5.9A load for about 5 hours if you use simple linear voltage regulators. You need some kind of regulation to drop the 19V down to the voltages you want. You have two options:

1. linear regulators - cheap, simple but for this application, grossly inefficient, especially for the 5V output. They convert the voltage drop into heat so the more you drop the hotter it gets. For example to produce 5V at 2A (10W total) it would produce 28W of heat!

2. switch mode regulators - much more efficient but more expensive and complicated.

You could consider pre-made modules, one for each output voltage. They are available from Ebay like these:
**broken link removed**

If you use these (I do) I would recommend you screw them down to a metal sheet to help dissipate the heat.

Brian.

Harryz said:

Sorry to confuse you guys.
My source is a battery, of course DC, that has an output of 19V 3.5A. I have to supply 4 circuits that need 12V 2A, 5V 2A, 12V 1A and 12V 0.9A. I divided 3A into 4 and I got 0.75A. I want that 0.75A boosted to become 2A for those loads.
Thanks for helping.

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In summary you need :
12v at (2+1+.9) 3.9A
5v at 2A

This is 56.8 watts.

Your supply @ 19v is capable of 66.5watts.

A 'normal' linear supply/ converter will not work, since the current WILL have to be divided.
However a more complex dual output switched supply would be able to do the job. No 'boosting' of current required.

This is what you want for both of your supplies (only one type of dc-dc converter)
http://www.linear.com/product/LTM8027 (around $30 on digi key)

It can output both 12V and 5V depending on how you configure it. See Table 2 on page 16 in the datasheet for the component values to set the output voltages.

There are other manufacturers like PowerOne (now Bel Power Solutions) that you can also look at.

kripacharya said:

A 'normal' linear supply/ converter will not work, since the current WILL have to be divided.

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I'm not sure what you mean by dividing the current will make it not work. The problem with a linear regulator will be the lousy efficiency with 19V-12V and 19V-5V (or even 12V-5V). With the large amount of wasted power you would need much more than the 3.5A from the battery when the loads are at maximum current draw.

If the OP needs the very low ripple then they should use the part I suggested and drop the voltage to say 6.5V and use an LDO to go from 6.5V to 5V.


BTW, I don't design the power supplies on products, but I do usually get involved due to the FPGAs I work on being the biggest power hog on the board. 8-O

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Harryz said:

The battery I am saying is a high power bank 30000mAH I found on the internet.
Yes the load will demand a continuous supply for at least 2 hours.

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If this is something from a no-name factory in China, you may have something that doesn't even come close to the quoted performance. Most of the power banks (external battery charger) I've seen on the internet especially those from well know manufacturers are in the 20000mAH range, I'm sure there might be some with more, but they all seem to come in packages that are about the same size, a 30000mAH would require a 50% bigger package. If it's not larger than the average external battery charger it's probably not going to give you 30000mAH and their quoted 2 hours of run time.