I'm using DC adapter 5.8 volt with 580 miliamp to produce 1.5 amp by using a pair of NPN transistor 2N3904 to be build as Darlington circuite.

Is it possible?

 

No

 

The immutable rule of ALL power conversion schemes is that:

The power out will ALWAYS be less than the power in. Sometimes it will be much less!​

Lets take your numbers:

5.8 V @ 580 mA ≈ 3.36 watts​

Assume your "process" is 80% efficient​

Ouput power will be 80% of 3.36 watts ≈ 2.7 watts​

Now 2.7 watts/1.5 Amperes is ≈ 1.8 volts​

​

I'm pretty sure that was not your intention, but that is the best you can do with any scheme you can come up with that is 80% efficient. Even if you had a better one you could only get to 2.24 volts.

If you want 5.8 V @ 1.5 Amperes, it will be better to start from the mains and buy or build the supply you need.

 

No.

 

I'm using DC adapter 5.8 volt with 580 miliamp to produce 1.5 amp by using a pair of NPN transistor 2N3904 to be build as Darlington circuite.

Is it possible?

This is somewhat like saying that you have a barrel with 55 gallons in it and asking if you put more holes in it can you get 100 gallons out of it.

 

The immutable rule of ALL power conversion schemes is that:

The power out will ALWAYS be less than the power in. Sometimes it will be much less!​

Lets take your numbers:

5.8 V @ 580 mA ≈ 3.36 watts​

Assume your "process" is 80% efficient​

Ouput power will be 80% of 3.36 watts ≈ 2.7 watts​

Now 2.7 watts/1.5 Amperes is ≈ 1.8 volts​

​

I'm pretty sure that was not your intention, but that is the best you can do with any scheme you can come up with that is 80% efficient. Even if you had a better one you could only get to 2.24 volts.

If you want 5.8 V @ 1.5 Amperes, it will be better to start from the mains and buy or build the supply you need.

yea i already knowed the output will be lower than actual but nevermind I just need to get a high amperage in within 1 top 1.5 amp

 

yea i already knowed the output will be lower than actual but nevermind I just need to get a high amperage in within 1 top 1.5 amp

What does "within 1 top 1.5 amp" mean? I can't decipher that.

You are not going to get 1.5 A from a supply that has a max output of 580 mA (which is a strange spec for a supply in the first place).

The only way to get 1.5 A out of it is to use a DC-DC converter to step the voltage down to no more than about 1.8 V.

What is the actual problem you are trying to solve?

 

yea i already knowed the output will be lower than actual but nevermind I just need to get a high amperage in within 1 top 1.5 amp

I already told you the answer in my previous post. I said: " If you want 5.8 V @ 1.5 Amperes, it will be better to start from the mains and buy or build the supply you need. " that is the one and only solution to your problem. Period, full stop.

 

The Short answer is NO!!
So now an adequate explanation of what it is actually that is the intended result is needed.
This is not a question that a "design engineer" in any area of engineering would be asking.

 

All of the answers of No are assuming two things that the TS has not stated:

That the output voltage remains the same

and

That the current is continuous

If either of those assumptions is not true, it IS possible to get 1.5A of current.

 

Not with a Darlington transistor.

(The parameter given)

 

All of the answers of No are assuming two things that the TS has not stated:

That the output voltage remains the same

and

That the current is continuous

If either of those assumptions is not true, it IS possible to get 1.5A of current.

I think I said as much in post #3, and mentioned a way to get the desired 5.8V @ 1.5A, but it did not involve a solution the TS was thinking of.

 

Exactly, the question has not been stated correctly.
But on the other hand, the mention of a Darlington makes everyone assume that the TS wants some sort of simple linear buffer which will deliver a higher current at the emitter than what is available on the collector.

Which it can’t.

 

Let's step back.
You stated:
(1) "5.8 volt with 580 milliamp" is your input
(2) " produce 1.5 amp"
But you failed to specify what the voltage will be at 1.5A.
Without complete spec hard to answer, others have assumed output voltage same.
But we can say:
Case (1) Output Voltage also 5.8V - not happening, period.
Case (2) Output Voltage less than 5.8V - also not happening with just Darlington pair.
Case (3) Output Voltage more than 5.8V - definitely not happening as you are violating conservation of power (see Post #3)

Now Case (2) can be solved with Buck DC-DC converter, but as mentioned in Post #3 you must conserve power and your output voltage would be around 1.8V @ 1.5A

Maybe your confusion comes from the description that a Darlington pair can amplify the "drive current" of a signal. So I could have an output of a device that can drive 5ma max, but with addition of Darlington pair I can increase "drive current" to 500ma BUT the power source supplying the Darlington pair must be able to provide at least 500ma at the desired output voltage.

 

I'm using DC adapter 5.8 volt with 580 miliamp to produce 1.5 amp by using a pair of NPN transistor 2N3904 to be build as Darlington circuite.

Is it possible?

To play the devil's advocate, the answer is yes, but you don't need the darlington circuit.
Do the math.

The DC adapter can provide 580mA @ 5.8V continuously.
This is 0.58A x 5.8V = 3.36W

If you had a 1.5V source feeding a 1Ω load, the current is 1.5A.
Power is 1.5A x 1.5V = 2.25W

Theoretically, if you can create a down converter to deliver lower voltage you could get 1.5A, depending on the voltage and the load.

 

We may have lost another one because we failed to confirm his cherished beliefs. Oh well, that's the way the Mercedes Benz.

 

All transistors have a datasheet that says its maximum allowed current.
The 2N3904 says 200mA (0.2A) and it works poorly above 100mA. It can boost 2mA to no more than 200mA.

A darlington transistor causes a voltage drop of at least 1.5V. Then 5.8V input creates 4.3V or less output.

Your 580mA adapter will smoke and maybe catch on fire if it is overloaded with 1.5A (1500mA).

 

All transistors have a datasheet that says its maximum allowed current.
The 2N3904 says 200mA (0.2A) and it works poorly above 100mA. It can boost 2mA to no more than 200mA.

A darlington transistor causes a voltage drop of at least 1.5V. Then 5.8V input creates 4.3V or less output.

Your 580mA adapter will smoke and maybe catch on fire if it is overloaded with 1.5A (1500mA).

I do decided to lower until 4.3 volt to charge my two 1.3 volt Li-ion batery ^0^ with a little bit higher amperage, Sorry to forget about to mention it.

 

I do decided to lower until 4.3 volt to charge my two 1.3 volt Li-ion batery ^0^ with a little bit higher amperage, Sorry to forget about to told this issue.

 

I'm using DC adapter 5.8 volt with 580 miliamp to produce 1.5 amp by using a pair of NPN transistor 2N3904 to be build as Darlington circuite.


Is it possible?

Sure...assuming three identical 580 mA adapters. Otherwise you are asking for extra current where no such surplus exists. Where are all of those amps going to come from, Uranus?