I made a very simple circuit as in the attachment
It's supposed to output 625mA, but only outputs 500mA in actuality....
any insights as to why this is happening? is it a reasonable marginal error?
I tried decreasing the resistance such that an output of 825mA is expected but only reached around 600mA...
I actually want it to reach it 1.2A, but at this rate it seems highly unlikely....

Thanks for the help

 

How are you measuring the current?
Are the measurements from a real circuit or simulation?

 

hi ah,
The formula is in the d/s, with a 2R you will get ~625mA.
Is your source input capable of supplying more that 600mA.??

E

 

Hello,

You show a simulation.
What is the actual 9 Volts source?
If it is a PP3 battery, the current will be limited by the internal resistance of the battery.
As the capacity of a PP3 battery is quite small, you will have a short lifetime.

Bertus

 

How are you measuring the current?
Are the measurements from a real circuit or simulation?

Measurements from a real circuit via multimeter

 

hi ah,
The formula is in the d/s, with a 2R you will get ~625mA.
Is your source input capable of supplying more that 600mA.??

E
View attachment 132497

Yes, I'm actually using an adapter which is capable of providing 2A and a voltage of 12V

 

Hello,

You show a simulation.
What is the actual 9 Volts source?
If it is a PP3 battery, the current will be limited by the internal resistance of the battery.
As the capacity of a PP3 battery is quite small, you will have a short lifetime.

Bertus

I'm actually using a 12V adapter... it's capable of providing 2A

 

OK,
So what happens when you use say a 1.5R resistor, measure the voltage across the 1.5R.
E

 

Hello,

What is the power rating of the resistor?

Does the LM317 get warm/hot?
You might need a heatsink.

Bertus

 

What is the measured input voltage and what is the voltage across the load?

 

Hello,

What is the power rating of the resistor?

Does the LM317 get warm/hot?
You might need a heatsink.

Bertus

It does get hot.. But in the datasheet it says it can handle 1.5A... so shouldn't be a problem right?

 

Abdulwahab.hajar.
Quickly measure (short-term) the short-circuit current of the battery (if it is not a rechargeable battery) without your current generator.
P=(9V-1.25V)*1.25V/2= 4.84375W !!!

 

What is the measured input voltage and what is the voltage across the load?

Input voltage is 12.6, voltage across the load is 0.7.....
Actually I just noticed that the voltage across the output and adjust isn't 1.25V it started out as 1.25V but Quickly dropped to 0.55-0.7V
Could it be the heat???

 

Hello,

When the LM317 gets hot, it will go into protection.
Use a big heatsink and try again.

Bertus

 

Hello,

When the LM317 gets hot, it will go into protection.
Use a big heatsink and try again.

Bertus

Thank you, I will try that

 

Abdulwahab.hajar.
Quickly measure (short-term) the short-circuit current of the battery (if it is not a rechargeable battery) without your current generator.
P=(9V-1.25V)*1.25V/2= 4.84375W !!!

it's 2.27A... I'm using an adapter not a battery

 

You need a load. P = (I^2)/R when in the limit R=0, P gets pretty big.

Don't try forcing 1e-12 Amps from a low Z source into nearly zero ohms like I did. V gets pretty big.

 

You need a load. P = (I^2)/R when in the limit R=0, P gets pretty big.

Don't try forcing 1e-12 Amps from a low Z source into nearly zero ohms like I did. V gets pretty big.

So what do you suggest, I'm trying to get a constant current for charging

 

Hello,

When the LM317 gets hot, it will go into protection.
Use a big heatsink and try again.

Bertus

I used a large heat sink, the voltage between the adjust and output is 1.23V..... However the current is still low

 

Hello,

Is the resistor getting warm/hot?
What is the power rating of the resistor?

Bertus