# designing current amplification using darlington pair

#### mahmoudsamneh

Joined Jun 25, 2017
8
hello
looking for a help i need to design a circuit that amplify the current
the input for it will be 0.6213 amp and its input voltage is about 24 volt
so i need to get an output to about 3 amp and 24volt approx
from where should i start calculation ?

#### absf

Joined Dec 29, 2010
1,968
First using P=I x V

Input power = 0.6213 x 24 = 14.9112W
Output power = 3 x 24 = 72W

No electronic components or circuits is able to do that. It's like asking a baby weighting 8 Kg to lift a 40 Kg stone.

Allen

#### mahmoudsamneh

Joined Jun 25, 2017
8
am designing a 48-24 v battery charger so i connected an ac source with 220v/50hz values with a transformer to step down the voltage to 48 v
after that i used full wave rectifier circuit(bridge) to convert the signal to dc measring the current yields to about 15 amp but after filtering process it drops down to 0.6213 at 24 v
that's why i'm thinking of using a darlington pair
any hints that could help me

#### MrAl

Joined Jun 17, 2014
9,757
hello
looking for a help i need to design a circuit that amplify the current
the input for it will be 0.6213 amp and its input voltage is about 24 volt
so i need to get an output to about 3 amp and 24volt approx
from where should i start calculation ?
Hi,

What made you think you need a darlington pair?

I ask because the inverse ratio of 0.6 amps to 3 amps is 30/6=5, so you only need a current gain of 5.
What is unfortunate however is that if you really need 24 volts out with only 24 volts in and that extra current, you might need a boost circuit not a transistor. If you can live with maybe a 22v output then a transistor might work ok.

If after filtering it drops down by so much though you should probably post the entire circuit because something else is happening also.

• absf and JoeJester

#### mahmoudsamneh

Joined Jun 25, 2017
8
Hi,

What made you think you need a darlington pair?

I ask because the inverse ratio of 0.6 amps to 3 amps is 30/6=5, so you only need a current gain of 5.
What is unfortunate however is that if you really need 24 volts out with only 24 volts in and that extra current, you might need a boost circuit not a transistor. If you can live with maybe a 22v output then a transistor might work ok.

If after filtering it drops down by so much though you should probably post the entire circuit because something else is happening also.
i made a simulation to the circuit using matlab
are you familiar with it ?

#### JoeJester

Joined Apr 26, 2005
4,390
i made a simulation to the circuit using matlab
are you familiar with it ?
Post a screenshot of the schematic.

#### MrAl

Joined Jun 17, 2014
9,757
i made a simulation to the circuit using matlab
are you familiar with it ?
Hi,

There are a lot of simulators and not everyone uses Matlab, so if you post your circuit as Joe suggested we can then all take a look at it. That would help us understand your problem better.

#### crutschow

Joined Mar 14, 2008
31,492
after that i used full wave rectifier circuit(bridge) to convert the signal to dc measring the current yields to about 15 amp but after filtering process it drops down to 0.6213 at 24 v
What you mean by that.
What is the "filtering process"?
What is the load at the output of the filter?
Is 0.6213 the amperes into the load?

#### mahmoudsamneh

Joined Jun 25, 2017
8
here is the full spec. used
ac voltage source 220v/50hz
transformer step down 220 to 48 v / nominal power 660 w / r1=r2=0 l1=l2=0 / rm=3 ohm / lm=2 h current at the primary side 74.7826 amps
at the secondary part 11.7134amps
4 diodes have 0.001 resistance and vf=0.7 v each
the capacitor = 500e-5 f
the current after capacitor 0.3442amps
the resistance 50 ohm
output voltage 29 volt aprox
battery 24 volt and 0.32 ohm
the circuit output should be 24 volt and 2 amps #### mahmoudsamneh

Joined Jun 25, 2017
8
What you mean by that.
What is the "filtering process"?
What is the load at the output of the filter?
Is 0.6213 the amperes into the load?
1 using capacitor to purify the dc signal
2. resistance 50 ohm and a battery to be charged 24 volt and 0.32 ohm
3.yes but some how it becomes 0.3442 things are getting messy
please take a look at the diagram above

#### crutschow

Joined Mar 14, 2008
31,492
Things are getting messy since you don't appear to understand Ohm's law.
How do you expect to get 3A through 50 ohms?
That would require 150V. If you are trying to limit the battery charge current, you need a much smaller resistor.

• absf

#### mahmoudsamneh

Joined Jun 25, 2017
8
i decreased resistance to 0.05 the current increased to 5.8 amps but the voltage equals to about 48 volt
why i used 50 ohm in order to make the drop voltage i know the current would be in e-3amps so i decided to amplify the current using darlington pair
or other solution that you guys might help me with

#### crutschow

Joined Mar 14, 2008
31,492
why i used 50 ohm in order to make the drop voltage i know the current would be in e-3amps so i decided to amplify the current using darlington pair
Why not use a resistor value that gives you the voltage desired?
Why did you use 50Ω?
Do you know Ohm's law relating voltage, current, and resistance?
If not, you need to learn it.

#### mahmoudsamneh

Joined Jun 25, 2017
8
Why not use a resistor value that gives you the voltage desired?
Why did you use 50Ω?
Do you know Ohm's law relating voltage, current, and resistance?
If not, you need to learn it.
did you see the circuit diagram?
am not trying to break ohm's law
am saying i have a 48 volt at the secondary part of the transformer
then i need at the output which in the circuit diagram is the battery to be 24 volt
so i did use a 50 ohm to make a drop voltage that leads to let the battery have a 24 volt
but the current become so small because of the resistance value
so my main topic was asking about darlington pair and how to start designing and doing the math in order to amplify the current

#### crutschow

Joined Mar 14, 2008
31,492
so i did use a 50 ohm to make a drop voltage that leads to let the battery have a 24 volt
but the current become so small because of the resistance value
so my main topic was asking about darlington pair and how to start designing and doing the math in order to amplify the current
48Vac rectified will give about 65Vdc.
If you want 2A @ 24V to the battery, then the needed series resistance (from Ohm's law) is (65-24) / 2 = 20.5Ω, so you need about a 20Ω, 100W resistor.
So how did you arrive at 50Ω? If you want to use transistors to generate a regulated output voltage, that's a different thing.

#### JoeJester

Joined Apr 26, 2005
4,390
Can you provide the exact assignment as written by the school or book?

What kind of battery are you charging? do you have a particular model in mind?

Have you visited Battery University?

#### crutschow

Joined Mar 14, 2008
31,492
The difficulty is, you want us to implement your solution (which may not be the correct one) rather than help you to solve your problem (which is still not clear).