hi,
i want to convert output pin of arduino (5 vdc , 40 mA) to 24 vdc, 200 mA .i can not use transistor for saturation problem , please suggest appropriate circuit ?

 

what saturation problem are you talking about?

 

what saturation problem are you talking about?

because:
Ib= (5-0.7)/2.15 = 2mA ,,, Ic = B*Ib=200 mA ,,, if load be 1k=> VCE= 24-200==SATURATION
so
VCE=0.2 => IC= 24-0.2/1=23.8 mA no 200mA

 

So use a Darlington,



or an NFet

 

so i must calculate parameters for that , in your opinion how much load can support Darlington or NFet for 24 Vdc,200 mA?

 

If the output of arduino is 5V/40mA and add a npn and pnp then you can get the current close to 4A.

 

If the output of arduino is 5V/40mA and add a npn and pnp then you can get the current close to 4A.

but this is not possible because occur satturation before it reach to higher current .

 

because:
Ib= (5-0.7)/2.15 = 2mA ,,, Ic = B*Ib=200 mA ,,, if load be 1k=> VCE= 24-200==SATURATION
so
VCE=0.2 => IC= 24-0.2/1=23.8 mA no 200mA

For my own curiosity, where did the value 2.15 come from when calculating Ib? From that point on, I am confused where the equations are coming from. Did you mean to divide by 1000 instead of 1? And Vce=24-200? Are you mixing units and precision?

 

For my own curiosity, where did the value 2.15 come from when calculating Ib? From that point on, I am confused where the equations are coming from. Did you mean to divide by 1000 instead of 1? And Vce=24-200? Are you mixing units and precision?

excuse me ,
Ib= (5v-0.7v)/2.15k = 2mA ,,, 5v is arduino output, vbe=0.7 ,Rb=2.15k ...because beta is 100 and i want to reach 200 mA
Ic = B*Ib=200 mA ,,, if load be 1k=> VCE= 24v-200v==SATURATION
so
VCE=0.2v => IC= (24v-0.2v)/1k=23.8 mA no 200mA

 

Go back and look at post #4

However, here is the Arduino pin driving a 2n2222 directly. Note that I set the base drive to be 20mA (within the capability of the Arduino pin). This guarantees saturation of the 2N2222, minimizing its power dissipation when on.

 

Beta is not a constant and gets lower with higher current. Usually you want to use Ib=0.1 Ic in switch application. So just use 120 ohm base resistor and be done with it.

 

Go back and look at post #4

thanks a lot MikeML

 

Beta is not a constant and gets lower with higher current. Usually you want to use Ib=0.1 Ic in switch application. So just use 120 ohm base resistor and be done with it.

thanks a lot kubeek , you right, i was in wrong.

 

but this is not possible because occur satturation before it reach to higher current .

When the bjt get into the saturation status, the hFE as 10 and then the current will be like this:
Arduino (40mA) → NPN (400mA) → PNP (4A)

 

So if you want to get a 200 mA current from Arduino 5V/40mA then you can do as this:
Arduino 5V/40mA → R_limit (2mA) → NPN (20mA) → PNP (200mA)

NPN can be use as 2N3904, PNP can be use as 2SA684(1A) or similar.

 

So if you want to get a 200 mA current from Arduino 5V/40mA then you can do as this:
Arduino 5V/40mA → R_limit (2mA) → NPN (20mA) → PNP (200mA)

NPN can be use as 2N3904, PNP can be use as 2SA684(1A) or similar.

thanks ScottWang, in your opinion, how much maximum load without any problem in this case?

 

thanks ScottWang, in your opinion, how much maximum load without any problem in this case?

The normally you will need to add some more current at least add about 20%, so the calculation will be as:
Arduino 5V/40mA → R_limit (2.4mA) → NPN (24mA) → PNP (240mA)
That will be enough for many applications.

You have to test how is the real current for the load needs and not just look at infos from the datasheet.

 

You have to test how is the real current for the load needs and not just look at infos from the datasheet.

ok, thanks.

 

What you need is just like the circuit shows in Fig-07, the values of resistors will be change according to the current.

 

thanks a lot for circuits ScottWang,