Author Topic: Current amplification  (Read 483 times)

0 Members and 1 Guest are viewing this topic.

Offline elec_gra

  • Contributor
  • Posts: 6
  • Country: mt
Current amplification
« on: March 07, 2018, 10:16:24 am »
Hi, I am trying to amplify a 2mA, 3.3V supply to 3.3V 50mA(pulse) as I need to drive a switching circuit with that specific current in order to work in saturation. I have tried a few circuits such as another switching circuit to have a higher current but this results in a larger voltage. Have also used an optocoupler but didn't quite work with that. Someone I know has suggested using a darlington transistor but I haven't got a good idea how to work with a darlington transistor. Any suggestions please? I would greatly appreciate!
I have also attached a switching circuit that I tried but didn't work(obviously as I need current not voltage).
 

Offline tecman

  • Frequent Contributor
  • **
  • Posts: 417
  • Country: us
Re: Current amplification
« Reply #1 on: March 07, 2018, 10:23:20 am »
Your circuit will work, but the big question is the required (desired) rise/fall times.  Transistor selection, drive current and base capacitance and transistor saturation call all effect responce times.

paul
 

Offline elec_gra

  • Contributor
  • Posts: 6
  • Country: mt
Re: Current amplification
« Reply #2 on: March 07, 2018, 10:25:02 am »
But I will be having a 12V output and 50mA, can I adjust to have 3.3V output?
 

Offline Hero999

  • Super Contributor
  • ***
  • Posts: 10843
  • Country: gb
Re: Current amplification
« Reply #3 on: March 07, 2018, 10:36:58 pm »
But I will be having a 12V output and 50mA, can I adjust to have 3.3V output?
What do you mean?

We need to see the full picture. Pleas post schematics of the surrounding circuitry.
 

Offline elec_gra

  • Contributor
  • Posts: 6
  • Country: mt
Re: Current amplification
« Reply #4 on: March 08, 2018, 12:01:32 am »
This is the switching circuit that I am using. In order to be able to operate the transistor(BC337) in saturation region for 500mA collector current I need to supply 3.3V 50mA at the base of this transistor.However my 3.3V supply is a microcontroller output which can only output a maximum of 2mA and I a trying to amplify this current to 50mA.But i'm not sure what kind of circuitry I can use.
 

Offline Hero999

  • Super Contributor
  • ***
  • Posts: 10843
  • Country: gb
Re: Current amplification
« Reply #5 on: March 08, 2018, 01:24:29 am »
You can probably get away with less than 50mA.

The data sheets specifies, a typical  VCE of 1V, when IC = 500mA and IB = 6mA.
https://www.onsemi.com/pub/Collateral/BC337-D.PDF

I suspect there's little point taking the base current any higher than 25mA.

What's with all the resistors, capacitors and the inductor? If the load is inductive, a simple free-wheeling diode is all that's required.

All that's needed is an emitter follower, on the output, see attached. As drawn, the base current for Q1 will be around 33mA. If you want it to be higher than that, reduce the value of R1.

« Last Edit: March 08, 2018, 01:50:09 am by Hero999 »
 

Offline elec_gra

  • Contributor
  • Posts: 6
  • Country: mt
Re: Current amplification
« Reply #6 on: March 08, 2018, 02:00:28 am »
I tried with a lower current than 50mA but I didnt get the 500mA I wanted.That's why I'm aiming for 50mA. I didn't see the Ib of 6mA when Ic =500mA.
The other resistors are part of the switching of the inductor.The inductor a solenoid actuator actually I'm trying to switch that actuator. The other other diodes are protection. What can I use for the input for Q2 ?
 

Offline Hero999

  • Super Contributor
  • ***
  • Posts: 10843
  • Country: gb
Re: Current amplification
« Reply #7 on: March 08, 2018, 06:34:33 am »
I tried with a lower current than 50mA but I didnt get the 500mA I wanted.That's why I'm aiming for 50mA. I didn't see the Ib of 6mA when Ic =500mA.
50mA is the current required to guarantee full saturation, with a collector current of 500mA, but it's overkill. Lower currents will ensure a low enough voltage drop, for most applications. See page 3 of the data sheet. Figure 3 states an Hfe of 90, when VCE = 1V and IC = 500mA, which would mean an IB of  500/60 = 5.56mA. This agrees with figure 4, which shows VC = 1V, when IB = 6mA and IC = 500mA.

https://www.onsemi.com/pub/Collateral/BC337-D.PDF

If you make IB more than 6mA, but less, than 50mA, VCE, should still be acceptable, hence why I suggested 25mA.

Quote
The other resistors are part of the switching of the inductor.The inductor a solenoid actuator actually I'm trying to switch that actuator. The other other diodes are protection. What can I use for the input for Q2 ?
The input to Q2 is the 3.3V logic signal. Q2's base current will be tiny, on the order of 100µA and is self-limiting: no additional base resistor required.
 

Offline elec_gra

  • Contributor
  • Posts: 6
  • Country: mt
Re: Current amplification
« Reply #8 on: March 08, 2018, 06:42:10 am »
Ok, I will try to work like that.Thanks!
 

Offline elec_gra

  • Contributor
  • Posts: 6
  • Country: mt
Re: Current amplification
« Reply #9 on: March 08, 2018, 09:39:25 pm »
BTW you were right I worked with much less than 50mA.Thanks a lot for your help :)
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf