Current Limter

Hi Aaimir,

Connecting the two diodes in series assures that there is sufficient voltage to forward bias Q1's emitter-base junction. Instead of ~0.7V at the base, the voltage is twice that amount, or about 1.4V. Then, moving around the base-emitter loop and taking into account the base to emitter drop, the voltage at the emitter will be about 0.7V. The emitter current, I_E, will be approximately 0.7V/R₂, as the equation indicates.

The single diode approach can make the I_E operating level less predictable because the forward bias characteristics of the bias diode and the transistor base-emitter junction can be quite different.

Regards, Thomas

PA - Linear Applications Engineering

Aaimir;

The diodes also attempt to temperature-compensate the emitter-base junction of the transistor; the TC match is poor but better than nothing.

I have to design a current limiter  for 1A, & charge a battery of 24V. Input volatge is 28V. I made ckt:

I put below deisgn steps, please check them:

-> Since max Vs=28V , I chose a transistor with Vce=40V max.

->Max current is 1A. Q1 transistor selected will have Ic max= 2A. (Safety factor included)

-> Noted Hfe from Q1 & calculated Ib= Ic/hfe

-> R2= 0.65 ohm, 1Watt because Vbe of Q2 is around 0.65V

-> How to calculate R1 & its wattage.?

-> Should Q2 same as Q1 or other can be selected because Ib of Q1 is in mA so Ic of Q2 is also low.?

Any other current limiter with better performance?

Aamir;

It is vitally important to also consider the power dissipation of Q1. In the case of a fully-discharged battery, you will have the full voltage drop across Q1 (28V) at the same time as it is putting out its full current (1A). You must calculate the junction temperature of Q1 when it is mounted on a big heat sink.

If you look at the LM317 voltage regulator data sheet, on page 18 there is a 1A current limiter circuit. It is very simple but it must also be mounted on a heat sink. You must do the power dissipation calculation to see if it can meet your requirement.

Another approach is to use a switching regulator. But, since I'm an analog guy, I'll leave that to someone else to follow up.