# Current Limter

Discussion in 'General Electronics Chat' started by aamirali, Feb 3, 2012.

1. ### aamirali Thread Starter Member

Feb 2, 2012
415
1
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?

2. ### crutschow Expert

Mar 14, 2008
13,509
3,385
You calculate R1 for the minimum hfe and maximum current, thus R1 = (28-24-0.7-0.7) / (1A/hfe-min).

It's wattage would be Ib$^{2}$*R1.

Q2 can be smaller since it carries less current.

The Q1 dissipation would be about 4W so Q1 will need a heat sink.

That limiter circuit should work fine for your needs. The limit is somewhat sensitive to the temperature of Q2 (limit varies at about -0.3%/C) but that should not be a problem for charging a battery.

Is this a lead-acid battery? To charge the battery properly you should has have a voltage limit circuit to avoid overcharging the battery.

3. ### aamirali Thread Starter Member

Feb 2, 2012
415
1
1. Why did minus 0.7V twice.

" R1 = (28-24-0.7-0.7) / (1A/hfe-min)"

2.My battery voltage can vary from 21.5V-26.4V i.e. full discharge to full charge (sry forgot to mention this earlier). How does R1 changes now.

4. ### crutschow Expert

Mar 14, 2008
13,509
3,385
1. The worst-case base-emitter drop of Q1 and the voltage across R2.

2. At the maximum battery voltage there will be very little voltage across R1 so it will be difficult for that circuit to maintain 1A of current. To avoid that problem place the battery in series with the collector of Q1 (between V1 and Q1's collector) and ground R2 and Q2's emitter. That way you will always have the full voltage available for R1 (R1 stays connected to V1).

For that case R1= (28V -1.4)/(1A/hfe-min).

5. ### Motardo New Member

Sep 21, 2011
20
2
Do you want to force 1A through the battery even when it's fully charged, or would it be better if the current tapers off as the battery charges up? I guess it depends on the type of battery...