lm 7805 short circuit protection

[Melfior_Ra]

Hi everyone.

i am trying to figure out how it is working this short circuit protection.

If i want to limit the current up to 4 A then Rsc should be 0.7/4 = 0.175 ohm. At this point Q2 starts conducting, the voltage at Q1 emitter is 11.3V, the voltage at Q2 collector is 10.6V which leads to voltage drop across r1 of 1.4V .The current through R1 is 1.4/R1 wich is arround 460 mA . To this current will add Q2 collector current and Q1 base current. All of this is taking by the LM 7805. How can i find this current (Ir1, IcQ2, IbQ1) ?

Please help me . Is this a good solution for current limiter?

 

[MrAl]

Hello,


The current through R1 is the voltage drop across Rsc plus the Q1 base emitter voltage drop all divided by R1:
iR1=(vRsc+vBEQ1)/R1
and at the point of current limit the current through Rsc is Q1 collector current plus Q1 base current, and Q1 base current is Q1 collector current divided by Q1 Beta.
Q1 collector current is the output current minus the current limit of the 7805. So if the current limit of the 7805 is iLim=1 amp then Q1 collector current is:
iCQ1=Iout-1
and the base current is:
iBQ1=iCQ1/BetaQ1

Now knowing iR1 and iBQ1 we add those to get iSum:
iSum=iR1+iBQ1
and this should be less than iLim of 1 amp, so we subtract iSum from 1 amp to get Q2 collector current:
iCQ2=1-iSum

This works OK as a current booster for the 7805 and similar, but dont expect super accurate current settings as the transistor BE drops change with temperature and so does the temperature characteristic of the 7805 to some degree.

 

[crutschow]

To be more specific, the transistor base-emitter voltage varies about -0.33%/°C, thus your current limit will vary by at least that amount with temperature. The transistor gain also increases with temperature. This means that the current limit will go down with an increase in temperature (generally a desirable feature).