Hello,
I am looking at Figure 23 from SNVS775J –MARCH 2000–REVISED MARCH 2013, for a slow turn-on circuit employing a LM317L-N. I wanted to know if there are any equations and graphs for this circuit available from TI, to find out the approximate time as needed for the application in mind.
Thanks!
Sorry, no equations, no tools, no graphs.
Circuit is pretty straight-forward.
Resistors R1 (240) and R2 (2.7k) set the final Vout.
Diode (1N4002) discharges C1 at power-down
Transitor PNP (2N2905) buffers the voltage on the delay capacitor back to the ADJ pin, and provides a current path to charge the delay capacitor.
At t=0:
Vin is applied ;
the voltage on the delay capacitor= 0V;
the voltage across R1= Vref=1.25V
the current through R1= Vref/R1= 1.25V/240= 5.2 milli-amp
the voltage across R2= Vbe+Vcap= 0.65V+0.0V= 0.65V
the current through R2= 0.65V/2.7k= 241 micro-amps
the current through R3= 0.65V/51k= 13 micro-amps
the current through 2N2905 emitter= 0.0052-0.000241-0.000013= 4.95milli-amp
the current through 2N2905 base= 4.95milli-amp/150 = 33 micro-amps (note the current gain here is a best guess)
the voltage at Vout= Vref+Vbe+Vcap = 1.25V+0.65V+0.0V= 1.90V
The final voltage across the capacitor will be the same as the final voltage across R2 = ((Vref/R1)xR2)= ((1.25V/240)x2.7k)= 14.06V
But at some point the voltage on the cap will be less than a Vbe away from the voltage across R2 and the charge current into the cap will start to drop off to zero.
To keep it simple I will just presume that the circuit will charge Cdelay in straight line (it's not) from 0V to 14.06V-Vbe= 13.41V
The average charge rate will be about 50% of the sum of the peak base current (33 micro-amps) and the peak current through R3 (13 micro-amps) = 46 micro-amps/2= 23 micro-amps
time = ((C / I) x dV) = ((25 uF / 23 uA) x 13.41V = 14.6s
Simulation shows that's pretty close ...
