Dear fellow Engineers,
I am currently developing a 3V to 48V step-up at 20mA-50mA around the LM3478. Since the webbench fails to provide an adequate solution (no transistor found), I tried to find a solution myself using the equations from the datasheet.
Long story short: the evaluation of the current limit after calculating the external slope compensation fails. The problem arises at the calculation of Vcs as Vcs = Vsense - D*(Vsl+dVsl). With Vsl = 0.092V and D = 0.94, dVsl = 40e-6*Rsl. Rsl is about 5.6k in my case. And Vcs goes negative, based on above formula. Actually, for D that high (0.94) Rsl would need to be very small to not let Vcs go negative.
My question now: is there any way around that? Do I miss something here?
Kind regards
Detailed design for Reference:
--- duty cycle
D = 1-((Vin-Vq)/(Vout+Vd))
Vin = 3V
Vout = 48V
Vq = 0.1V
Vd = 0.3V
D = 0.94 (and here might be the problem)
--- inductor size
L > D(1-D)Vin/(2Iout*fs)
D = 0.94
Vin = 3V
Iout = 0.02A
fs = 1e3
L > 4.23uH
Chosing 6.8uH (4.7 is even worse)
--- peak inductor current
Ilpeak = avgIl + dIl
avgIl = Iout/(1-D)
dIl = (D*Vin)/(2*fs*L)
Iout = 0.05A
D = 0.94
Vin = 3V
fs = 1e6
L = 6.8uH
avgIl = 0.84A
dIl = 0.21A
Ilpeak = 1.05A
Rf2 = (1.26*Rf1)/(Vout-1.26V)
Rf1 = 100k
Rf2 = 2.7k
--- current limit and sense resistor
Rsen = (Vsense-D*Vsense*Vslratio)/(ISWlimit)
ISWlimit = Iout/(1-D)+(D*Vin)/(2*fs*L)
Vsense = 156mv (135 to 180)
D = 0.94
VSLratio (VSL/Vsense) = 0.49 (0.3 to 0.7)
Iout = 0.02A
Vin = 3V
fs = 1e6
L = 6.8u
ISWlimit = 0.854A
Rsen = 0.0985
Use Rsen = 0.1R
--- slope compensation
check if Rsen < (2*Vsl*fs*L)/(Vo-(2xVin))
Vsl = 0.092
-> 0.03
Rsen is larger than limit. Slope compensation needed!
chose Rsl > ((Rsen*(Vo-2Vin))/(2*fs*L)-Vsl)/40uA
Rsl > 5.42k
Use Rsl = 5.6k
--- reevaluation of current limit etc
Vcs = Vsense - D*(Vsl+dVsl)
dVsl = 40uA*Rsl
ISWlimit = Vcs/Rsen
Vsense = 156mv (135 to 180)
Vsl = 0.092
dVsl = 0.224
Vcs = NEGATIVE VALUE!
