General equation for the capacitor voltage:

The general equation for the capacitor voltage may be written as

 V_c (t ) = A + B e^{- t/ T0}

where

T₀   =   C₀  R_A  R_B /(R_A   + R_B  )

If there were no control, at t = ∞ ,

V_c (t ) =   (R_B /(R_A  + R_B ) )Vcc

Substitution of this in Eq. (105) gives

       A + B = (2/3 )Vcc

Also, in infinite time (that meanst = ∞ ) capacitor voltage shall attain a steady value of R_B /(R_A  + R_B ) Vcc , because of which from Eq. (105), we obtain, A =         (R_B ) /(R_A  + R_B ) V_cc , therefore, one finds

               B = (2/3 ) V _cc   - (R_B )/ (R_A   + R _B )   V                  _cc = (2R_A  - R_B ) V_cc / 3 (R _A + R _B  )

The required equation is, thus,

V  (t ) = (R_B )  / (R_A+ R _B ))      V_cc = ((2R_A  - R_B ) V_cc / 3 (R_A+ R _B )) e ^{- t /T0}                           

Now, for t = T_L, V_c (t) =   V_cc /3 , therefore, substituting these values in Eq. (105), we determine

V _cc {(1/3) -  (R_B ) / ( R_A+ R _B   )     =   (2R_A   - R_B  ) V_cc / 3( R_A+ R _B   ) e^{- TL/ T0}

which simplifies to

 (R_A  - 2R_B ) =(2R _A  + R_B   ) e ^{- t/T0}

e ^{- t /T0}  = 2R_A  - R_B / R_A  - 2R_B

∴          T _L  = (C₀ R_A R_B  /C_{RA +} R_B ) ln (2R_A   - R_B  / R_A  - 2R_B) ; R _A> 2R_B

Furthermore, T_H = C₀ R_A ln 2 (as usual), therefore, 50% duty cycle shall be attained if

 (R_B )  / (R_A+ R _B )) ln (2R_A   - R_B  / R_A  - 2R_B) =     ln 2

(2R_A   - R_B  / R_A  - 2R_B) = 2 ^{(1+ (RA/RB))}

2 ((R_A  /  R_B  -1))/ ((R_A  /  R_B  -2)) = 2 ^{(1+ (RA/RB))}

2K - 1/ K - 2  = 2^{1+ K}

where

R_A/ R_B    = k

Clearly, Eq. (7.110) does not have a closed solution, but it can be solved by hit and trial, by picking up some value of k and then seeing that RHS equals LHS. This happens at k = 2.3, hence, if we take RB = R then Ra should be 2.3 R.