Yes, I realized later that I could use D1 as the jumper for the regulator.
I probably have this all wrong but with the diode dropping .25V and a cell at 2.8V and with R1 and R2 at 19.1k and 4.7k pin 7 will see (2.8-.25)(4700/23800)= .5V
To get low voltage regulation to work I need to be able to show pin 7 - .5V when Vb=5.6V.
If the voltage drop across the diode is some value other than .25v then .5V is the wrong target for the divider.
Do I ignore the diode in the calculation? 2.8 x .1975= .55V.
5.6(4700/R1+4700)=.55
5.6(4700)=.55(R1+4700)=.55R1+.55(4700)
(5.6(4700) - .55(4700))/.55 = R1 = 43155ohms
Using R1 = 43155 and accounting for the same .25v diode drop pin 7 should actually see .5V. Let’s try it.
(5.6-.25)(4700/(43155+4700)=.53V not .5V.
This was assuming the voltage divider was left connected after the diode and the voltage drop across the diode ignored.
Try again with a target of .5V and connecting again after the diode.
(5.35(4700) - .5(4700))/.5=. R1 = 45684 ohms.
And one more before the diode
(5.6(4700) - .5(4700))/.5 = 47940 ohms .
I have the feeling I’ve missed something crucial but at this point I figure I’ll just try some different values for R1 and see what happens but to me it looks like with R2 at 4.7k and
Divider connected after the diode, R1 = ~ 45.7k ohms.
Divider connected before the diode, R1 = 47.9k ohms
Whether you put the divider before or after, I don’t see how the diode can be ignored.