Added rev. C

[challenge-bot] / pcb / reset_notes.txt

diff --git a/pcb/reset_notes.txt b/pcb/reset_notes.txt
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+Patches:
+
+1) Swap U1 Vin/Vout -- 3 cuts + 3 wires
+
+Notes:
+
+1) U1 Vin/Vout pins are swapped
+2) Add a reset circuit
+
+Use a SNLVC2G06 (dual open collector inverter):
+
+At VCC = 5V:
+
+   VIH = .7 * VCC = 3.5V
+   VIL = .3 * VCC = 1.5V
+   VOL = .55V
+   IOL = 32mA
+
+Assuming that the capacitor is clamped to VOL most
+of the time, the R*C that corresponds to charging
+from VOL to VIH in T seconds:
+
+The basic capacitor charging equation:
+
+  V(t) = VOL + (VCC - VOL) * (1 - e(-t/R*C))                   (1)
+
+Replacing V(t) with VIH and T:
+
+  VIH = VOL + (VCC - VOL) * (1 - e(-T/R*C))                    (2)
+
+Solving for RC:
+
+  (VIH - VOL)/(VCC - VOL) = 1 - e(-T/R*C))                     (3)
+
+  e(-T/R*C) = 1 - (VIH - VOL)/(VCC - VOL)                      (4)
+
+  -T/R*C = ln(1 - (VIH - VOL)/(VCC - VOL))                     (5)
+
+  R*C = -T/ln(1 - (VIH - VOL)/(VCC - VOL))                     (6)
+
+Substituting in for T=1ms, VIH, VOL, and VCC:
+
+  R*C =  -.001/ln(1 - (3.5 - .55)/(5 - .55))                   (7)
+
+  R*C = -.001/-1.08744
+
+  R*C = .009196                                                        (8)
+
+Setting C = .1uF
+
+  R = .009196 / .0000001 = 9196
+
+Using an R near 10K should do the trick:
+
+The maximum discharge current is IOL.  Use Ohm's law
+to determine determine discharge resistor:
+
+   V = I * R
+   R = V / I
+     = 5 / .032
+     = 156.25
+
+Setting Rdischarge = 180 should be good enough.
+
+The charge/discharge ratio is:
+
+  Rcharge/Rdischarge = 10K/180 = 55.5
+