Commit | Line | Data |
---|---|---|
468dbff9 WG |
1 | Patches: |
2 | ||
3 | 1) Swap U1 Vin/Vout -- 3 cuts + 3 wires | |
4 | ||
5 | Notes: | |
6 | ||
7 | 1) U1 Vin/Vout pins are swapped | |
8 | 2) Add a reset circuit | |
9 | ||
10 | Use a SNLVC2G06 (dual open collector inverter): | |
11 | ||
12 | At VCC = 5V: | |
13 | ||
14 | VIH = .7 * VCC = 3.5V | |
15 | VIL = .3 * VCC = 1.5V | |
16 | VOL = .55V | |
17 | IOL = 32mA | |
18 | ||
19 | Assuming that the capacitor is clamped to VOL most | |
20 | of the time, the R*C that corresponds to charging | |
21 | from VOL to VIH in T seconds: | |
22 | ||
23 | The basic capacitor charging equation: | |
24 | ||
25 | V(t) = VOL + (VCC - VOL) * (1 - e(-t/R*C)) (1) | |
26 | ||
27 | Replacing V(t) with VIH and T: | |
28 | ||
29 | VIH = VOL + (VCC - VOL) * (1 - e(-T/R*C)) (2) | |
30 | ||
31 | Solving for RC: | |
32 | ||
33 | (VIH - VOL)/(VCC - VOL) = 1 - e(-T/R*C)) (3) | |
34 | ||
35 | e(-T/R*C) = 1 - (VIH - VOL)/(VCC - VOL) (4) | |
36 | ||
37 | -T/R*C = ln(1 - (VIH - VOL)/(VCC - VOL)) (5) | |
38 | ||
39 | R*C = -T/ln(1 - (VIH - VOL)/(VCC - VOL)) (6) | |
40 | ||
41 | Substituting in for T=1ms, VIH, VOL, and VCC: | |
42 | ||
43 | R*C = -.001/ln(1 - (3.5 - .55)/(5 - .55)) (7) | |
44 | ||
45 | R*C = -.001/-1.08744 | |
46 | ||
47 | R*C = .009196 (8) | |
48 | ||
49 | Setting C = .1uF | |
50 | ||
51 | R = .009196 / .0000001 = 9196 | |
52 | ||
53 | Using an R near 10K should do the trick: | |
54 | ||
55 | The maximum discharge current is IOL. Use Ohm's law | |
56 | to determine determine discharge resistor: | |
57 | ||
58 | V = I * R | |
59 | R = V / I | |
60 | = 5 / .032 | |
61 | = 156.25 | |
62 | ||
63 | Setting Rdischarge = 180 should be good enough. | |
64 | ||
65 | The charge/discharge ratio is: | |
66 | ||
67 | Rcharge/Rdischarge = 10K/180 = 55.5 | |
68 |