RC timing circuit

Question

I'm reading AOE 3rd edition, book say below:

Here’s how the circuit works: the voltage divider R3R4 holds the (−) input at 37% of the supply voltage, in this case about +1.8 V; let’s call that the “reference voltage.”

So if the circuit has been sitting there for a while, C1 is fully discharged, and the comparator’s output is at ground. When you push the START button momentarily, C1 charges quickly (10 ms time constant) to +5 V, which makes the comparator’s output switch to +5 V; see Figure 1.39. After the button is released, the capacitor discharges exponentially toward ground, with a time constant of τ = R2C1, which we’ve set to be 1 minute. At that time its voltage crosses the reference voltage, so the comparator’s output switches rapidly back to ground. (Note that we’ve conveniently chosen the reference voltage to be a fraction 1/e of V+, so it takes exactly one time constant τ for that to happen. For R2 we used the closest standard value to 6 MΩ; see Appendix C.) The bottom line is that the output spends 1 minute at +5 V, after the button is pushed.

My question is:

1. What is between resistor R1 above and the connection point below in Figure 1.38.
2. If R1 and R2 are considered as a voltage divider, can R2 and C1 be swapped? Will the circuit be the same as the original one after swapping?
3. In caption of Figure 1.39, my problem point is : R1C1 ≈10 ms in "The voltage VC1 has a rise time of R1C1 ≈ 10 ms."

For question 3 confused: I calculated that the product of resistor R1 and capacitor C1 equals an accurate value of 10 ms, $$(10^3)(10 * 10^{-6}) = 10^{-2} second = 10 ms $$ so I think it should be "The voltage VC1 has a rise time of R1C1 = 10 ms.", simply say, should use R1C1 = 10 ms, not use R1C1 ≈ 10 ms.

And if R1 and R2 are considered as a voltage divider, then the total resistance Rth of R1 and R2 in series is $$ \frac{(10^3)(6.2 * 10^6)}{(10^3) + (6.2 * 10^6)} = 999.8387Ω ≈ 1000Ω $$ The Rth ≈ 1k(approximately equal to R1 that its value is 1kΩ), then calculate $$ Rth*C1 ≈ (10^3)(10 * 10^{-6}) = 10^{-2} second = 10 ms $$

I see https://artofelectronics.net/errata/ say Page 24 modify change "1 ms" to "10 ms" in caption of Fig. 1.39, please see below picture.

4. The last question is about: "After the button is released, the capacitor discharges exponentially toward ground, with a time constant of τ = R2C1, which we’ve set to be 1 minute", I calculate $$ R2C1 = (10 * 10^{-6}) (6.2 * 10^6) = 62 second $$ not one minute, because one minute is 60 second, so why author say one minute?

Answer 1

1. That's the pushbutton (momentary) switch.
2. Swapping R2 and C1 has no functional effect, since they're in parallel.
3. It's approximate because of R2. Even though it has a very large value compared to R1, it does throw off the time constant slightly, as you calculated.
4. That's why the text says, "For R2 we used the closest standard value to 6 MΩ; see Appendix C." In other words, you can't buy a 6 MΩ resistor, but you can buy a 6.2 MΩ resistor — this was deemed to be "close enough".