| Electronics > Beginners |
| Schmitt disappointment |
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| PerranOak:
Excellent, thank you all. There is a 100nF cap near-ish to the supply pins but the pins are on the diagonal opposites of the chip - mental! What would be a low ESR? My one has ESR = 15Ohms but I have no way of measuring the inductance of it. How do you know what the impedance of the co-ax is? Is it as simple as measuring it in the normal way? Cheers. |
| wraper:
--- Quote from: PerranOak on October 04, 2018, 04:22:43 pm ---My one has ESR = 15Ohms but I have no way of measuring the inductance of it. --- End quote --- It's useless to measure ESR of ceramic capacitors. At 100 kHz you won't get any meaningful figure. |
| Mr. Scram:
Why do many chips have power and ground on opposite sides of the chip? Isn't that exactly what you don't want from a loop area point of view? |
| Zero999:
--- Quote from: Mr. Scram on October 04, 2018, 06:31:40 pm ---Why do many chips have power and ground on opposite sides of the chip? Isn't that exactly what you don't want from a loop area point of view? --- End quote --- It's probably a relic from back when decoupling wasn't essential. Many of the older CMOS ICs, such as the CD40106 will work perfectly with little or no decoupling capacitors. |
| StillTrying:
--- Quote from: PerranOak on October 04, 2018, 04:22:43 pm ---There is a 100nF cap near-ish to the supply pins but the pins are on the diagonal opposites of the chip - mental! --- End quote --- On a breadboard I use a 100n with leads long enough so that the cap is above the IC, the breadboard's power lines are too many connections and uHs away to supply power for ns edges. :) |
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