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I have circuit that includes various 74HCxxx series logic chips. Four of the logic lines represent a 4 bit number and I have a 7 segment display used to show the value of that 4 bit number. I use an MC14495 chip to convert from the 4 bits to the 7 segment display lines. This is setup and works fine on a breadboard as a prototype. But I am only clocking it manually by toggling a switch. My worry is that a real version is clocked a 1Mhz and maybe the fast switching 7 segment display will need a bypass capacitor?

(Some of the time the values shown will change so fast at 1Mhz that you will need to be able to make us of it. But there will be times when the value is static for several seconds and hence it does provide a use.)

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  • \$\begingroup\$ Why would you not simply include the $.01 part and be done with it? \$\endgroup\$ – Olin Lathrop Nov 10 '16 at 12:14
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A breadboard tends to have high capacitance due to its construction as multiple parallel contacts, and will often act as bypass capacitors for ICs. Transferring to a PCB without adding caps tends to leave some designs not working.

That said, the 7 segment display is just bare leds, and those do not need a bypass capacitor. The ICs you are using will. 0.1 µF caps as close to the VCC and Gnd pins for your logic ICs and the binary coded decimal led driver chip.

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It depends on the "internal resistance" / impedance of your power supply, that is supplying the LEDs, since they will be drawing most of the current in the setup you describe.

But another issue would be whether the current switching spikes could falsely trigger your logic devices, by disturbing the "ground" voltage due to the return current.

Hence a good policy might be to have separate wires / feed tracks ("branches" below) to the logic and the LED drive, in a "star" rather than "daisychain" arrangement - both for Vcc/Vdd (+ rail) and 0V.

Then use a low-impedance (eg ceramic or polymer) 0.1uF cap on each drive chip, to the ground pin of the LEDs, and (say) a 10uF capacitor across the power on the device on the LED branch closest to the power supply.

So the 0.1uF's supply spikes of current, and the 10uF smooths things out generally, supplying current to the local branch and effectively reducing the impedance of the power supply for the LED branch.

You should adopt a similar approach for the logic branch, but since the current is presumably lower, you could use say, 0.01uF devices.

Basically the capacitors are there "on hand" to supply local short-term pulses of current where it's needed, rather than that spike of current having to flow up and down the wires to the power supply there and then. Hence they should be as close as possible to where they are needed - both on the Vcc AND 0V side .... and the capacitor leads kept as short & straight as possible to reduce effective inductance,

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