Does this schematic need so many pull down resistors?

Wouldn’t 1 pull down resistor on COM pin be enough?

Why not try it and see if it works?

I don’t have it, just curious

Perhaps look at the internals of the device.

Note that my knowledge of electronics is very rusty but my logic would be that floating inputs will reflect on the COM pin; so yes, you will need resistors on all pins.

I think the single pull down on Pin 3 would be OK as this is an analog bidirectional multiplexer it only connects any addressed pin to Pin3. the pins are not input but input and output at any time, this means you can make pin3 as a common input and vise versa so when addressed 1 of 8 pins the if the switch is not pressed microcontroller will read 0 and if switch is pressed a 1 would be read by micro controller.

https://www.ti.com/lit/ds/symlink/cd4052b.pdf?ts=1639198708690&ref_url=https%3A%2F%2Fwww.ti.com%2Fproduct%2FCD4052B doesn’t seem to contain any internal circuitry

Only 1 input is connected at a time, so only 1 out of 8 resistors is pulling the MCU input at any time, and they all identical value, might as well be one on the other side

Hi,

It is an application for a keypad, common practice to use pull down resistors on each key.
Also without the resistors, you will have open circuit CMOS input pins when key not pressed, not a good idea.

Tom... :grinning: :+1: :coffee: :australia:

Why CMOS? It is because it shows 3.3v?

The CD4051B is a CMOS device, read the top of the page on the link you provided.

Hi,

Because it is CMOS....
Even though it is a 4051B meaning, the B, the input is buffered.
This is to protect the high impedance of the gate.
Which static electricity can damage, hence the conductive foam and tubes in storage.

Leaving a CMOS input open can lead to unwanted oscillation or damage due to ESD conditions.

Even though buffering is included to protect the input it is considered to be Best practice not to leave it open circuit.

Tom... :grinning: :+1: :coffee: :australia:

so what do you do if analog signal is fed to those inputs, via capacitor? do you pull them down too?

Hi,

You can't capacitively couple to the input, that would mean a signal at the input pin that goes positive and negative with respect to its gnd.
4051B cannot work with negative input, like most digital ICs.

Tom... :grinning: :+1: :coffee: :australia:

I don’t understand what you mean. the capacitor is in series, but even if it was in parallel why this implies negative input?

Hi,

If you put a capacitor in series with the input, as the voltage on the source side increase and decreases, the capacitor will charge and discharge.
This means the current effectively through the IC input will be oscillating in and out, this translates to a positive and negative potential on the input pin.

The same reason you cannot place an AC signal on an AVR controller input pin, analog or digital.

Tom.... :grinning: :+1: :coffee: :australia:

OMG guys do you know CD405x is an analog MUX? There is no digital input buffer on the Chn and COM pins. There is no problem with floating voltage there. The voltage may be anywhere between VDD and VEE - you may provide negative supply and the chip will handle even negative "input" voltage.

I don't know why they show pull-downs on all the buttons. Maybe a brainless copy of another design?

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What a funny - and thoroughly muddled - discussion. :roll_eyes:

Yes, a single pull-down on the common - or very much preferably, a pull-up with the switches going to ground - is the proper way to implement this function.

Yes, there is no digital input buffer on the "switch" pins(, the "B" suffix refers to full buffering on the digital control pins), so floating voltages are of no concern. Indeed the very opposite - this is actually an analog multiplexer so any voltage between the supply rails is precisely as valid as any other.

The voltages on the switching terminals must stay within the range of VEE to VDD and indeed are presumably so constrained by protection diodes, not necessarily the same as the protection cited for the digital control signals.

CD4000_protection

That sounds as good an explanation as any. :grin: More a "concept" circuit than a genuine design recommendation. Like the bungled potentiometer connection on the HD44780 designs.

Note the CD4051B as a member of the CD4000 series is essentially obsolete except for a design requiring more than 10 V span of the switched voltages (±5 V from VSS given VEE is -5 V). Apart from that exception, you would use the higher-specced 75HC4051.

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It seems to be okay since it's an analog mux and as you can see in the block diagram its pretty simple, which is enhanced by the fact that this is also a bidirectional mux. However, the convention is to have pull resistors on all the input pins (and not on the output pins), especially with something that are high-speed like a keyboard.
If I were you, I won't be using a mux at all. I will be using a shift register.

And just which convention might this be? :rofl: