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I'm trying to understand part of an envelope generator circuit. Here's the part of the schematic for the GATE IN, marked by the K, which can accepts a signal of between 0V and +5V.

I understand that the 470k resistor is a pullup resistor which will keep the NOT gate (a CD4069 hex inverter with a +5V supply voltage) input pin high if there is no signal from the gate.

What's the purpose of the 330k resistor though? Why shouldn't I just connect K directly to the input pin?

Update: More details on the GATE IN signal.

This circuit is intended to be used in a modular synthesizer. The GATE IN control voltage is expected to trigger the attack stage of the envelope generator at +5V and the release stage at 0V.

There shouldn't be any other values presented to GATE IN. However, other voltages present in modular sythesizers typically range from -15V to +15V, and with some effort (to break things) could be presented to GATE IN.

enter image description here

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    \$\begingroup\$ It's impossible to say for certain without knowing the signal applied at "Gate in", and without knowing the gate's specifics. Probably, this solves as voltage level shifter. \$\endgroup\$ – Marcus Müller Apr 21 '20 at 21:42
  • \$\begingroup\$ Thank you. I added some more details about the "Gate in". \$\endgroup\$ – donturner Apr 21 '20 at 21:53
  • \$\begingroup\$ what model / kind of device is the NOT gate? \$\endgroup\$ – Marcus Müller Apr 21 '20 at 21:54
  • \$\begingroup\$ It's a CD4069UBE hex inverter: ti.com/lit/ds/symlink/cd4069ub.pdf \$\endgroup\$ – donturner Apr 21 '20 at 21:56
  • \$\begingroup\$ ah, that is variably supply voltage: What's the supply voltage of your CD4069UBE? \$\endgroup\$ – Marcus Müller Apr 21 '20 at 22:10
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It's probably to protect the gate input from voltages outside the range of 0~5V, including ESD.

The relatively modern 4069UB on your linked datasheet is rated to withstand +/-10mA at the input, which would theoretically represent more than +/-6kV at the input. In practice the resistor would probably flash over and the supply voltage might get lifted to a destructively high voltage for a high positive input voltage via the two resistors shown.

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  • \$\begingroup\$ Sorry, what do you mean by "flash over"? \$\endgroup\$ – donturner Apr 22 '20 at 7:12
  • \$\begingroup\$ I think I understand. A very high positive input voltage might arc the 330k resistor causing the 4069 supply voltage to be forced to a "destructively high voltage". Could ESD do that? Are you saying the 330k resistor is probably unnecessary? (sorry for all the questions, really just trying to get a good grasp of audio circuit design) \$\endgroup\$ – donturner Apr 22 '20 at 9:12
  • \$\begingroup\$ Yes, spark over or arc. It depends what the resistor construction is like- a smallish 0402 or smaller resistor might be rated for 10-50V. only, and actually fail at a few hundred V. It is not unnecessary- it makes the input far more robust, but it's not necessary to functionality. Leaving off protection on a CMOS IC input connected to the outside world would be very bad practice generally. \$\endgroup\$ – Spehro Pefhany Apr 22 '20 at 14:31
  • \$\begingroup\$ Thank you. Is there a common term for this resistor, like "current limiting resistor" or "protection resistor"? And how would one choose a suitable value for it? \$\endgroup\$ – donturner Apr 22 '20 at 15:14
  • \$\begingroup\$ I would say "current-limiting resistor" or "protection resistor" depending on whether it's intended to be deliberately driven outside the range. My ca. 1970 First Edition McMOS (Motorola) CMOS Handbook does not use a consistent term, but refers to them. The maximum value is limited by leakage from the input (and PCB), typically worst at maximum temperature (specified as 1uA max), and by how much you can allow the input to be slowed due to input capacitance (15pF max). So 1uA * 330k = 0.33V. The 1uA is also a very loose spec (typical is much less). \$\endgroup\$ – Spehro Pefhany Apr 22 '20 at 15:26

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