cy8ckit: disable 32.768kHz crystal

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RoAl_1309171
Level 2
Level 2

Hi,

I have a cy8ckit-059. I would like to use pins P15[2] and P15[3] as GPIOs, but they're wired to the 32.768kHz RTC crystal. Is there any way to do that, short of desoldering the crystal and its capacitors?

Thank you,

  Bob

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1 Solution
Len_CONSULTRON
Level 9
Level 9
Beta tester 500 solutions authored 1000 replies posted

Bob,

JoMe (Bob) is correct.  The two 22pF caps (C41 and C42) should be of little consequence.

He is correct in saying that if your signals on these pins aren't too fast, they will be fine.  This applies to these pins as inputs or outputs.

Let's make the following assumptions:

  1. Your worst-case use for each pin is as an output.
  2. Output source current needed (Ioh) = 4mA.
  3. Your output source voltage is VDDIO - 0.6V.
  4. This means your output drive resistance/impedance = 0.6V/4mA = 150 ohms.
  5. The maximum capacitance on the circuit is 30pF
  6. Therefore your RC (tau) of the circuit is 150 ohms * 30pF = 4.5 x 10^-9 secs = 4.5ns.
  7. 1/RC = 222MHz

Therefore, your probably safe using it as an input or output.

If you're using it as an input, check your output drive impedance and substitute that value in 4.

If you're got more than 30pF total capacitance on the circuit and using the pin as an output, consider using an external logic element if a digital signal or an opamp if analog.

Len

Len
"Engineering is an Art. The Art of Compromise."

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5 Replies
Bob_Marlowe
Level 10
Level 10
First like given 50 questions asked 10 questions asked

The schematic for the CY8CKIT-059 shows a "No Load" for the 32kHz crystal. The 22pF caps should make no harm when the switching frequency is not too high.

Bob

Len_CONSULTRON
Level 9
Level 9
Beta tester 500 solutions authored 1000 replies posted

Bob,

JoMe (Bob) is correct.  The two 22pF caps (C41 and C42) should be of little consequence.

He is correct in saying that if your signals on these pins aren't too fast, they will be fine.  This applies to these pins as inputs or outputs.

Let's make the following assumptions:

  1. Your worst-case use for each pin is as an output.
  2. Output source current needed (Ioh) = 4mA.
  3. Your output source voltage is VDDIO - 0.6V.
  4. This means your output drive resistance/impedance = 0.6V/4mA = 150 ohms.
  5. The maximum capacitance on the circuit is 30pF
  6. Therefore your RC (tau) of the circuit is 150 ohms * 30pF = 4.5 x 10^-9 secs = 4.5ns.
  7. 1/RC = 222MHz

Therefore, your probably safe using it as an input or output.

If you're using it as an input, check your output drive impedance and substitute that value in 4.

If you're got more than 30pF total capacitance on the circuit and using the pin as an output, consider using an external logic element if a digital signal or an opamp if analog.

Len

Len
"Engineering is an Art. The Art of Compromise."
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RoAl_1309171
Level 2
Level 2

Thank you both for your very helpful answers!

- Bob

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Bob,

You're welcome.  Thank you for the 'Correct Answer" credit but JoMe_264151 answered correctly first.

Len

Len
"Engineering is an Art. The Art of Compromise."
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I wish I could mark both your answers as correct.

- Bob

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