Why is this pin connected like this? Isn't it usually left floating?

拿得起铁

Why is this pin connected like this? Isn't it usually left floating? [Copy link]

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Why is this pin 1 connected like this? Isn't it usually left floating?

频率元件专家

U6 is an active crystal oscillator. Pin 1 is the enable pin (low active).

拿得起铁

If it is an enable pin, how is it connected to the power supply through a resistor? It is clearly marked as low level enable.

频率元件专家

It must be a mistake.

拿得起铁

But isn't pin 1 of the crystal oscillator usually left floating?

频率元件专家

You need to see what kind of crystal it is. This is not an ordinary crystal. This is an active crystal.

拿得起铁

Oh, what is that crystal? What's so special about it? It's for FPGA.

频率元件专家

If pin 1 is not enabled, there is no 50MHz frequency output. This crystal has an oscillation circuit inside.

拿得起铁

I mean, what's the difference between this and a normal crystal oscillator? Is it just an extra enable pin?

频率元件专家

Can work independently

拿得起铁

Yes, there is an active crystal oscillator.

频率元件专家

Ordinary crystal oscillators are resonators that need to work with the oscillation circuit in the main chip.

拿得起铁

I mean, what is the difference between this crystal oscillator with an enable pin and an active crystal oscillator with pin 1 generally left floating? Is the difference in operation just that there are multiple enables?

频率元件专家

Yes

拿得起铁

Oh, how is the resistor size determined? 10K

频率元件专家

It plays the role of low standby current consumption. This is a simple question of the high and low levels of the circuit. If you are not afraid of high standby current, you can choose 1K or smaller.

拿得起铁

I understand it roughly, but I don't know if there is an internal circuit. Can you tell me how to reduce the current?

lx773533

VCC pulls the level down to /OE through a 10K resistor, and the current is very small at this time

ming0303

study