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Why we have to use a 6-metal process for RF?

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yikwon1

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I have some questions.
When we design the RF tranceiver, we usually use a 6-metal layer.

I slightly think that the reason is only inductor because inductor have to get
a high Q. is that right?

There are no methods to design for RF using 4-metal?

I'm wondering now.

Please give me some answers...

Thanks.
 

There are RF designs in 2 layer metal. I don't know why you would need 6
 

Why only 6? The more you have, the further is the inductor above the substrate, which is good.
 

The top metal and sometimes the top metal -1 layers are used for Inductors and MIM caps and usually use large minimum metal line widths as a consequence. So these layers cannot be used for dense metal routing.
This means RF designs suffer from restrictions on 2 metal layers. So the 6 layers comes from allowing 4 layers for dense interconnects.
Normally a Foundry will allow 6+ metal layers for RF as long as the last 2 layers conform to the larger min metal line widths and spacings.
To go lower than 6 say to 4 would mean you only have 2 dense metal layers to xork with. This might be OK for your design but could have other consequences.
As mentioned before, the inductor (and MIM) needs to be far enough away from the Si surface to guarantee performance.
Also, some foundries allow Active circuitry under pad. This can only be allowed if there is sufficient height above the circuitry before large stress (metal features) exist.
For RF, with large top metal features, 4 layers would not be enough.
 

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