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Diodes used for Antenna Error correction

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nandam

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Hi,

I am in the confusion of using diodes for Antenna error correction.
For a PMOS gate,
1.When I put diode in nwell, the P-type is connected to the net and n-type to VDD(the well supply),
the charges (accumulated during fab) at the gate will forward bias the diode there by damaging the gate oxide connected to net.
2.For the same PMOS, I put diode in P-sub, the n-type is connected to the net and p-type to VSS(the sub supply).
During etching, the charges accumulated on the gate will reverse bias the diode, and there is no path for discharging of charges to sub.
Then how does it help.

The second point is confusing can anyone plz help me on this.

Thanks,
Nanda.
 

Hi Nanda,

Diodes can be used for Antenna correction which are in reverse bias condition. So that, in normal mode wont affect the functionality.
1. PMOS gate
============
-->Add diode in Nwell, tie Nwell ( n terminal of diode) to VDD and gate node to signal (p terminal of diode). So now diode is in reverse bias condition.
-->During the Etching process or fabrication, charge accumulated on Metal strip or gate will be discharged through diode by protecting the gate oxide.
-->In normal function this diode will add the reverse biased cap on that GATE node.

Similarly for NMOS gate.

Thanks and regards,
Basu
 

What you miss, is that these diodes become conductive
during the etch process and do not charge-pump the
way you assume. The input current is low and the UV
/ soft X-rays cause photoconduction. The only goal is
to ensure that every gate oxide has a parallel silicon
path to its opposite side, that will not leak objectionably
when the part is powered up.
 

Thanks Basu/dick_freebird.
I agree with previous post...
1.When charges accumulated during etching might be so high that it might be more than VDD there by forward biasing the diode, which again would damage the gate oxide.
2.Its always advisable to put diodes in P-sub. How its going to help, either we dont have a path for charge diffusion as the diode is reverse biased.

Thanks,
Nanda.
 

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