## October 17, 2014

### Latch-Up in CMOS

CMOS device is often portrayed to be an impeccable device, especially in the textbooks. There are some innate problems in the CMOS device and one of them is the latch-up. We're gonna talk about it in detail.

Consider the cross-section of a CMOS inverter. Please note that I have skipped drawing some metal layers and contacts for the sake of simplicity. My focus is on explaining the problem of latch-up and not the layout design rules!

 Figure 1: CMOS with parasitic BJTs

In the above cross-section, note that 1-2-3 form a parasitic pnp type bipolar junction transistor, while 4-3-2 form a parasitic npn bipolar junction transistor. Since parasitic transistors would be present in every CMOS device! A simplistic figure depicting these parasitic transistors is given below:

 Figure 2: Simplistic Figure depicting parasitic BJTs

Here, the npn and the pnp transistors are depicted with 2 and 3 being common between the two transistors. Also note that the n-well layer and the p-substrate ate lightly doped layers, and hence offer greater resistance as compared to the n+, p+ drain and source regions. The n-well resistance of PMOS is depicted by the resistor R2 and the resistance of p-substrate is depicted by the resistor R1.

Let's say, we've got a spike at the output of the CMOS inverter.

1. This is a negative spike (or a bump), which decreases the potential of VOUT below ground potential by 0.7V.
2. As a result, the npn transistor 4-3-2 gets turned ON, and the emitter (n+, 4) starts emitting electrons, which eventually get collected at the collector (n-well, 2) and go into VDD.
3. The current hence flows in the reverse direction from the n+ body towards the n-well region.
4. We will have a voltage drop in the direction of current, as a result the potential at the n-well can reach 0,7V below VDD. This turns the pnp transistor ON, because p+ region emitter is at VDD!
5. The current is collected at the collector (p-substrate, 3) and flows into the ground through the p+ body region.
6. Going opposite to the direction of current, the potential difference increases, and the voltage at p-substrate (just below the n+ of NMOS, might reach 0.7V, thereby injecting more current!
 Figure 3: Sequence of events leading to latch-up
As evident from the above steps, just one spike at the output initiates a chain reaction resulting in current flow through the device incessantly, and hence the device gets worn out in a very short span of time! This is the latch-up!

How to mitigate the problem?
Well, as we just noticed, the main cause of latch-up was the resistance of n-well and the p-substrate layers. Hence, the most logical solution would be to increase their doping concentration by ion implantation. But that would deteriorate the transistor operation! Remember, we need to keep the doping of wells and substrate low; while those of source and drain high yo ensure a good transistor operation.

What do we do now?
Well, we shall stick with ion implantation, but instead of doing it near the silicon surface and hence near the drain and source, implant a deep n-well with high doping concentration, and similarly, implant a deep p-well inside the p-substrate to reduce the resistance and hence kill the parastic BJT action!

References:
• Latch-up in CMOS, NPTEL lectures by Dr. Nandita Dasgupta.

1. How does low resistance path kill BJTs action ?

2. if the resistance become low, it will reduce the voltage drop. so BJT will not get sufficient voltage at the base to turn it on,so this will kill the BJT action

3. how does the npn transistor gets on for the negative spike??
and also pnp transistor's current flow???

4. Pls check this blog also
http://backendesign.blogspot.in/

5. can you please explain how tap cells help in avoiding latchup?