Voltage inside an semiconductor chip?

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Discussion Overview

The discussion revolves around the behavior of voltage within integrated circuits (ICs), particularly focusing on how a standard voltage of 5 volts can be applied to very small semiconductor areas without causing damage or overheating. The conversation touches on the implications of voltage density and current flow in semiconductor materials.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how a 5-volt input to a nano-sized IC does not lead to excessive voltage per area, expressing concern about potential overheating.
  • Another participant explains that voltage is an "across" quantity and that the thickness of the semiconductor junction is crucial, suggesting that current density is low enough to prevent damage.
  • A different participant seeks further justification for the voltage factor, indicating a need for clarity on how voltage is managed in such small areas.
  • One participant mentions various factors that determine voltage in ICs, noting that breakdown conditions and gate potential limits in CMOS circuits are important considerations.
  • It is highlighted that in digital circuits, internal voltages can be switched rapidly, and lower voltages can reduce power dissipation, with examples like watches using low voltages and currents.

Areas of Agreement / Disagreement

Participants express differing views on how voltage is justified in small areas, with some agreeing on the importance of junction thickness while others seek further clarification on the implications of voltage density. The discussion remains unresolved regarding the justification of voltage in this context.

Contextual Notes

There are unresolved questions about the relationship between voltage, area, and current density, as well as the specific conditions under which breakdown occurs in semiconductor junctions.

shankar
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i have a basic question about IC

we give 5 volt in general into an IC but the size of the IC is in nano measure if this 5 volt is given to such a small area the volt/area will become very large and how the IC still functions without burning.

let me know how this happens .
 
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shankar said:
i have a basic question about IC

we give 5 volt in general into an IC but the size of the IC is in nano measure if this 5 volt is given to such a small area the volt/area will become very large and how the IC still functions without burning.

let me know how this happens .

Voltage is the "across" quantity, so the important thing is the thicknes of the semiconductor junction. It doesn't need to be very thick to withstand 5 volts. The "through" quantity is current, which is usually in the miliampere range. So the current density ne charge flux is not that great.
 
ya you are right but how to justify the voltage factor. As you said, yes it is across factor but till we nned to justify voltage if not volt/area. howto do it.
 
chip voltages

There are several factors determining voltage -- but not area --
in cases where normal transistors are used there are breakdown conditions to think of since some junctions are reverse biassed and they are not very thick.
In Cmos circuits there are limits to the gate potental before 'punch through'
of the gate.
But there is also a question of circuit speed and power, in digital circuits such as CMOS the internal voltages are switched from rail to rail if this voltage is smaller and for given drive conditions ( size and doping defined ) the transit times of the nodes can be shorter and dissipation reduced.
Watches are a good example employing thousands of transisitors with voltages as low a 1 volt and currents < 1 microAmp, but the clock is only
a few 10's of kilohertz.
 

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