Is Venus in the habitale zone of a robot (transistor CMOS brain, metal body) ? Any definition by NASA? So the robot's information-processor works based on CMOS technology, the energy is powered by lithium-ion battery/solar and the body is made of stainless steel.
For reference you can think of the Mars rover.
So Mars seems to be obviously in the habitable zone for robots. How about venus?
Any official definition for the habitale zone of robots based on CMOS AI by NASA?
 A: It depends on your definition of 'surface'. Yes, on the hard rocky surface, temperature and extreme acidity makes it very hard for any sort of electronics to last more than a few minutes
But floating at 50km of altitude in Venus the temperature is barely tropical (around $30$ ºC), and sulphuric acid makes about 2% of the atmosphere. That's definitely on the 'habitable zone' for robots, and even humans (they still need protection from the atmosphere though)
A: As highlit by Diffeomorphism's answer, one can't really make sense of the notion of "habitable" zone for an individual planet. What I think you mean to ask is "is Venus inhabitable by robots / humans.
The habitable zone of a star, in contrast, is the zone where a possible planet might support life (or, by analogy with your question, robots). This is often taken by astrobiologists to mean the zone where there can be water in its liquid state. By this definition, Venus is almost certainly within the Sun's habitable zone; we can reason this as follows.
By the Stephan-Boltzmann law and the inverse square with orbit radius power relationship, the intensity of sunlight reaching a planet with orbit $R_p$ is:
$$\sigma\,T_\odot^4\,\left(\frac{R_p}{r_\odot}\right)^2\tag{1}$$
where $T_\odot$ is the temperature of the surface of the star in question, $r_\odot$ its surface's radius, $R_p$ the orbit radius for the planet in question and $\sigma$ the Stephan-Boltzmann constant (although this cancels out below). The planet gathers light over  a cross-sectional area of $\pi\,r_p^2$, where $r_p$ is its radius, so, if it is a blackbody, the incoming power it absorbs from the star is: 
$$\pi\,r_p^2\,\sigma\,T_\odot^4\,\left(\frac{R_p}{r_\odot}\right)^2\tag{2}$$
Assuming again that it is a blackbody it re-radiates power, from the Stephan-Boltzmann law:
$$4\,\pi\,\sigma \,T_p^4 \,r_p^2\tag{3}$$
at steady state, incoming and outgoing powers are equal, so, in the absence of greenhouse effect, the assertion of steady state means the equality of (2) and (3), whence:
$$T_p=\frac{1}{2}\,T_\odot\,\sqrt{\frac{r_\odot}{R_p}}\tag{4}$$
Plugging the numbers in for Earth  - Sun ($T_\odot=6000{\rm K}$, $R_\odot = 6.958\times10^5{\rm km}$, $R_\oplus=1.496\times10^8{\rm km}$ I get $T_e=204{\rm K}$. This is rather colder than the surface of the Earth (but the right order of magnitude), and the difference between this and Earth's real surface temperature is accounted for by albedo and greenhouse effects. Now look at the square root dependence in (4) - this is a weak dependence and, given that Venus orbits at about 2/3au, this means that, in the absence of greenhouse effect and assuming albedo of 1, a planet in Venus's orbit would have a surface temperature of $\sqrt{3/2}$ times the value for the Earth. In other words, a surface temperature of $250{\rm K}$. This is still below Earth's surface temperature. So a different atmospheric composition would have made Venus eminently habitable.
Thus, by the potential liquid water definition, Venus most definitely would be in our Sun's habitable zone, either for life or robots.
