5 ms·

Yes, robots in Blindsight were wire-operated with relatively brittle optical links (see "aperture leakage"). You can't power robot through optical link. Ionizi

by thesz 8y ago

Yes, robots in Blindsight were wire-operated with relatively brittle optical links (see "aperture leakage"). You can't power robot through optical link. Ionizing radiation damages batteries, look at the research. The solution is to present power, sensor output and control input through high-current wires. >Also, what do you mean by "associated high currents"? If you apply 12V instead of 5V to a logic gate you will get current that is 12/5 higher, isn't? With speeder transition, lower jitter, etc.

varjag 8y ago
Robots used in Chernobyl weren't battery operated, that's what I was saying. Look at historic footage, they all have quite thick power/signal harnesses attached. They were meant for rather brute work, like bulldozing the contaminated fragments. Even today you'd be hard pressed to do that with batteries. > f you apply 12V instead of 5V to a logic gate you will get current that is 12/5 higher, isn't? The necessary current is determined by the load you drive, which is tiny in semiconductor applications. Electric power is current times voltage, that means with higher voltage you need less current.
- thesz 8y ago
  >Electric power is current times voltage, that means with higher voltage you need less current. It just do not work like this. Logic gates are constructed from voltage-controlled variable resistors, to simplify things. When input voltage is 0 (Vgnd), some transistors have high resistance and some other low one. When you switch the voltage to Vcc (5V/12V), they revert roles and these that had high resistance now have low one. The resistance they have at different input voltages is fixed in the manufacturing process and cannot be changed. Thus if you put power voltage Vcc=12V you will have (for same input voltage) 12/5 higher current than for Vcc=5V.
  - varjag 8y ago
    The PN boundary (assuming that's what you are trying to explain) doesn't work like that. The current at given voltage is determined by the work which has to be done for switching the load. For signal electronics, we're talking microamperes here, that's not anywhere near "high current". Even the TTY loop I mentioned, a 1920s tech used to drive large relays and solenoids directly is just 40mA.
    - thesz 8y ago
      The load is either 0V or Vcc. Which is 0 and 5V and 0 and 12V and you can easily guess everything else. Also look here: https://en.wikipedia.org/wiki/Low-power_electronics https://en.wikipedia.org/wiki/Low-power_electronics Energy dissipated by change of voltage in the capacitor is E=CU^2/2. Now I stop.

by thesz 8y ago

varjag 8y ago

Robots used in Chernobyl weren't battery operated, that's what I was saying. Look at historic footage, they all have quite thick power/signal harnesses attached. They were meant for rather brute work, like bulldozing the contaminated fragments. Even today you'd be hard pressed to do that with batteries. > f you apply 12V instead of 5V to a logic gate you will get current that is 12/5 higher, isn't? The necessary current is determined by the load you drive, which is tiny in semiconductor applications. Electric power is current times voltage, that means with higher voltage you need less current.

thesz 8y ago
>Electric power is current times voltage, that means with higher voltage you need less current. It just do not work like this. Logic gates are constructed from voltage-controlled variable resistors, to simplify things. When input voltage is 0 (Vgnd), some transistors have high resistance and some other low one. When you switch the voltage to Vcc (5V/12V), they revert roles and these that had high resistance now have low one. The resistance they have at different input voltages is fixed in the manufacturing process and cannot be changed. Thus if you put power voltage Vcc=12V you will have (for same input voltage) 12/5 higher current than for Vcc=5V.
- varjag 8y ago
  The PN boundary (assuming that's what you are trying to explain) doesn't work like that. The current at given voltage is determined by the work which has to be done for switching the load. For signal electronics, we're talking microamperes here, that's not anywhere near "high current". Even the TTY loop I mentioned, a 1920s tech used to drive large relays and solenoids directly is just 40mA.
  - thesz 8y ago
    The load is either 0V or Vcc. Which is 0 and 5V and 0 and 12V and you can easily guess everything else. Also look here: https://en.wikipedia.org/wiki/Low-power_electronics https://en.wikipedia.org/wiki/Low-power_electronics Energy dissipated by change of voltage in the capacitor is E=CU^2/2. Now I stop.