4 ms·

Got this question in an interview at National Semiconductor (out of MSEE school). I didn't think it was too paradox-y: we didn't work the math through fully bu

Got this question in an interview at National Semiconductor (out of MSEE school). I didn't think it was too paradox-y: we didn't work the math through fully but my answer of "start with the circuit with a resistor with value R and the energy will burn up in R; now recalculate as R→0 and the energy will still burn up in R (even though R is zero-y...". That seemed to satisfy the interviewer (who had not heard that answer before).

gnramires 6y ago
I think an interesting question following it up is, how do you transfer energy between capacitors then, without losing so much energy? (someone else asked if this is a fundamental limitation) There are many alternatives. A good start is noting that the inefficiency is actually lower the lower the starting voltage difference: V1 = V, V2 = V-dV V' = (V + (V-dV))/2 = V+dV/2 We can define efficiency as the ratio of energy lost by the first to energy transferred to the other. dU1 = U1-U1' = CV^2/2-CV'^2/2 = C/2 (V dV-dV^2/4) dU2 = U2'-U2 = CV'^2/2-C(V-dV)^2/2 = C/2 (-V dV+dV^2/4+2VdV-dV^2) = C/2 (V dV - 3dV^2/4 ) n = (dU2/dU1) = ( V - 3dV/4 ) / ( V - dV/4) ~ 1 - 3/4 dV/V (for small dV) as dV → 0, n → 1. You lose 3/4 of the fractional difference in efficiency, so for 10% difference your efficiency is ~92.5%, pretty great. Now, there are indeed devices that change voltages without energy losses! (transformers, for example). So if you plug in a variable transformer that keeps the voltage close to the target, your (dis)charging efficiency can be arbitrarily high. Of course, if your second voltage is 0, the efficiency must start at 1/3 no matter what (which can seem to imply this cannot be changed) -- but as soon as you have a small voltage you can start tracking it and keep efficiency high. Challenge to the reader: Use quantum mechanics and thermodynamics to derive a fundamental limit of efficiency (which must be less than 1 at positive temperatures) :)
- mindslight 6y ago
  Inductor in series with a diode. The voltage differential stores energy in the inductor's magnetic field, which comes back out when the destination capacitor is at a higher voltage than the source capacitor. The diode keeps the process from repeating in reverse (oscillation).
  - gnramires 6y ago
    Cool, but what if you want the voltages to be exactly equal (while keeping no losses)? :)
    - mindslight 6y ago
      Dump the charge into a third capacitor, connect the two you want equal, and dump the charge back.

gnramires 6y ago

I think an interesting question following it up is, how do you transfer energy between capacitors then, without losing so much energy? (someone else asked if this is a fundamental limitation) There are many alternatives. A good start is noting that the inefficiency is actually lower the lower the starting voltage difference: V1 = V, V2 = V-dV V' = (V + (V-dV))/2 = V+dV/2 We can define efficiency as the ratio of energy lost by the first to energy transferred to the other. dU1 = U1-U1' = CV^2/2-CV'^2/2 = C/2 (V dV-dV^2/4) dU2 = U2'-U2 = CV'^2/2-C(V-dV)^2/2 = C/2 (-V dV+dV^2/4+2VdV-dV^2) = C/2 (V dV - 3dV^2/4 ) n = (dU2/dU1) = ( V - 3dV/4 ) / ( V - dV/4) ~ 1 - 3/4 dV/V (for small dV) as dV → 0, n → 1. You lose 3/4 of the fractional difference in efficiency, so for 10% difference your efficiency is ~92.5%, pretty great. Now, there are indeed devices that change voltages without energy losses! (transformers, for example). So if you plug in a variable transformer that keeps the voltage close to the target, your (dis)charging efficiency can be arbitrarily high. Of course, if your second voltage is 0, the efficiency must start at 1/3 no matter what (which can seem to imply this cannot be changed) -- but as soon as you have a small voltage you can start tracking it and keep efficiency high. Challenge to the reader: Use quantum mechanics and thermodynamics to derive a fundamental limit of efficiency (which must be less than 1 at positive temperatures) :)

mindslight 6y ago
Inductor in series with a diode. The voltage differential stores energy in the inductor's magnetic field, which comes back out when the destination capacitor is at a higher voltage than the source capacitor. The diode keeps the process from repeating in reverse (oscillation).
- gnramires 6y ago
  Cool, but what if you want the voltages to be exactly equal (while keeping no losses)? :)
  - mindslight 6y ago
    Dump the charge into a third capacitor, connect the two you want equal, and dump the charge back.