4 ms·
> But what lower resistance means is that you have higher power consumption as: p=iv & v=ir -> P=V^2/R So we see low resistance-> higher current. and higher cur
by Majromax 7y ago
> But what lower resistance means is that you have higher power consumption as: p=iv & v=ir -> P=V^2/R So we see low resistance-> higher current. and higher current -> more power.
I think you're forgetting time in this equation. If two otherwise identical MOSFETs have different gate resistances, the lower-resistance MOSFET will charge faster and draw switching current for less time.
Take this all the way back to a simple linear circuit: a 1-ohm, 1-farad RC circuit will have a time constant of 1s, and if supplied with 1V the current will be I(t) = exp(-t). The power lost to heat in the resistor is then I^2R = exp(-2t), and integrated from 0 to infinity the net energy loss is 0.5J.
If instead the circuit has a resistance of 0.5 ohm, the resulting current is I(t) = 2exp(-2t). Power loss is I^2R = (0.5)*(4 exp(-4t)) = 2 exp(-4t), and integrated from 0 to infinity gives a net energy loss of again 0.5J.
In the meantime, the lower-resistance MOSFET is likely to save power from other effects. Most notably, faster switching implies the transistor spends less time in its linear region where it itself acts as a resistor between source and drain.
- Buraksr 7y agoYou bring up an excellent point, I had neglected to show the effect of time in my explanation as you said. And it is true that this has real impact as to the magnitude of what I was talking about. But so long as the switching time != infinity, the power consumed is not equal. Using your example, 1 Ohm Case : P_res = integral exp(-2t) from 0 to tao -> .4323J 1/2 Ohm Case : P_res = integral 2*exp(-4t) from 0 to tao -> .4908J Of course the effect diminishes greatly (to zero as you showed) as the switching time increases. But we are dealing with a great number of transistors switching very many times so the effect is still noticeable given timings are tight enough. The thought on power savings as you mention is interesting but i am unsure off the top of my head of it.
- lpmay 7y agoNo, the power consumed is equal. The capacitance doesn't stop charging at time tao as you imply, it continues to charge until it reaches the final driving voltage, which is why integrating to infinity is a (very) much closer approximation
- Buraksr 7y agoWhat I am meaning is the input to the transistor on a chip is highly volatile, so the aproxmation of its power consumption using infinite or long periods of time of a single value is misleading. If a transistor is only reciving 5v than it is fair to say the power consumed is equal. But realisticly the input could change every clock cycle and we would see the cases vary. The reason I only integrate to tao was to show if the input switches before t=infinity then a diffrence between cases clearly manifest. But as you say, when it does not switch for time t>>tao the diffrence is small.