??? 10/03/06 23:26 Modified: 10/03/06 23:37 Read: times |
#125786 - proportional versus f Responding to: ???'s previous message |
Darren said:
If I have 10Hz, 10V r.m.s. signal feeding a 10R resistor, the power or heating effect being dissipated in that resistor is 10 Watts??, assuming my voltage generator has zero source resistance, i.e., 10Vr.m.s. is dropped across the load resistor.
Now, I increase my frequency to 1MHz, the amplitude is still 10 Vr.m.s., what is the power being dissipated in load resistor now. According to MR OHM, the power being dissipated is still 10 Watts, in fact it remains 10 Watts, irrispective of the frequency. Electrical power is proportional to current^2 or voltage ^2, it is not proportional to its frequency, for that we simply convert time varying signals to their equivalent d.c. value and hence r.m.s. equation. Yes, of course, you are right, that's just the definition of RMS, that the heat dissipation is identical. But what we mean when talking about power consumption with CMOS gates has nothing to do with this equation, but has to do with the simple fact, that whenever a CMOS gate is toggling a spike current is flowing, charging the very small but unavoidable gate stray capacitance. And that the more often this takes place per second, this the more adds to the overall supply current, resulting in an increase of current consumption of CMOS gate with frequency. Let's analyze what happens when a simple CMOS inverter is driven by another CMOS gate, whichs output toggles at the rhythm of a square wave signal, just like it happens within each micro thousands of times. Have a look at this picture, which I draw for a different thread, but is good enough to demonstrate what I mean: If the first gate (on the left) toggles from low to high, the gate stray capacitance Cstray becomes charged. So, as the voltage across its pins changes from 0V to Vcc, a current must have flown through its "terminals", moving the charge Q = C x Vcc. As Cstray and the source impedance of driving PMOSFET are rather small, charging-up of Cstray goes very quickly. Assuming Cstray being about 1pF and source impedance of PMOSFET being about 50Ohm, time constant is about 50psec, which is even less, than the switching time of CMOS inverter, which is about 1nsec, depending on CMOS technology, of course. So, whenever a series of CMOS inverters within a micro is toggling, stray capacitances are charged for a very brief period, where the charging current follows the equation Q = dI x dt, where Q is the moved charge. We can estimate the current dI as follows: Assuming a Cstray of 1pF, a charging period of about 1nsec and Vcc = 5V, dI = Q / dt = C x Vcc / dt = 5mA. Of course, this value is an average current, which is in real a spike current. But this doesn't matter here at all, we only need to know, that for a period of about 1nsec a certain current is flowing into Cstray, which is of course nothing else than the supply current of this CMOS inverter, a supply current, which only flows during the toggling event of CMOS inverter. When the CMOS inverters toggles back from high to low, Cstray becomes discharged through the NMOSFET, but no additional supply current is flowing through the Vcc terminals of this gate: Additionally to that supply current another supply current is flowing into the Vcc pin of inverter, the so called shot through current, which shall not be discussed further here. Now assume, that the output of first CMOS inverter toggles from low to high at a frequency f, means that the output toggles from low to high every T = 1/f. Then, the total current consumption of this CMOS inverter, means the average current consumption, is (dI x dt + 0 x (T - dt)) / T = dI x dt / T = dI x dt x f. And, as dI and dt are nearly constant, current consumption of CMOS inverter is proportional to f! As the power consumption is P = Vcc x dI x dt x f, also the power consumption is proportional to f, just what this picture was expressing: Kai |