On Mon, Aug 20, 2001 at 11:34:04AM +0100, laurence@orchards.org.uk wrote:
At 11:39 20/08/01 +0300 you scribbled:
Hi
Correct... And even more so, because the fact that with higher core voltage the rise/fall times decreases, and thus consumes more power, generates more heat, and generate more RF- and EMI
Jaska.
Don't understand about the rise/fall times decreasing?
Rise/Fall times actually increase with increase in temperature. Hotter transistors have higher on-resistance, as the atoms are bouncing around much more, getting in the way of the electrons (that's an oversimplified explanation, but it's good enough. Higher resistance driving the same capacitance equals slower edge. In theory, the higher resistance transistors will dissipate less power, but the slower edge rate means that the following transistors will spend more time in the semi-on state, which is where all the power is used.
Using the standard formula watts = volts * amps if the required power remains the same then if you increase the voltage the current must fall!
True, but the required power is not a constant. What you are saying is true in the case of electrical transmission - to deliver the same amount of power to the destination, a higher voltage means lower current, and therefore less heat is dissipated in the wires. This is why power is distributed along lines at hundreds of kV. In the case of chips, most of the heat is being dissipated the transistors, and their resistance remains approximately constant. So, the equation P=V^2/R applies here, i.e power is proportional to the *square* of the voltage. Trust me I know. Look at my .sig :-) -- David Smith Tel: +44 (0)1454 462380 (direct) Silicon Design Engineer Fax: +44 (0)1454 617910 STMicroelectronics TINA (ST only): (065) 2380 1000 Aztec West Home: 01454 616963 Almondsbury Mobile: 07932 642724 Bristol Work Email: Dave.Smith@st.com BS32 4SQ Home Email: David.Smith@ds-electronics.co.uk