I am using 3 of the 4 outputs on the ADCLK944. Must I terminate unconnected outputs?
They can be left unconnected.
However, some customers add weak (i.e., large) pull-downs so that they have a test point. it's something to consider.
For this series of chips ADCLK944, 946, 948, 950, and 954, how much current consumption there is for each individual PECL output stage?
For example, if we want to use ADCLK948, but weakly pull-down (very large resistors) 6 of the 8 PECL output pairs, and only use the other two. How much power supply current, then will be needed for the whole chip?
Thanks a lot.
> For this series of chips ADCLK944, 946, 948, 950, and 954, how much current consumption
> there is for each individual PECL output stage?
It depends on whether you're powering the LVPECL fanout at Vcc=2.5 or 3.3V.
I'm going to walk you through an ADCLK944 calculation at 3.3V:
From the datasheet, the negative supply current of 37 mA (typ) is the amount of current passing through the ADCLK944, and not through the pull-down resistors. The positive supply current is listed at 138 mA, so we can see that 25 mA per differential pair is passing through the pull-downs.
The common mode voltage is 2V for 3.3V LVPECL, so R=V/I = 2/0.0125 = 160 ohm pull-downs must have been used. (The other case that yields the same results is 50 ohm termination to Vcc-2V).
Anyhow, in order to compute the total power consumed when using an ADCLK944, do this:
1. Multiply the negative supply current of 37 mA (typ) * 3.3V.
2. Compute the current out of each driver. Let's say two differential outputs are terminated
with 180 ohm pull-downs, and two are weak 1.8k pull-downs.
2a. For the 180s, there are four drivers total: 4*(2V/180 ohm) = 4 * 11mA = 44mA.
2b. For the 1.8k's, there are four drivers total: 4*(2V/1800 ohm) = 4 * 1.1mA = 4.4mA.
2c. Total current through the p/d resistors: 48.4mA. (This current passes through the ACLK944 Vcc pins.))
3. Total ADCLK944 positive current: 48.4mA + 37 mA = 85.4mA.
3a. Total power: 85.4*3.3V = 281.8 mW. Of that, 96.8 mW is burned in the p/d resistors. (48.4mA*2V).
It's very sensible and helpful. Thanks.
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