The data sheet for the AD835 analog multiplier mentions that its inputs have a "Differential Voltage Range" of ±1V and a "Differential Clipping Level" of at least ±1.2V (typ. ±1.4V).

What is the difference in meaning between the two figures? Is ±1V simply the "design" input range, but the chip is actually guaranteed to handle a bit more without distortion?

In my actual circuit, two of the chips seem to indeed perform fine (i.e. no clipping/distortion on the output) above ±1V (actually, up to about ±1.6V), but of course I'd rather not take a bet on that…


1 Answer 1


Your performance on one unit at room temp may be better than typical but it should never be used > +/-1V max as clipping may occur @ 1.2 V on some units under some condition.

  • \$\begingroup\$ Also transistors are sometimes sorted by hfe (gain). Some days your new machine just don't work because the gains, leakages and resistances all line up just right. Calculate the worst case and give a little headroom. \$\endgroup\$
    – Spoon
    Nov 23, 2013 at 18:16
  • \$\begingroup\$ Sure, but my question is more or less why they would give the ±1 V figure in the first place. Analog Devices guarantee that their chips can at least do ±1.2 V, so I factor in an appropriate amount of headroom (that depends on the requirements of my application) and look that I stay within that. Where would the ±1 V figure come in? \$\endgroup\$
    – dnadlinger
    Nov 24, 2013 at 0:30
  • \$\begingroup\$ A lot of signals in the Audio Video world are 1V pk-pk and at relatively high frequency (for me) so I'm out of my depth. However maximintegrated.com/app-notes/index.mvp/id/734 quote The unit of measure for the amplitude is in terms of an IRE unit. IRE is an arbitrary unit where 140 IRE = 1Vp-p. quote and en.wikipedia.org/wiki/Composite_video. The advantage of having the headroom (ic supply = +-5V)is the circuits work better away from the rails and the small amplitude of the signals means they change slower so are less distorted than if they had been 5Vp-p. \$\endgroup\$
    – Spoon
    Nov 24, 2013 at 12:45
  • \$\begingroup\$ @Spoon: Thanks for your efforts, but honestly I don't really see how this is related to my question. I know what headroom is and why RF applications usually require rather heavily biased transistors, but that still doesn't explain the reason for having the ±1 V figure in addition to the ±1.2 V one. \$\endgroup\$
    – dnadlinger
    Nov 24, 2013 at 18:27
  • \$\begingroup\$ How do you prove you have the headroom in the IC? You make it capable of driving a bit more and then you document it in the datasheet. Your document is by stating just how far the extra bit is. \$\endgroup\$
    – Spoon
    Nov 24, 2013 at 22:49

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