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Bit of a basic tech question, but having looked through all the datasheets and also having used AD633 for years now, I am not sure what its 'base unit' is?

What I mean is, I want to now multiply millivolts, not volts as before, should this still work? (I am assuming yes)

So will it treat 60mV*60mV as such thus = 3600mV, which by means of the transfer function is then divided by 10V = so chip output should be 360mV? (Assuming all other pins grounded, using X1 and Y1 as inputs)

Last thing I want is it performing (0.06V*0.06V)/10=0.00036V

If I had time I would build this but components are in transit and existing setup that is using my 633s is working and valuable, so I rather not disturb it.

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  • \$\begingroup\$ Link to datasheet. Question doesn't make a whole lot of sense in its current form, you're mixing mV and V - 60mV*60mV/10,000 mV is the same as (0.06V*0.06V)/10V. \$\endgroup\$ – Brian Drummond Mar 8 '17 at 16:53
  • \$\begingroup\$ Look at the paragraph in the datasheet titled "Functional Description". \$\endgroup\$ – The Photon Mar 8 '17 at 17:00
  • \$\begingroup\$ You may have to amplify your signals to minimize your errors. Last thing U want is unfortunately what you get. Your latter calculation is correct. But mV²= μV \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 8 '17 at 17:19
  • \$\begingroup\$ Thank you guys - I completely in my haze missed that the AD633 does not simply divide by 10, but divides by 10V As I said to Jack, and as kindly advised by Tony, I think a non-inverting opamp with a gain of 5 would suit my application, keeping within a supply voltage of 12V and and (with gain) having a maximum possible input of 10V for X1 and 9V for Y1. Will have to watch for accuracy at the low end of the spectrum... Thank you \$\endgroup\$ – Rendeverance Mar 8 '17 at 17:46
  • \$\begingroup\$ 60 mV times 60 mV is not 3600 mV. It is 3600 microvolt. Check your math. \$\endgroup\$ – WhatRoughBeast Mar 8 '17 at 18:38
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The AD633 always divides by 10V. It says that at the very beginning of the datasheet, in figure 1. 10V is not the same as 10 without units.

When doing calculations in mV and V (or indeed any other mixture of units: meters and feet; kilos and grams; seconds and hours) you should convert everything into the same units before doing your maths.

So for two 60mV inputs, you have 60mv\$\times\$60mV/10V. You can re-write that as either 60mV\$\times\$60mV/10000mV=0.36mV or as 0.06V\$\times\$0.06V/10V=0.00036V. Both of these answers are the same. So you're probably going to have to change your design a bit.

In fact, when you're used to working with different units, you'll start to notice that they always balance. Look at the units in the two correct equations above. If you forget about the numbers, and just look at the units on the left hand side, you have mV\$\times\$mV/mV. Cancel these as you would in algebra, and you have mV, which is the unit on the right hand side. Similarly, V\$\times\$V/V simplifies to V. If you look at the original equation though, you have mV\$\times\$mV/V, so your answer is in weird units, so you know something is wrong. This handy trick is called dimensional analysis.

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  • \$\begingroup\$ Thanks Jack - I think I am having an off day. While waiting for an answer I concluded if I amplify my signals by a factor of 5 with a non-inverting opamp, I should be able to extract something meaningful from the multiplication... Thank you for your time \$\endgroup\$ – Rendeverance Mar 8 '17 at 17:40
  • \$\begingroup\$ (I had missed that the divisor was not just 10 but 10V, hence how I got confused... thanks again :) ) \$\endgroup\$ – Rendeverance Mar 8 '17 at 17:41

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