I'm designign a circuit using the AMC1200 isolation amplifier from TI.


As you can see the output is differential, and I'd like too feed it to a schmidt trigger to catch the 2.5V transitions. The problem is that the trigger is single ended, so here is my question: is there an easy way to transform the fully differential output in a single ended output i.e. having the difference between the outputs? I know I can use just one of the outputs but I basically loose half of the dynamic range, and that might be a problem.

  • \$\begingroup\$ How fast is the signal you want to convert? \$\endgroup\$ – Gustavo Litovsky Oct 7 '13 at 17:49
  • \$\begingroup\$ good question. the signal is practically DC. \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 0:47

Here's what the output from the AMC1200 looks like: -

enter image description here

As you can see either output will swing from about 1.3V to about 3.9V and if you used a normal differential to single end converter, the minimum voltage out would be 2.6V to 7.8V.

I suspect you have a 5V supply and don't want to over-complicate things with adding a 10V positive supply (and possibly a small negative supply) so that you could use an instrumentation amp like the AD8221: -

enter image description here

But, if you do think this complication is necessary then that's fine. You don't need a gain set resistor but you do need to set an offset on pin 6 to set the "mid-point" or neutral position of your output. The AD8221 is good for over 100kHz so no problem there.

On the other hand just use one of the single-ended outputs from the isolation amp and I'm sure it will be just fine. You're only feeding a comparator so any noise can be easily filtered and using a Schmitt trigger should avoid multiple threshold triggering.

  • \$\begingroup\$ thanks for your reply. I actually have a 12V supply line and I'll obtain the 5V with a zener or a 7805. the use one output way is the easiest for sure, but the width of the trigger hysteresis should be less than 100mV that is not very much... \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 0:49
  • \$\begingroup\$ Maybe if you have 12V then use a differential amp but on its output reduce the signal by 2:1 with a potential divider with two resistors. Then feed the schmitt. \$\endgroup\$ – Andy aka Oct 8 '13 at 7:11
  • \$\begingroup\$ yeah that's an option, but since the application is a huge battery pack monitoring on an automotive application I'd rather keep the chip count as low as possible. \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 9:35
  • \$\begingroup\$ @VladimirCravero are you keeping the chip count low to save energy in the battery? \$\endgroup\$ – Andy aka Oct 8 '13 at 9:45
  • \$\begingroup\$ nope it's just the "what is not there can not break" philosophy. and saving power is a good thing too, althought the system can deliver approx. 65kW continuosly to the wheels so the mWs really are not an issue \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 10:10

There are chips called differential line receivers that do exactly what you are asking about.

  • \$\begingroup\$ yep I know that, I did not really look into it because I was hoping for a simple "tie some passives togheter" solution. I'll evaluate your idea too. \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 0:50
  • \$\begingroup\$ I believe this guy: maximintegrated.com/datasheet/index.mvp/id/3248/t/or/pt/… can do the job. \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 10:30
  • \$\begingroup\$ Actually a line driver won't work since the output from the ins. amp. is always positive and never changes sign... \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 12:54
  • \$\begingroup\$ @Vlad: There are differential line receivers that work with both signals always positive. \$\endgroup\$ – Olin Lathrop Oct 8 '13 at 13:04

The reason the output of that device is differential is noise immunity. Current measurements suffer greatly from noise and using differential can shield them. This device was made to interface directly with a differential input ADC that will take care of properly recovering the data. If, however, you don't need fast measurements and don't care about this, a differential to single ended converter can be used.

A simple way to do this is to use an Op Amp.

As you might know, the output of the Op Amp is directly proportional to the difference in the two. You will need to design the Op Amp with the right gain depending on the voltage swing of the device.

Note, however, that the bandwidth can be rather limited, and that you will lose noise immunity and some accuracy.

  • \$\begingroup\$ maybe you provided the wrong link, since they talk about a single ended to differential converter that is exactly the opposite of what I need. \$\endgroup\$ – Vladimir Cravero Oct 8 '13 at 0:51
  • \$\begingroup\$ @VladimirCravero: Whoops you're right. \$\endgroup\$ – Gustavo Litovsky Oct 8 '13 at 1:38

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