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# THS4532: Vid Specification

Part Number: THS4532

The THS4532 states that the absolute maximum Vid is 1V. This to me implies that the difference between the (+) and (-) inputs cannot exceed 1V. I am trying to understand this, as I would think that the differential voltage could be as large as need to saturate Vout according to how Vs+/ Vs- is set. For example figure 36 shows the Vs at 5V and the gain set at 1V/V. It then shows a Vout of 2Vpp. If the limit were 1V then how would this be possible. My understanding of the differential amp is the difference voltage between the output pins should equal Vid * G. In this case this implies that Vid is 2V as Vout is 2Vpp. Am I wrong in my assumptions or is the data sheet specification incorrect.

Thanks for your help with this!

Richard Elmquist

• Hi Richard,

It is true that the absolute maximum input differential voltage (Vid) between the (+) and (-) inputs cannot exceed 1V. In normal application, such as the standard circuit shown below, both the (+) and (-) inputs are at small signal and they usually track each-other such that there is no difference voltage between these pins. The example in Figure 36 with gain set at 1V/V is not calculated from the (+) and (-) inputs, but instead from the input voltage (Vin) applied before RG or Vo = G*Vin. I would recommend you to go through the app note for more information on this topic: .

Best Regards,

Rohit

• Rohit,