Hello,
I am using the DAC7760 in a device where the user can select whether he wants the analog output to be 4-20 mA or 0-10V. This was implemented by having the DAC7760's current and voltage outputs active at the same time, and then using a jumper to decide what is connected to the connector pins.
When the jumper is in voltage position, I can change the load with the board powered, including short-circuiting it completely, and no damage occurs to the IC. However, if I remove the jumper from the board (from either the V or mA poistion) and connect to the V position when there is power on the board, the IC short-circuits my VIN and burns the IC. It's happened to me 3 times already. It happened when the load on VOUT was a short circuit and when it was connected to an opamp input (very high impedance), so it seems to be unrelated to the load.
I don't understand why inserting the jumper "live" is any different than changing the load on the connector. The only difference I see is that it adds a 470 pf capacitor instantly and that the insertion of the jumper can cause momentary rapid connects/disconnects. Any ideas what can be causing this, or want can be done to fix it.
Attached is my schematic. Vin = 24V
EDIT: I tried adding a permanent load and VOUT (10k) that is there even when the jumper is removed. I then had another 10K load connected to my AOUT1 signal, making the load 5k. When I removed the jumper, effectively changing the load from 5k to 10k, the IC blew.... I up to 5 dead ICs! What is wrong with my design?