I have a circuit and am trying to understand it better. It takes a 0-4V signal and translates it to a 4-20mA output. I believe the offset resistor is supposed to generate the 4mA current when Vsig is 0V and the transistor portion is supposed to supply the additional signal current (up to 16mA) when Vsig is at 4V. The feedback current in addition to the sense current gives me my output current of 4 to 20mA.
I've been told that node A is always going to be 0V due to the virtual short. I have 2 problems with this and I think this is where my confusion is coming from.
1) This seems to me to be an open loop configuration. Do the opamp inputs in this configuration act as a virtual short, therefore making node A always 0V?
2) If both opamp inputs are always 0V, shouldn't the opamp output always be 0V? Without any base current, wouldn't the opamp always be operating in shut off?
If you could give me some insight as to why this works, I would really appreciate it.
I've been told that node A is always going to be 0V due to the virtual short. I have 2 problems with this and I think this is where my confusion is coming from.
1) This seems to me to be an open loop configuration. Do the opamp inputs in this configuration act as a virtual short, therefore making node A always 0V?
2) If both opamp inputs are always 0V, shouldn't the opamp output always be 0V? Without any base current, wouldn't the opamp always be operating in shut off?
If you could give me some insight as to why this works, I would really appreciate it.
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