ET 510 Digital Electronics

You are working at an engineering company designing a circuit to amplify the signal from a biological assay. The assay measures the concentration of malaria in a sample of blood. Your instrument needs to be able to measure concentrations even in patients who are heavily infected as well as patients who only have a trace infection. Your instrument relies on the output of a photosensor which produces a current of µA per nmole/liter of malaria. Assume this is a weak current signal which needs an ideal current input and expects a OV input voltage (virtual ground).

In order to be sensitive to low concentrations but also able to measure the full range of infection,your design should have two outputs,one of the outputs produces a signal from 0-1V for concentrations of 0- 100 nmole/liter (0- OOµA). The other output produces a voltage of  V for 100 nmole/liter up to 3V for 8100 nmole/liter (lOOµA -> 8. mA).

Your system has an accurate 3.3V source as well as a -3.3V. Your output do not matter when being operated outside their useful range (as in,you don't care about the 0->1V output for inputs greater than 100 µA and you don't care about the 1->3V output for input currents less than 100 µA but neither output should ever go negative.

Design and draw a circuit with the limit of 5 op-amps or fewer. Here is important information to guide you.

You have a -IOmV to lOmV differential voltage signal,you want a gain of 0.4 A/V. To achieve this,you must build a circuit that drives au LED with lmA of cmTent when the differential input is - IOmV, 5mA of current when the differential input  is OmV, and 9mA of current when the differential input is +1OmV. The LED can either be driven  with a floating or grounded configuration. The common mode of the differential signal varies from between about -2V to +2V, and you have +IOV and -IOV voltage supplies. All conditions described have to be in a unified design, i.e not 3 seperate circuits. Again, no more than 5 op-amps are allowed for this circuit design.