subthrehold operation

When the transistors operate in subthreshold region, in this case, the transistors are not fully conducted, the current flowing through the transistors are pretty small.

Therefore, my question is:
1, for the subthreshold operation, what is the range of operating current?
2, is it possible for the circuit which is operating in subthreshold region to drive a large capactitive load? or how can I optimize this circuit to enable it to drive a large load?

Re: logarithmic current to voltage converter

the sub-threshold current is depend on the device size. If it can drive a large cap load is depend on how long you need the voltage settling and the cap leakage current level.

Re: logarithmic current to voltage converter

you can add in series of this configuration a low noise amplifier which boost the small amplitude of your signal so you still get the logarithmic conversion with output signal boosted.

Re: logarithmic current to voltage converter

You should consider a diode as an alternate.

Diodes working from pA to some tens of uA so about 7 decades. Only the series resistance make a derivation from the log(i)->v principle. In MOS you have the subthreshold transition region, substrate effects, gate leakage and drain voltage dependence. All complicate a circuit for an accurate log(i)->v conversion.

With BiCMOS and an NPN you can get 9-10 decades depending on the process. A further option is to use the substrate vertical P+/NWELL/PSUB bipolar. Because of the vertical current gain only a fraction of the emitter current into the P+ is flowing out of the NWELL base. That lowers the effect of the NWELL base resistance of the log(i)->v accuracy. If the current gain in a non-epi CMOS process is between 5-20 you can extend the range by a further decade.

logarithmic current to voltage converter

And to drive a load, you will need some sort of buffer between this stage and the next. For example a simple source follower (either PMOS or NMOS) will only present a tiny capacitive load to this stage (to keep it fast) but can drive a much larger capacitive load on the output.