I am in the process of designing a simple multiplier using a diode. For biasing the diode at knee voltage should I use a current source or voltage. I have experimentally plotted the IV curve for the diode. So I know the knee voltage and the respectively current.
My argument is voltage source should be used. Why?
1. Usually in digital circuits a voltage source is used for biasing.
2. If I know the current and knee voltage, I can use the voltage source.
The main idea is to bias the diode at the knee point.
Secondly, the biasing is by a photodiode. Will it be better to use it as a voltage source or current source?
Current source of course!
The diode current follows an exponential law.
Having the same forward voltage across device and temperature variations may easily result in large variation in current.
Without mentioning the intended multiplier operation principle, it isn't a clear question, I fear. Basically, you'll need more than one diode to design a multiplier.
If you mean a diode frequency mixer, you preferably would increase the LO voltage to a level, that doesn't need a bias. With low level operation, I would use a current bias for it's temperature independence, as already said.
Actually, all contributors suggested a current source bias. But the problem is about optimizing the conversion gain for a specific semiconductor device. So instead of passing it off as a theoretical problem, you should better perform a conversion gain versus current (or voltage) bias level measurement.
It should be noted, that various literature links provided in response to your multiple diode mixer threads contain a lot empirical results about DC bias. In a short, the picture is more complex than just "bias at the knee voltage". A special point to consider with mixers: Rectified RF will affect the bias, changing the bias circuit impedance from zero (pure voltage bias) over resistive bias to infinite (pure current bias) will most likely result in different conversion gain versus level characteristics.