I need to convert from 4 to 3.3V. So I was thinking of a diode, ...
Presumably you are considering using the forward voltage drop, Vf of a silicon diode placed in series with the load to drop about 0.7 V to bring your voltage down.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Using a diode to reduce supply voltage.
This works but you need to be aware of a few things:
- It is not a regulator. It just drops voltage and the voltage drop depends on the current. At more than a few mA it typically drops about 0.7 V but at lower currents the voltage drop will decrease. The I-V curve for the diode will help you calculate this.
- If V1 increases or decreases the the load voltage will increase or decrease with it. Again, it's not a regulator.
- The fact that you're looking for 3.3 V suggests that you want to power a micro-controller. You need a regulator.
... or voltage regulator. Voltage regulators, I know that it uses about 5 mA to do the job.
- Regulators do require some current to operate.
- Regulators also require some "headroom" - the difference between the input and output voltage - to operate. If you are thinking of a 78xx series regulator then these need about 2 V or so. You don't have that much headroom so you would be looking for a "low drop-out" (LDO) regulator.
How much current a diode uses?
A diode doesn't "use" current. It passes current but reduces the voltage. The power dissipated in the diode can be calculated from \$ P = VI = 0.7 \times I \$.