circuit-designdemodulationdiodesschottky
I was wondering how this demodulator works. The input voltage is on the left at C2 and the output voltage is on the right before C3.
What is clear: C2 filters out low frequencies at the input, C3 filters out high frequencies at the output, R2 limits the current coming from the reference voltage V+.
What is unclear: the function of the two Schottky diodes D1 and D2.
What happens is:
As the base voltage rises, the transistor begins to turn on and it's collector voltage drops (assuming it has a collector resistor or similar current limiting element)
Normally a typical bipolar transistors saturation voltage is around 200mV or less. When the collector voltage, Vce drops below Vbe - Vschottky though, the schottky starts to conduct (now being forward biased) and the base current starts to flow through it into the collector. This "steals" current from the base, preventing the transistor turning on more and the collector reaching it's saturation voltage.
The system will reach a state of equilibrium, since the transistor can't turn on any more without it's base current dropping (you could see it as a form of negative feedback) and will settle just around Vbe-Vschotkky (e.g.~700mv-450mV as opposed to ~200mV)
So, to clarify things, the formula for Vce is:
Vce = Vbe - Vschottky
If we have this circuit and apply a ramped voltage from 0-2V:
We get simulation results like this:
Note that when Vcollector drops below ~700mV, the Schottky begins to conduct and the collector voltage levels out at around 650mV.
If we remove the Schottky, then:
We can see the collector drops all the way to 89mV (I used the cursor as it's hard to see from the graph)
Your analysis is correct. The idea is that the current flows through the diode and into the power supply. What happens to it in the power supply depends on the design of that part of the circuit.
If the current is a transient pulse (from EMI or something like that) then the circuit's decoupling capacitance and the output capacitance of the power supply will easily absorb it. If it is a small constant current then it will be used by the chips in the circuit instead of drawing current from the power supply. If it is excessive, it may cause the power rail to rise and destroy the rest of the circuit as most power supplies are not bi-directional.
The key thing is to ensure that the last situation cannot happen, and this is one of the reasons you normally see a series resistor before the protection diodes. If the potential current is still too high it is better to use a disconnection FET or shunt clamp design instead.
Best Answer
R2 does a bit more than 'limit the current coming from Vref'. It biasses the two diodes with a small current, so that they are conducting. It also recharges C3 when the input signal is removed.
When a positive signal peak is input to C2, the voltage across D2 changes little, because it's already conducting, so C2 charges up to the input voltage peak.
When a negative peak is input to C2, D1 conducts with its voltage changing little, pulling the voltage on C3 down, not just by the amount of the negative peak, but also by the amount of the previous positive peak that C2 is charged to.
This is termed a 'voltage doubler'. C3's voltage drops by twice the peak input voltage.
Once the input signal has been removed, C3 slowly charges from the current flowing through R2. The speed of this recharge limits the rate at which this demodulator can follow fluctuations in the input amplitude.