What if we decrease/increase the resistance 3kohm to a higher/lower value? Can someone explain this behaviour to me in a very simple way? 
Will the current then flow through the silicon diode?
Electronic – Current flow through diode
circuit analysisdiodessemiconductors
Related Solutions
Let me redraw the central part of the circuit. Without the diode being present, current will flow clockwise. When D1 is inserted, it is reverse-biased, because V2 is higher than V1 (they are in parallel). Reverse-biased means no current flows.
From then on, you can treat this entire thing as its Thevenin equivalent - no matter what you do the path V1-D1 will always be inert, because V2 makes sure D1 stays reverse-biased.
simulate this circuit – Schematic created using CircuitLab
Just to clarify this issue, let's make a thought experiment.
- The circuit of V1, V2, D1 is clearly a sub-circuit of the circuit in question.
- Voltage between NODE1 and NODE2 is guaranteed to be 20V, since V2 is connected directly between the two. (V2 being an ideal voltage source, voltage across it cannot change by definition.)
- Since V1-D1 and V2 are in parallel, they share the same voltage across them.
- When you subtract 5V from 20V (this is perfectly legal with voltage sources), you're left with 15V between D1's anode and cathode.
- Diode will act as an open circuit, since the potential at its cathode is higher than its anode = it is reverse biased.
KCL for the junction is $$I_1=I_2+I_3$$ The voltage on the junction is \$V_1\$ Then $$I_1=(4.7-V_1)/40$$ $$I_2=V_1/50$$ $$I_3 = (V_1-0.6)/20$$
Solving: $$ (4.7-V_1)/40 = V_1/50 + (V_1-0.6)/20 $$ giving us $$V_1=1.55V$$ Hence $$I_3=(1.55-0.6)/20=0.0475A=47.5mA$$ Close enough to \$45.3\$..`.
Best Answer
A very simple answer with another dodgy water analogy.
Figure 1. Parallel pressure relief valves. Relief pressure P1 < P2.
If we take these two pressure relief valves on a pneumatic or hydraulic circuit it should be clear that when the pressure rises enough to open the lower pressure valve that the pressure above the two valves can no longer increase (because it has been relieved) and P2 will not open. The parallel diodes with different \$ V_f \$, forward voltages, will behave in a similar manner.
I used this behavior in my solution to How to replace a green LED with a red one in a simple transistor switch? yesterday.
Note that if the pressure is high and the resistance of P1 is high enough relative to the downstream resistance then the pressure drop will increase to the point that P2 begins to open. This is analogous to reducing the resistor in your question to such a low value that very large currents flow through the germanium diode, the voltage rises due to its internal resistance and when it reaches 0.7 V the silicon diode will turn on. They'll both die quite quickly.