gyratorinductornpn
According to this link, the following schematic is equivalent to an inductance R1*R2*C placed between input and ground. Under which conditions is this true? As a guideline, we assume Vcc=9V. See this post for the context of the question.
simulate this circuit – Schematic created using CircuitLab
Application with R1=390 Ohm, R2=22 kOhm, R3=2.2 kOhm, R4=70 kOhm, C1=100 nF, C2=200 uF.
L4 = 10 H is shorted. So it can be removed from the circuit.
You have to calculate parallel inductance of
which is 3 H. Then you add the series 4 H (L1) and you get 7 H equivalent inductance.
The typical gain of a 2N3904 is 200-300 at a couple mA collector current (more as it warms up due to not being saturated)
Even with 560K, that's half a mA or so, which will give plenty of light from a modern LED, but you should be able to see that it's not as bright as when a 10K resistor is used.
Do not use the hfe for this calculation if you want the transistor saturated hard on, use a forced beta of something like 20 to 50, if the typical hfe is 200 or so and the minimum 100 or so. If you use, say 30, in your equation you get a resistor value of 8.8K, so you might use 10K or 8.2K.
The reason is that you won't likely have a guaranteed hfe for the current you're using, and the hfe decreases at temperature extremes. It's still only "wasting" a few percent of the LED current, so no big deal.
To prove this to yourself, take a voltmeter and measure Vce of the transistor when it is on. If it is something like 50-100mV it is saturated.
Best Answer
There are better gyrator circuits so let me give you the downsides of this circuit first.
The idea behind this gyrator circuit is that at the emitter there is a voltage that connects back to the input via R1. If the emitter voltage is phase shifted to the input, it will take a current (via R1) from the input that appears to be reactive i.e. it looks like an inductor's current.
However the input is also feeding the network (C1 etc.) which does the phase shifting and so this "capacitive circuit" is in parallel with the "intentional" inductive current via R1. This makes it a band-pass filter but, it can look like an inductor across a range of frequencies.
Better gyrators use an op-amp or another transistor to buffer C1, Anyway, the analysis: -
At point B (base) the AC voltage relative there to the input voltage is: -
\$\dfrac{R_2}{R_2+\dfrac{1}{sC_1}}\$ and this voltage is also at the emitter (the emitter voltage is fractionally less in AC terms but this can be largely ignored). The emitter also acts as a reasonably good ideal voltage source so we don't have to worry about its output impedance of a few ohms.
The current into R1 is the voltage across it divided by R1 (I = V/R): -
Current = \$\dfrac{V_{IN}}{R_1}(1-\dfrac{sC_1 R_2}{sC_1 R_2+1})\$
The impedance, Z into R1 is \$V_{IN}\$ divided by current: -
Z = \$\dfrac{R_1}{1-\dfrac{sC_1 R_2}{sC_1 R_2+1}}\$ = \$\dfrac{R_1+sC_1 R_1 R_2}{1+sC_1 R_2-sC_1 R_2} = R_1+sC_1 R_1 R_2\$
In other words the impedance looking into R1 is an inductance of C1*R1*R2 in series with a resistor of R1 ohms. Remember there is current through the capacitor but this can be ignored if R2 is a lot bigger than R1 and the gain of the transistor is high.