Calculating the Total Current in a Mixed Circuit
First determine the resistors which are in parrallel and series.
In this case, the 10K is connected to the positive and the negative rail - making it parallel to the 10K variable Pot - which is also connected to the positive and negative rails. (the 3rd pin of the LM386 Amp is positive).
The 1K Pot is also connected parallel.
The circuit basically looks like this
As you can see, there are three parallel resistors. The total resistance is
1/r1 + 1/r2 + 1/rn
= 833 Ohms (Assuming the variable resistors are set to maximum resistance
To calculate the current we can use ohms law
I = 9v / 833 = .01 Amp
.01 x 1,000 = 10mA
However, we still have to take into account the capacitors. How do we measure the resistance of capacitors ? Capacitors are reactive components, so their reactance or resistance must be measured when a AC signal is applied. The formula is:
Xc = 1
Where Xc is the Capacitive Reactance, f the Frequency in Hertz and C the farad.
The following capacitors are in parallel in the circuit.
TOTAL:320.1uf / 1,000,000 =
C =.00032 FARADS
Xc = 1
2 * 3.142 * 125,000,000 * .00032
Xc = 0.00000398 Ohms
The following are in series:
Total= 220uf = .0045uf / 1,000,000
C =.000000004.5 FARADS
Xc = 1
2 * 3.142 * 125,000,000 * .0000000045
Xc = 3.97 Ohms
Now we have the series bunch of caps, and the parallel bunch of caps, in series - so calculating the capacitance and hence the impedance of the two "bunches" gives us approx:
TOTAL ~ 3 ohms
as you can see with this high frequency, the reactance of the capacitors is quite minimal
There are two or three that i dont now whether they are parallel of series. Lets leave them out.
So lets say the total resistance in this circuit is then about 835 Ohms which gives us approximately 10mA consumption
We also have to calculate the Inductive reactance.
The formula for Inductive Reactance is:
Xl = 2 * 3.142 * f * L
Where L is in Henries and f is in Hertz
So given our tank inductor to be approx .211 uH this equates to
.211 / 1,000,000
One can either calculate the individual impedances for each of the inductors (like for L1 below), or calculate the individual henries.
Xl = 2 * 3.142 * 125,000,000 * .00000021
Xl = 164 Ohms
For L2 9.3mH
Now L2 is connected from a positive to a negative, whilst L1 is connected to the positive rail so we can
say that these two inductors are connected in Parallel.
L1 = 1 / .211 = 4.74
L2 = 1 / 9.3 = 0.107
L = 1 / (4.74 + 0.107)
L = .206 / 1,000,000 (to convert from macroHenry to Henry)
L = .0000002
So Z (impedance for both the inductors is)
Z = 2 * 3.142 * 125,000,000 * .0000002
Z = 162 OHMS
So we now have
162 + 4 +833
= 998 Ohms
I = 9 / 998
I = 9mA drawn from the circuit not taking into account the FET (say 5mA) and IC 386 (4mA)
So total current drawn