Secret Diagram

Initially don't apply any sort of RF input, simply use the power and check with a meter the voltage at point 1. Calibrate R3 as you get the reading near about 0.7V. Use the antenna with a 2 x 100 Ω 0.5W resistors in parallel at the RF output. After this attach the rf source that may be supposed to boost and link up this RF input to the output.
Carefully set up C1 so as to extract the most optimal potential magnitude on the rf link. Next up, rotate R3 once more to acquire 0.7 V at point 1. Finally set up C5 and C6 for ensuring the highest output voltage (quite between 12V to 18V).

Verify the dissipation of T1′s heatsink, if you see it to be just fine switch off the power, isolate and pull of the 2 resistors of 100 Ω and connect back the antenna (make sure the the probe stays connected). Connect again the power and yet again allign C1, C5 and C6 for highest voltage read on the probe.

You could think of using a an ammeter for enabling reading the current incourse via T1. This must be restricted below 150mA at 12V and 100mA at 24V otherwise the transistor may well fry of. L2 and L3 coils must be installed at an angle of 90 degrees between the two. Don’t run the 1W rf fm amplifier in case you witness your TV set is getting jammed and perhaps the laws of the your country prohibits the use of such FM transmitters.

R1 = 100 Ω
R2 = 2.2K Ω for 12 V and 4.7 kΩ for 24 V
R3 = 10 KΩ
R4 = 100 Ω
C1 = C5 = C6 = 10 – 60 pF
C2 = C4 = 1 nF
C3 = 10 uF
D1 = 1N4148
L1 = 20 turns of 0.2mm super enameled wire over R4
L2 = 7 turns of 0.8mm super enameled wire with 6mm diameter on air
L3 = 4 turns of 0.8mm super enameled wire with 7mm diameter on air
T1 = 2N4427, KT920A, KT934A, KT904, BLX65, 2SC1970, BLY87 (2N2219, output of 0.4W) at 12V
T1 = 2N3866, 2N3553, KT922A, BLY91, BLX92A at 24V

Small 1.5V to 5V or 12V DC/DC converter with LT1073 chip. The IC is available in three different versions, depending on output voltage. Two with fixed output voltage of 5V and 12V, and the most interesting that can be adjusted. The adjustment is done through a voltage divider with two resistors, of mass, output and Terminal 8, internally connected to the voltage comparator IC, which is responsible for stabilizing the output voltage. 

Regulator Handles Two Input Voltages circuit in Fig 1 supplies both 3.3 and 5V to transitional circuits that employ both the new 3.3V and older 5V devices. Additionally, because the regulator accepts either 3.3 or 5V inputs, you could plug it into either a new 3.3V system or an old 5V system.The circuit consists of two sections: a dc/dc converter and a double-pole, double-throw (dpdt) switch. The dpdt switch comprises a pair of dual n-channel MOSFETs (Q2 and Q3) and their associated high-side drivers.

Upon power-up, the comparator in IC2 determines the state of the circuit. The comparator’s output, IC2 pin 6, goes to the input of the MOSFET driver, IC1. The driver internally generates a gatedrive voltage 8.8V above the device’s supply voltage. This high voltage drives the appropriate MOSFETs in Q2 and Q3.

IC2 is also the heart of a flying-capacitor, buck/boost dc/dc converter. Unlike other switching-regulator schemes, this topology needs no transformers. Transistor Q1 controls this section’s output voltage, VS. When VIN is at 5V, Q1 is off, forcing the section to operate as a step-down converter. In this mode, the section produces 3.3V, which goes to the output through Q3B. Also in this mode, 5V power goes directly through Q2A, and Q2B and Q3A are both off.

When VIN is 3.3V, IC1 turns on Q1, shorting out the 140-kΩ resistor and forcing the dc/dc-converter section into step-up mode. In this mode the converter section generates 5V at VS, powering the 5V output via Q2B. Also in this mode, 3.3V goes directly from the circuit’s input to the output via Q3A. Q2A and Q3B are both off.No-load quiescent current consumption is approximately 500 μA.

Lower-frequency converters would reduce power consumption at the expense of a larger inductor. The efficiency of the dc/dc-converter section is 73% in either mode. But because this power accounts for only half of the circuit’s output power, the circuit’s overall efficiency is approximately 80% with VIN=3.3V and 86% with VIN=5V.