0
\$\begingroup\$

I am examining my own radio circuit now for several days, so I included a pic:

circuit

When the circuit has been powered up, it works fine without fingers or equal grounding methods, for example by getting use of wall-socket etc. I tested this circuit without the component C3 (330n) and I was frustrated, because it did not work. I tried to ground the circuit by touching it with fingers and I must say by the way, it is not advisable, but it worked and I got a good sound from receiver. The thing was, how can I leave this circuit for a longer distance without touching it. I got the idea to add a capacitor, parallel to R2. I have seen 100nF gave me a weak sound as result, so I was encouraged to increase this 100nF something up to 330nF and finally, there I had a good result without using external grounding stuff. Well, I know the C3 capacitor is blocking DC voltage. Why did it work?

Note:

V1=source voltage V2=input wave

L2,C1=tank circuit ~107MHz

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

The feedback via C2 can only work if the base has a low RF impedance to ground. If there is no C3, the voltage changes introduced at the emitter do not modulate the base current enough. A value starting from 1 nF is good enough for frequencies in the FM radio band.

The value of R2 is a bit low, I use 47 kΩ there.

If you want to modulate the oscillator, you need a decoupling capacitor at the input. Without it, the DC operation point of the transistor is undefined.

You acheive a better antenna match If you center tap the coil and connect the antenna there.

\$\endgroup\$
2
  • \$\begingroup\$ I upvoted the answer, you gave me. I have one more question, can you tell me how you calculated the 1nF for FM and the 47Kohm? Then, I will accept it as complete answer. \$\endgroup\$
    – lastime
    Oct 8, 2023 at 20:44
  • 1
    \$\begingroup\$ XC of 1 nF at 100 MHz (1/(2 pi f C)) is below 2 ohm, higher capacitor values will not improve the circuit. The circuit works good with an emitter voltage around 2 V. With 220 ohm emitter resistor the current would be around 10 mA. The DC gain of the transistor is, say, 100. So you need a base current in the range of 100 µA. Assuming a base voltage of 3 V and a supply of 12 V the maximum resistor value is 9 V / 100 µA = 90 kohm. My circuit runs from a 9 V battery and 47 kohm is the best value there, found by experiments. Don't forget a 100 nF decoupling capacitor across the supply voltage. \$\endgroup\$
    – Jens
    Oct 8, 2023 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.