# Radio transmitter using a crystal oscilliator

I am on a mission to control my RC car using a Arduino. The simplest way I thought of doing this is to add Arduino outputs into the transmitter. Long story short I wanted to learn more about how these transmitters worked.

I have done some research and managed to find a schematic that is similar to the transmitter circuit that I currently have posted below. My transmitter is using a 27MHz crystal oscillator to create a carrier wave.

I do not understand how the crystal oscillator is biased and how the carrier wave is mixed with the transmission signal (message left, right, forward, reverse) from the IC.

Sorry if I have not used the terms correctly I am new to RF and want to learn.

• en.wikipedia.org/wiki/Capacitive_coupling Commented Jan 1, 2014 at 1:34
• I know what that is , but when a signal is mixed isn't it added in series, this cap is in parallel.
– subz
Commented Jan 1, 2014 at 2:07
• Hazarding a guess that if it's typical R/C, your IC is producing something called pulse position modulation (PPM). This is a simple on/off signal that starts with an index pulse, and then follows with a series of more pulses each of which represents one R/C 'channel'. I would say Q1 forms the oscillator, and that the 'EC' signal from the IC is applying the on/off modulation to that via Q2. Commented Jan 1, 2014 at 3:08

The voltage from pin 8 of SCTX2BC changes the DC component of the voltage on the base of Q2 transistor, modifying its quiescent IC current and its gain. So this is how the RF signal amplitude is modulated.

The RF signal from the Q1 based oscillator is connected via the capacitive voltage divider consisiting of C5 and C6.

The bias of Q1 may be found by the following equation:

$$V_{bias} = R7\,\frac{I_E}{\beta+1}+0.65V+I_E\times R8 = 9V$$

$$V_{CE} = 9V - I_E \times R8$$

$$I_C = \frac{\beta}{\beta + 1}I_E$$

C4 assures the positive feedback.

• just curious about how it is a positive feedback , cheers
– subz
Commented Jan 2, 2014 at 12:34
• The signal is routed from the collector to the emitter, so from the point of view of this signal this is a common base amplifier, which doesn't invert the signal.
– wzab
Commented Jan 2, 2014 at 18:41
• thank you very much, time to look up on common base amplifiers and its uses :)
– subz
Commented Jan 4, 2014 at 4:57

I'd like to expand on the oscillator bias schema shown.

The collector of Q1 is biased using a choke inductor which has 0V drop at DC but presents ~370Ω impedance at 27MHz (Z_L=ωxL). The base is biased with 100kΩ (though it's in parallel with βxR_emitter, which is still in 10s of kΩs). The emitter is degenerated by 100Ω. Based on basic BJT equations (already in other answers) and C945s datasheet (for values of beta), we can estimate that the Q1 Vb and Vc are at about 4.5V, 3.8V respectively, and collector/emitter current is about 5-8mA.

Also because the Q1 collector sees a much lower impedance at 27MHz than the base, I believe that the main oscillation signal is generated at the base, then transferred to the emitter. Then a fraction of the emitter signal (about 1/6 or so, based on current division between R8 & C4+L3) is coupled by C4 to the collector and then adds back to the base signal through the crystal which is very high impedance at resonant frequency of 27MHz (due to the inherent property of the resonance tanks).

I believe that this oscillator is a base-to-emitter current amplifier. At risk of responding to comments in the answer, the idea proposed in one of the previous comments that this is a common-base amplifier with input at collector and output at emitter is likely not feasible. In that configuration, the current would be barely amplified (1.001 times?), and voltage signal would be significanly attenuated, thus not meeting the criteria for oscillations given the losses in aforementioned 1/6 current division in the feedback path.

• If you include the transistor's base-emitter capacitance in your analysis, this looks like a classic Clapp oscillator. Commented Dec 16, 2023 at 20:36
• @JohnDoty could you pls help explain which key property makes this a Clapp oscillator when we consider the Cbe?
– S.C.
Commented Dec 17, 2023 at 21:12
• A Clapp oscillator has a CE capacitor, a BE capacitor, and a CB series-tuned resonator. Commented Dec 17, 2023 at 22:50