# What determines the transmit frequency in this 555-based transmitter?

I'm a general class ham, and am just starting to do my own homebrewing. I was planning on basing a QRP transmitter off of this instructables project, seeing as I have a veritable plethora of 555 ics in my shack toolbox.

The problem is, this transmitter is designed to operate a bit below the commerical AM band. How is the general operating frequency being set? I'd like to get it set around the 1.8 mhz band (160 meters).

Much help is appreciated, as I'm going to use this to teach the other kids in the ham radio club at my school how to build basic transmitters. Picture of circuit:

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What are the boundaries of the 1.8 MHz band? –  AndrejaKo Mar 26 '11 at 12:58
1.8 to 2.0 MHz. –  technowizard12 Mar 26 '11 at 20:02
I'd go with something a little better; this transmitter almost certainly has FCC issues. –  Brian Carlton Sep 13 '11 at 20:37
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## 1 Answer

From what I can see form the circuit, the timer is running in astable mode. The frequency is controlled by the equivalent resistor made by adding resistances of P1 and R3, resistor R1 and capacitor C1.

If you want to experiment, go to 555 calculator and take a look at bottom schematic. Your P1+R3 are its R1, your R1 is its R2 and your C1 is its C.

UPDATE: I'll try to make it a bit clearer how this transmitter gets its frequency. First, read through the whole instructable. There is a nice explanation related to harmonics in it.

This transmitter controls antenna output using the Q1 transistor. The transistor is triggered by the output of the 555 timer. Therefore there is direct relation between 555 frequency and transmission frequency.

The timer itself is controlled by two resistors and a capacitor. Timer monitors situation on the C1 capacitor. When it is $\frac {2}{3}$ full, timer will emit high output and start discharging the capacitor. When it is $\frac {1}{3}$ full, timer will start emitting low output and start charging the capacitor. When the capacitor is charging, current is going through resistors (P1+R3) and R1. They limit the charging current and modify the time it takes to charge the capacitor. When the capacitor is being discharged, current goes from C1 through resistor R1 into discharge pin which is connected to ground during discharge. This way, R1 controls the discharge time.

Now about the 1.8 MHz band. You may be able to directly transmit at that band by using proper timer settings. For example TS555 timers made by STmicroelectroncs can provide up to 2.7 MHz frequency in astable mode. To get the 1.8 MHz frequency, you can use formulas from the 555 timer. Basically, you should pick the resistors, potentiometer and capacitor so that $((R3+P1)+R1)*C1=8.05*10^{-7}$. If you for example take a 22 pF capacitor (they are commonly used for microcontroller crystals oscillators), resistors added together should be around 37 $k\Omega$. You can take for example R1 to be 8.2 $k\Omega$ and then set the P1+R3 to be 20 $k\Omega$. After that, you can calculate exactly what kind of potentiometer and resistor you need for the transmitter to work correctly using the calculator.

I recommend that you do some more research before making the circuit with the values I recommended. Capacitors usually have high tolerances, so its impact on the circuit should be minimized. Resistors with 1% can be very cheaply obtained, but precise potentiometers or rheostats may be expensive. For example at local stores here, a good multi-turn potentiometer costs between 10€ and 20€, while cheap single turn one costs about 2€.

The point of the above paragraph is that there may be other set of values which could make it much easier and cheaper to set the correct frequency and provide higher precision. I unfortunately don't have enough experience to provide a better set of values.

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Thanks so much. This is exactly what I needed! –  technowizard12 Mar 26 '11 at 19:55
Where do you figure the 8.05*10^-8? –  technowizard12 Mar 28 '11 at 4:55
@technowizard12 From the formulas on the site I linked to. $\frac{1}{\ln(2)}=1.442695...\approx1.45$. Next we have $f=\frac{1.45}{(R_1+2R_2)C}$. If we take that f=1.8 MHz, we get $\frac{1.45}{1.8*10^6}=(R_1+2R_2)C$. The left side of the equation is $\approx 8.056*10^{-7}$. I don't know how I got $10^{-8}$ back then. The rest of the calculation is correct. –  AndrejaKo Mar 28 '11 at 6:00
I'm not so sure about the rest of the calculation being correct, but the 555 calculator does give frequency of 1798201 Hz for the component values I provided. Strange... –  AndrejaKo Mar 28 '11 at 7:30
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