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).

Help is much help 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: http://www.instructables.com/image/FD38QHUGIKL42D3/Schematic.jpg">

  • 1
    \$\begingroup\$ What are the boundaries of the 1.8 MHz band? \$\endgroup\$
    – AndrejaKo
    Commented Mar 26, 2011 at 12:58
  • \$\begingroup\$ 1.8 to 2.0 MHz. \$\endgroup\$ Commented Mar 26, 2011 at 20:02
  • 3
    \$\begingroup\$ I'd go with something a little better; this transmitter almost certainly has FCC issues. \$\endgroup\$ Commented Sep 13, 2011 at 20:37

3 Answers 3


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.

  • \$\begingroup\$ Where do you figure the 8.05*10^-8? \$\endgroup\$ Commented Mar 28, 2011 at 4:55
  • \$\begingroup\$ @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. \$\endgroup\$
    – AndrejaKo
    Commented Mar 28, 2011 at 6:00
  • \$\begingroup\$ 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... \$\endgroup\$
    – AndrejaKo
    Commented Mar 28, 2011 at 7:30

This circuit is going to generate square waves into your antenna. As a licensed ham, you are responsible for transmitting a clean signal. I'm all for trying stuff, but it's best that you put this into a dummy load instead of an antenna, and see if you can hear it on your receiver. Then tune around and look for harmonics and other spurious output. As a matter of fact, the article states that the 555 is running at a fraction of the desired frequency, and you are picking up one of the harmonics on your radio.

This is not as off-the-wall as it sounds. It's a very common practice to "multiply" the frequency to get a higher frequency that the oscillating circuit can't reach by itself. It's also common to build RF amplifiers that are anything but linear. What's needed, though, is a filter to remove all but the desired signal. What you'll need is at least a tuned tank circuit before the antenna. The result won't be a very strong signal, but it may be legal.

Note that a perfect square wave has only odd harmonics. It should be interesting to tune through the bands and see if the odd harmonics are stronger than the even ones. This also suggests that you don't try to, say, run at 500 kHz trying to get 2 MHz. You'd be better off with 400 kHz or 667 kHz. Now, your square wave won't be perfect, so there will still be some energy at the even harmonics. But still, there's a lot to play around with here.

One final note. Almost everything about this circuit will be unsuitable for AM. The original article even said so. But if you get that far, I think that should be discussed in another question.

  • \$\begingroup\$ I totally agree with your analysis (+1). The circuit (link) is more akin to a spark transmitter than an AM system. Its modulation (?) looks more like narrowband FM or PWM - using pin 4 (the reset pin) as a control input, certainly not AM as I would recognise it. The output transistor is totally redundant and will not work as an emitter follower. Fortunately the really poor aerial should limit the range to within a few feet of the offending item. Its a shame that folks put up these projects for beginners. \$\endgroup\$ Commented Oct 4, 2014 at 10:08
  • \$\begingroup\$ Funny that--in my original answer, I had the sentence, "This is marginally better than a spark gap", but I removed it, as there was enough to worry about already, and the SE community isn't all that thrilled with these sorts of remarks. As to the modulation, I think the idea is that you talk into the mic, and yeah, I can tell that's my voice coming out on the radio, although it doesn't sound too good...It's kind of a proof of concept (that mostly fails). \$\endgroup\$
    – gbarry
    Commented Oct 4, 2014 at 19:56

AndrejaKo, I agree with 99.99% of your answer, so my point is a bit of a nitpicker, but maybe apprpriate when instructing. I agree with you that the charging and discharging paths of the timing capacitor, C1, are different: Rcharging= R3 + P1 + R1, while Rdischarge = R1. Depending upon your timer (I'm going by the datasheet of theLM555, which I'm not sure is appropriate for 1.8MHz opn, but it shows the output to be HI when C1 charges from 1/3Vcc to 2/3Vcc, and LO when C1 discharges from 2/3Vcc to 1/3Vcc. This means Time-ON = ln(2)C1(R3+P1+R1), while Time-OFF = ln(2)C1(R1). Adding these two times together gives the total period of one cycle of the output astable wave : T = ln(2)C1(R3+P1+2R1). Since f = 1/T, we can next solve for the circuit parameters: 1.8 x 10^6 = 1/[ln(2)C1(R3+P1+2R1)] C1(R3+P1+2R1) = 8.015 x 10^-7

And then from here, without my having any first hand experience with this communication technique, I yield to Mr. AndreaJo's advice concerning choosing of the individual component values.


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