Simple AM transmitter

[George L.]

hello,

I am trying to make a VERY simple AM transmitter that will transmit a signal (push a botton and hear a beep) :wink: . I would like the transmitter to reach up to 10 meters (across a room).

I have attached a circuit that I though was suitible for this purpose. I am a newbie ... so I would really appreciate it if someone could tell me if the circuit will work ... and if it will reach about 10 meters.

Please help

Thanks,

George

 

[audioguru]

Hi George,
A radio transmitter needs an RF oscillator. You sketched what might be a tuned RF amplifier, if the transistor wasn't turned-on so hard that it is saturated all the time.

You can change the transistor circuit and make it an RF oscillator by adding feedback. Look-up a Hartley or Colpitts oscillator.

You can fix the transistor circuit by reducing the value of its emitter resistor, so most of the supply voltage won't be wasted across it.
You can stop the transistor from being saturated by looking-up its current gain on its datasheet, and increasing the value of the base resistor accordingly.

Then you will have an RF transmitter that will transmit a carrier at the frequency of its tuned coil and capacitor. It won't make an amplitude-modulated beep until you capacitive-couple an audio-frequency tone oscillator to its base. :lol:

 

[zachtheterrible]

You can change the transistor circuit and make it an RF oscillator by adding feedback. Look-up a Hartley or Colpitts oscillator.

To do this just add a capacitor across the collector and emitter of the transistor.

If your frequency is going to be around 100MHZ, the capacitor can be somewhere around 10pf or so. It doesn't really matter. It can be two wires twisted together :lol:

You do know that this circuit transmits FM AND AM. So you could just as easily use an FM radio. This would be easier because I can tell you the values of everything :lol:

 

[stevez]

Your statement says you want to push a button and hear a beep. Your schematic shows something very simple - and something that contains few parts such as this could transmit but without some means of modulation it will only be a continuous RF signal - unmodulated. Depending on the reciever, you may hear little more than a hiss, if you hear that unless the reciever has the right equipment to add a tone. The tone could be a BFO or something to react to the RF on the input and sound a tone. The little FM transmitters with an oscillator attached might serve well - within the limitations of that device. Small AM (broadcast band) transmitters are similar - you might get a heterodyne tone if the transmit frequency was offset from recieve slightly so you could skip the modulation.

 

[mstechca]

You definitely need an external capacitor connected between collector and emitter of the transistor. I find that a value around 4.7pF is optimal. It seems that higher values reduces the transmitters bandwidth. In simple terms, bandwidth = amount of dial movement in one direction.

In my design, I connect a capacitor in parallel with the resistor. I think these two make some important frequency. My design uses 20K and 0.1uF.

and connect emitter to ground. A resistor there will limit the distance.

 

[audioguru]

You definitely need an external capacitor connected between collector and emitter of the transistor. I find that a value around 4.7pF is optimal...... and connect emitter to ground....

The transistor's emitter is its input for positive feedback through the cap between collector and emitter so it oscillates.
If you ground the emitter then how can it oscillate? I know, I also saw the circuits like that on the web. They probably don't work.

You are talking about tiny capacitor values suitable for a VHF 120MHz aircraft receiver to pick-up. I don't think George has an airplane on the other side of his room, probably just an ordinary 1Mhz AM broadcast band radio. So your 4.7pF feedback cap becomes about 330pF. :lol:

 

[mstechca]

330pF???

haha. maybe for AM, but for VHF, not for me!

I think if anything, the resistance I use from emitter to ground is < 1 ohm.
I tend to think of the feedback capacitor as the bandwidth capacitor.

 

[audioguru]

The capacitive reactance of a 4.7pF cap at 100MHz is 340 ohms.
At 100MHz, the transistor has a voltage gain of about 5.
Therefore if the emitter resistor is less than about 85 ohms, the transistor won't oscillate.

Of course the 4pF of the transistor and a few more pF for stray capacitance increase the value of the feedback capacitance, so the transistor will barely oscillate with an emitter resistor of about 34 ohms and won't oscillate with less resistance.

 

[George L.]

thanks everyone....

I don't think I made much sense when I said "push a botton and hear a beep..." The transmitter is not doing the beeping, the receiver is. The transmitter transmits a signal to an AM radio that will either make an LED light up or make a distict sound.

I am trying to use this circuit for discrete communication between two people in a room, so the device should be as small as possible, therefore as simple as possible. I would like to transmit in AM because it is easier to make an AM receiver.

This is what I will use as a receiver: https://www.aaroncake.net/circuits/radio.htm

so a simple solution (as Audioguru said) would be to just put a 330pf capacitor between the emitter and collector of the transistor???

the goal is for this transmitter to be prortable so I can't connect it to "real" ground.

one more thing... how do you calculate the capacitance of the cap. across the emitter and collector??? (4.7?, 10?, 330? :? )

thanks,

George

 

[audioguru]

Hi George,
Try making the Op Amp Radio first. I don't think it will work since the opamp isn't biased correctly and the detected audio feeds the dead-short of the virtual ground at its inverting input.

How do you expect to make the opamp light an LED or make a distinctive sound?

 

[ljcox]

George,
I agree with audioguru, the circuit taken form the internet and posted by him above needs improvement. I suggest you connect a capacitor of about 10nF fro the cathode of the diode to ground and relpavce the capacitor in series with the - input with a resistor of say 10k. Then you can adjust the gain of the op amp by selecting the feedback resistor.

The gain will approximately = Rf/Ri where Rf is the feedback resistor and Ri the 10k I mentioned. So for example, if Rf = 100k, the gain would be = 10.

Len

 

[audioguru]

Hi Len,
The opamp isn't biased correctly with its non-inverting reference input connected to ground. It doesn't have a divider nor a negative supply!

Your proposal to replace its input capacitor with a resistor would have it trying to drive its output negative for the detection of any modulation of a carrier that the crystal radio part receives.

The crystal radio part will pickup many local AM stations all jumbled together since it has only a single tuned circuit. Your proposal for it to have a load as low as 10k will reduce its selectivity even more.

I think the opamp should be a non-inverting amplifier that will have a very high input impedance so that the filter cap's value can be small and so that the tuned circuit can have a high Q because it will be unloaded for its best selectivity.
Then the opamp's input can be biased at mid-supply or a negative supply can be added.

Before showing George how to do it like this, let's see (hear) what he says about the original radio's performance. :lol:

 

[ljcox]

I agree, I missed the point about the bias. It is essentially an amplified crystal set.

However, in the dim past when I made crystsal sets, we listened with earphones that had a resistance of about 2k. So the 10k should be adequate. Certainly, the selectivity won't be brilliant, but he only wants a short range receiver.

The simplest way to bias it would be to connect two 22k resistors in series across the supply rails and connect the + input of the op amp to the mid point.

Also, the gnd line should be connected to earth and the antenna cut to length according to the frequency.

Len

 

[audioguru]

In "the good old days" weren't there only a few local AM stations?
Can you find a blank spot today on the dial of a poor-selectivity crystal radio for it to pickup a weak transmitter?

I think George is going to need all the selectivity that he can get, unless he is in the middle of an ocean, and doesn't use it at night. :lol:

 

[mstechca]

The capacitive reactance of a 4.7pF cap at 100MHz is 340 ohms.
At 100MHz, the transistor has a voltage gain of about 5.
Therefore if the emitter resistor is less than about 85 ohms, the transistor won't oscillate.

Ok, so we got 340 ohms going from Collector to Emitter.
But for my transmitter, I use a 0.1uH inductor with a capacitor between 12 and 27pF.

So then the impedance between + and collector will then be < 10 ohms.

If we add, I get 340 + 10 = 350 ohms. I still can't figure out why 85 ohms is the minimum resistor (between emitter and ground) required to start an oscillation.

Does the power supply "choke" or go down quickly if it is connected in parallel with a small resistor? (between 200 and 1000 ohms)

 

[audioguru]

But for my transmitter, I use a 0.1uH inductor with a capacitor between 12 and 27pF.

So then the impedance between + and collector will then be < 10 ohms.

How do you calculate less than 10 ohms? An inductor in parallel with a capacitor are a very high impedance at their resonant frequency!

It works like this:
A parallel tuned circuit is a low impedance below its resonant frequency because the inductive reactance is low (getting closer to being just a piece of wire). A parallel tuned circuit is a low impedance above its resonant frequency because its capacitive reactance is low. At the resonant frequency the inductive and capacitive reactances cancel, resulting in a very high impedance.

Don't you understand how an oscillator works?
It needs to have positive feedback and a gain of one or more.

The transistor is an amp with a voltage gain of 5.
The feedback capacitor is feeding the emitter resistor and make a voltage-divider with an attenuation of 1/5.
Therefore the overall gain = 1 and it oscillates at the frequency of the parallel tuned circuit. Increase the value of the emitter resistor to about 100 ohms and the oscillator will have more positive feedback and will oscillate more reliably.
This is an approximation because the transistor's emitter has an impedance which is its internal resistance multiplied by it beta (if I remember correctly), which makes the positive feedback less because it is in parallel with the emitter resistor. Therefore the oscillator will be more reliable with an emitter resistor of about 220 ohms. :lol:

 

[mstechca]

several months ago, impedance was a completely new term to me. That's why I don't know as much as you.

I guess gain must then equal Xc / Re

In my transmitter circuit, I rounded my resistance to 0 ohms because it is a 1 inch wire from emitter to ground. I'll up it to one ohm.

So if impedance = 340 ohms, and I use a 1 ohm emitter resistor, wouldn't the attenuation be 1/340 and the gain be 340? ;-)

 

[audioguru]

I guess gain must then equal Xc / Re

No, the attenuation is Re/Xc+Re.

In my transmitter circuit, I rounded my resistance to 0 ohms because it is a 1 inch wire from emitter to ground. I'll up it to one ohm.

A 1" piece of wire has a resistance of about nothing.

So if impedance = 340 ohms, and I use a 1 ohm emitter resistor, wouldn't the attenuation be 1/340 and the gain be 340?

The attenuation will be so high that there won't be any positive feedback, so it won't oscillate. Also, don't forget that an "audio" transistor has a max voltage gain of only about 5 at 100MHz.
If you use a 1 ohm emitter resistor the the max gain of the transistor at 100MHz must be at least 341 for the circuit to oscillate. Impossible. :lol:

 

[mstechca]

No, the attenuation is Re/Xc+Re.

lmao. Its interesting how no emitter resistor gives 0 attenuation!

0 / Xc + 0 = 0 / Xc = 0

The attenuation will be so high that there won't be any positive feedback, so it won't oscillate. Also, don't forget that an "audio" transistor has a max voltage gain of only about 5 at 100MHz.
If you use a 1 ohm emitter resistor the the max gain of the transistor at 100MHz must be at least 341 for the circuit to oscillate. Impossible.

so basically I'm looking for a capacitor that can produce a ridiculously low reactance ( < 10 ohms). I think I can now see why I stay under 10pF for the collector to emitter capacitor

 

[audioguru]

lmao. Its interesting how no emitter resistor gives 0 attenuation!

0 / Xc + 0 = 0 / Xc = 0

No! 0/X = 1/infinity. Way too much attenuation.

so basically I'm looking for a capacitor that can produce a ridiculously low reactance ( < 10 ohms).

No, then the transistor can't drive it, since it isn't a power transistor operating at a huge current. Also, a very low reactance capacitor will load-down the tuned circuit. The total impedance of the attenuator must be high enough that the transistor can drive it.

I think I can now see why I stay under 10pF for the collector to emitter capacitor

A 10pF cap plus about 4pF for the transistor plus about 3pF for stray capacitance= 17pF which has a reactance at 100MHz of 94 ohms, which is too low. A little transistor operating at a reasonable current will have a hard time driving it, and it will load-down the tuned circuit too much. :lol: