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Radio transmitters - where does the voltage come from?

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jamesinnewcastle

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Hi

I'm looking at a CB transmitter and it states that it can deliver 4W into a 50ohm load while running from a 13.5V supply.

It is a one transistor amplifier and there are no internally generated voltage sources, so the 13.5V is all there is.

If you do the maths it appears that you need a 39.9V peak to peak sinewave across the 50ohm load to cause it to dissipate 4W. Where does that voltage come from??

I'm guessing that there is a resonant circuit but does anyone know the correct answer? I've seen other amps promising up to 200W, surely the Q needed there would be very difficult to achieve? Or is my maths just completely wrong?

Many Thanks
James

Yes you are right. The voltage is changed using a passive LC (inductor-capacitor) circuit between the output transistor and the 50 ohm load. This is called a matching circuit and is commonly used in RF amplifiers. You can estimate the actual impedance that this circuit presents to the transistor, possibly something around one half an ohm which theoretically might allow it to generate about 8 watts if all was ideal, but I guessed that the transistor would only be 50% efficient at its job.

Thanks Ron

I'll go home tonight and do some research - because now, what I would have considered as a simple 'filter' circuit before has gained a life of its own and become almost active.

An initial question that occurs is that there is no damping resistor in the circuit I have so how do they control the Q and hence the output voltage and power? In a production environment I would have thought this important.

Cheers
James

Be aware that some CB radio manufacturers, the ones geared toward the illegal market, are not filtering their poorly biased cr@p radios.

Phew - the maths is daunting - the stuff I've read about matching circuits is very deep - I'm sure that the output voltage is tied up with a dB figure somewhere.....

Anyone got a link to a site with a more gentle description of how the thing works? I know this amplifier is the class where the conduction is less than 180 degrees, this is effectively a pulse output - I am guessing that it is a pulse that hits the tuned matching circuit and makes it resonate by hitting it at the right PRF. To my mind thats not really an amplifier as such, more a larger hammer than the last stage!

ke5frf - I don't have the amplifier - just the circuit and an on-going discussion at work. The radio spectrum is safe!

James

An initial question that occurs is that there is no damping resistor in the circuit I have so how do they control the Q and hence the output voltage and power? In a production environment I would have thought this important.

Cheers
James

We may have to see the schematic to comment much further.

Hi Ron

Attached are the circuit and parts lists - I've worked out the inductance values, well, ish simply from the physical size and some formulas - so exact values are suspicious but near. (Don't have them with me at the moment)

L4 with C6 and C7 resonate around 27MHz, what I assume is a Pi filter L2 and C3 C4 and C5 is around 23 MHz (possibly should be higher than 27MHz in practice?>, L3 has a reactance of around 60 ohms. All seem about right.

While I can sort of guess which each bit does I still can't see how you can work out the output voltage/power and what controls it. Not sure what the diode pairings are all about either.

James

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General observations first. You called it a transmitter, I call it an add-on amplifier. A transmitter is usually a more complete combination of modulator and amplifier. But I'm nit-picking. On my initial look, the receive path is switched through S1 but I don't see anything to drive S1, so perhaps it is just a manual bypass switch?

Alternatively, the path through L1, C2 and C9 may be the receive path. Clearly the diodes are there to perform a crude form of rx/tx switching. For analysis, assume that rx signal voltages are too low to bias the diodes, whereas tx signals always are large enough to bias the diodes.

Bad design is shown by allowing the bias voltage through R1 to appear on the output antenna connector when S1 is in bypass. Then again, maybe this is intentional to provide bias to something further up the antenna line? I also believe it is bad design to couple back to back rectifiers to an output network to any degree as these will generate harmonics and harmonics should be avoided near the antenna terminal.

One way to conveniently analyze the design approach is to understand the mosts common design procedure for matching circuits such as these, and that is using the Smith Chart. You can find many online references to how to use this tool. Analysis of the LC networks may start by assuming that they are trying to make the S22 of this system as low as possible in a 50 ohm system. In other words, perhaps you can assume that the matched impedance at ANT is 50 ohms. Using a Smith chart, you can work backwards through C6+C7, then L4 then L3 to arrive at a possible output Z for the transistor. Same sort of procedure can be used for the input network, only in the opposite direction (on a smith chart, direction is important).

Again, I caution that looking at a matching network as some sort of filter can be misleading as we are not so much interested in resonances as we are in the ability of each reactance to shift the impedance. Make no mistake, ultimately the matching network does perform some filtering as well, but that is not its main purpose. It is an impedance shifting network first and foremost.

Last edited:
Hi

I'm looking at a CB transmitter and it states that it can deliver 4W into a 50ohm load while running from a 13.5V supply.

It is a one transistor amplifier and there are no internally generated voltage sources, so the 13.5V is all there is.

If you do the maths it appears that you need a 39.9V peak to peak sinewave across the 50ohm load to cause it to dissipate 4W. Where does that voltage come from??

Many Thanks
James

Last I remembered that power = Voltage squared/impedance. Transposing and plugging in the values 4W x 50Ω= 200. √200=14.14 volts. Where does the 39.9V come from?

Last I remembered that power = Voltage squared/impedance. Transposing and plugging in the values 4W x 50Ω= 200. √200=14.14 volts. Where does the 39.9V come from?

Power = voltage squared divided by impedance. So, if you had a much lower impedance (like maybe 1 ohm) presented to the transistors collector, you could generate 4 watts of power at that point? Yes you could. Then it is just a matter of using reactive components to trade voltage for current and vise versa. That is what the output matching network does.

Last I remembered that power = Voltage squared/impedance. Transposing and plugging in the values 4W x 50Ω= 200. √200=14.14 volts. Where does the 39.9V come from?

I was working thus - the 14.4 V you have calculated is the equivalent heat producing value - that is 14.4 V RMS.

Vpk = root(2) VRMS and Vpk-pk is 2 x Vpk

thus Vpk-pk is 2 * root 2 * 14.4 = 40 odd volts (pk-pk)

That's a lot of volts, a normal amplifier would need a 40V supply to produce that as Class A. Even more as rail to rail would be unlikely.

James

I've looked at Smith Charts etc - and I have a long way to go in learning the whole theory. I can understand that you need to match impedances to transfer max power and that you don't want to reflect power back into the transmitter. But none of the stuff I have seen there leads to prediction of the voltage, it's all dB, impedance, etc.

Impedance matching doesn't suggest voltages nor filtering or dB values.

I'll have to keep on looking - I guess I need the Dummys explanation!

James

I've looked at Smith Charts etc - and I have a long way to go in learning the whole theory. I can understand that you need to match impedances to transfer max power and that you don't want to reflect power back into the transmitter. But none of the stuff I have seen there leads to prediction of the voltage, it's all dB, impedance, etc.

Impedance matching doesn't suggest voltages nor filtering or dB values.

I'll have to keep on looking - I guess I need the Dummys explanation!

James

OK, consider it this way. If you present a lower impedance to the collector (drain) of the transistor you can generate more power with a given supply voltage, with class A operation. So the voltage need not be greater to get more power.

I was always under the impression that transmitter power was not watts output at the antenna but the DC power (DC amps times DC volts) into the final amplifier. At least that's the way it is with amateur radio. You guy sure that you're not applying your math skills to the wrong set of parameters?

Amateur radio practices are not necessarily the same as mainstream commercial practices. In the world of two-way radios, cellphones, wireless gadgets of all sorts, military radio, and just about everything else except perhaps high power broadcasting (not sure what they do), transmitter power is defined by the output power fed to a specified load resistance and measured with a watt meter. I think perhaps the measurement of DC input is a hold-over from the days when it was hard to measure RF output, but easy to measure DC input. The knowledge that the efficiency of converting the DC into RF could not exceed 100% puts an upper limit on the RF output even if you are only measuring the DC input, so its not a bad method for those without fancy test gear.

OK, consider it this way. If you present a lower impedance to the collector (drain) of the transistor you can generate more power with a given supply voltage, with class A operation. So the voltage need not be greater to get more power.

That makes sense, at that point in the circuit the load resistance might be a few ohms and so only require a few volts to generate 4W - At the output though, the load will be 50 ohms, and at that point the peak to peak voltage must be 40V (given 4W output power). I'm after any sort of formula or even 'spoken' words that says "it will be 40V because...."

James

That makes sense, at that point in the circuit the load resistance might be a few ohms and so only require a few volts to generate 4W - At the output though, the load will be 50 ohms, and at that point the peak to peak voltage must be 40V (given 4W output power). I'm after any sort of formula or even 'spoken' words that says "it will be 40V because...."

James

If you are a fan of the film "Blast From the Past" you might recognize the finish to this line....it will be 40V because...it must.

In RF work, we do our network design using power and impedance and let the volts and amps work themselves out. There is no doubt that at the antenna terminal, the peak to peak voltage must be 40V to give 4 watts at 50 ohms. As I mentioned in a previous post, the components used between the drain of the transistor and the antenna output are all reactive, so they don't use up power, but they do shift voltages and currents (voltage goes up, current goes down, power stays the same). If we pay attention to the impedance going from 1 ohm at the input of the matching network to 50 ohms at the output of the matching network, then the voltages and currents will also transform from 5.6 volts and 5.6 amps peak to peak at the input, to 40 v and .28 amps peak to peak at the antenna terminal (if i've done my arithmetic correctly that is).

RadioRon, I thought power level was supposed to be tested using a EM field meter at a specific distance from the transmitter? If I'm not mistaken transmitters/receivers have to be tested with their antennas as a unit to receive validation. I've read many warnings that even though the antenna's on many devices are removable changing the antenna can void the FCC certification. Mind you that's a whole lot of they gotta catch ya doing it =)

RadioRon, I thought power level was supposed to be tested using a EM field meter at a specific distance from the transmitter? If I'm not mistaken transmitters/receivers have to be tested with their antennas as a unit to receive validation. I've read many warnings that even though the antenna's on many devices are removable changing the antenna can void the FCC certification. Mind you that's a whole lot of they gotta catch ya doing it =)

Good point, but this only applies, in the USA, to some "part 15 devices". These are unlicensed transmitters operating at low power and are a special case. My comments about measuring conducted power applied in general to licensed transmitters and some unlicensed devices and I regret not making this distinction in my last email. It would be interesting to know how many devices of each category are out there. I would guess that there are many millions of unlicensed part 15 transmitting devices out there. I would also guess that there are several hundred million licensed devices out there. Cell phones are licensed devices of a special sort and there are many more cellphones than there are all the other types of licensed transmitters combined.

Hrrm, here's a question then. If cell phones are licensed on their power output through a resistor load could I hook a micro yagi or biquad to my cell phone and still be legal?

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