SSTRAN ouput power

I have an older Gizmo transmitter, which has an estimated RF output of 10-15 mW with the 100mW DC Power input limit. I've been looking at ways to legally inch that up a few dozen mW, and here's a summary of what I've got so far.

First, I remember reading quite of bit of information Keith Hamilton presented on another forum when the Rangemaster first came out, to the effect that he had developed his transmitter specifically to get the very most RF out of the 100 mW power input limit. So one aspect of this is the efficiency of the transmitter circuitry in getting the most out of that 100 mW input.

A second factor that Ernie at PanAxis discussed in the documentation in his AM100 plans was modulation techniques. Effective modulation techniques make sure the RF you're putting out carries the most audio information possible, making the signal sound louder in the receiver and slightly increasing range, too.

So if you have an efficient transmitter with good modulation you're going to get the most usable signal over the longest distance, all other things being equal (which they never are!)

As far as modulation goes, that's going to be pretty hard to hack, it seems to me. there might be some mods for certain transmitters like the Wentzel and Wild Planet DJ and help, but that is probably fairly set.

Similarly, efficiency is going to be hard to mod after the fact.

I'm looking at adding a buffer/final amplifier to the output of the Gizmo that might help pull a little more output power without going over the 100 mW limit. I'm doing that by ordering this amplifier plan from PanAxis:

*AMA5 3W AM AMPLIFIER ($5.00) on this page here: http://www.panaxis.com/P-index.html

WHile this amp plan is sold as a three watt amp (EEK!) there is a mod in the plan to limit power input for part 15 use. An alternate schematic is offered in the plan showing the part 15 use with the input power limited to 100 mW using a resistor.

Since the regs state "100 mW input power to the final amplifier" I hope that by adding this circuit to the Gizmo that I'll be able to get 60 -70 mW RF out of my final - a bit more power but still within the regs.

More discussion on this thread here - http://part15.us/node/858

Since the regs state "100 mW input power to the final amplifier" I hope that by adding this circuit to the Gizmo that I'll be able to get 60 -70 mW RF out of my final - a bit more power but still within the regs.

Any amplifier added to a tx that already has amplitude modulation (as in Part 15 AM) will need to operate in a linear mode in order to avoid creating a lot of distortion of that r-f waveform. The problem is, most linear amplifiers are not very efficient, perhaps around 35% at best.

So the output level of your linear amplifier with 100 mW input power probably will be not as much as you are hoping for, and possibly not even worth the cost and effort -- as far as "coverage" goes.

If the modulated r-f output stage in a Part 15 AM tx is operating Class C or D and uses "high-level" modulation, then depending on the circuit design and the load impedance the tx sees, the power output of its unmodulated carrier could be around 80 mW when its DC input power is 100 mW.

Amplitude modulation to +/- 100% by an audio sinewave increases the total power output to 1.5X the unmodulated value.
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Thank you Rich - I didn't realize that about linear amps and efficiency - that really puts my idea in a better light. It really wouldn’t be worth the trouble to build the 100mW booster if I’m only getting 35% of that – that’s just the long way around to where I already am 🙂

Looks like I better look for a new XMTR!

I have been able to make a tuning network to achieve about 60mW into a true 50 ohm load. 1 caveat tho- it cannot be properly modulated at that power output level, mostly because of the modulator circuit used; it's very similar to 'cathode modulation' in the tube transmitter world. However, I've got a few more tricks to try to get proper 100% modulation. I'm hoping to get a true 100mW output (or something close) into a true 50 ohm load. Obviously, the problems we've all been having with AMT-3000, is that, by itself, it will not load into 50 ohms! I've come up with a way it can be done, but it involves an external matching network. If anyone is interested, let me know and I'll give you the details I have so far...schematic and all.

Bob
[email protected]

Just a quick note that Part 15 rules specify a limit of 100 mW DC input into the final RF output amplifier, not 100mW RF output from the transmitter. So achieving a 100mW output would be a tough hill to climb.

That being said, any improvement to the utilization of those precious 100 mW of DC input power would be magnificent!

Allow me to mention again that the test results I reported in the original post were done with no attempt to "tune" the AMT-3000 so Bob's approach using a tuner makes sense. The AM-25 will drive a load between 20 and 100 ohms according to my bench tests but I found the high power output results from a DC input power which greatly exceeds 100 mW. (see: http://part15.us/node/637 and http://part15.us/node/938)

Bob, I don't know if you are aware but a while ago on this board we had a discussion about the output matching and PhilB offered an explanation. It's in a long thread but you might get some good insight from it:

http://part15.us/node/883

From PhilB's comments and my own experience, the unit expects a load of about 800 ohms resistive at the collector of the output transistor and this is done by setting the antenna slightly off resonance so an inductance appears in series with the antenna.

I have also found when the unit is presented with a different load that the audio degrades and the RF waveform distorts. With the proper load, both are excellent. I mention this so you have some things other than power to look for.

I also measured the DC input power to my AMT 3000 and it is around 93 mW. I will post later the method I used to do this.

Neil

Here's an article I wrote some time ago about this. Hope it is useful.

Description of measurement of the SSTRAN AMT3000 AM transmitter final input power.

For those who just want the results, I measured the final input power to be 91 milliwatts which was not affected by the load.

The 100 mW. limit applies to the DC input power to the final RF stage. It has been common practice to measure transmitter input power as the DC input power. So for this unit, we need to measure the voltage across the final and the current through the final. Power is then P = VI.

My description is keyed to the AMT3000 schematic rev. 05-1. The SSTRAN schematic is copyright SSTRAN, so if you need one, contact them. I will not provide it.

There may be some interpretation problems about the final stage (see below). I used Q5 as the final stage and did not include the base bias current. We can indirectly measure the current through Q5 by measuring the voltage across R8 (VR8) and, using ohm's law, calculate the current through Q2. This current is the sum of the currents through Q1, Q4, and Q5. The Q1 current is obtained by measuring the voltage across R3 (VR3) and applying ohm's law. This, subtracted from the current through Q2 gives the current through Q4 and Q5. Since Q4 and Q5 are symmetrically biased, IQ4 = IQ5 and the current through Q5 is one half the total current.

Summary:

IQ5 = 1/2 ( VR8/R8 - VR3/R3 ) = 7 milliamps in my unit.

The voltage measured at the emitter of Q5 ( VQ5E ) to ground = 2.03 volts.

The voltage across Q5 (C to E) = +15 volts - VQ5E.

The power is P = ( +15 - VQ5E) x IQ5 = 12.97 volts x 7 milliamps = 91 milliwatts.

If one wants to count Q2 as being part of the final stage, P = 15 volts x 7 milliamps = 105 milliwatts (excluding the bias current from Q1). I exclude Q2 from being part of the final stage because it does not process the RF signal. It serves as a constant current sink which is controlled by the audio. Common practice is to measure transmitter final input power under conditions of no modulation, therefore Q2 power does not add to the RF signal. Q2 provides a constant current bias for Q4 and Q5 which is included in the calculation for final stage power. The only power Q5 has available to produce output power is the voltage across Q5 x the current through it and it is reasonable to use this figure. The power used by Q2 does not produce RF output and can be excluded. (This is similar to excluding the filament power in a tube since the filament power, though necessary for the tube to function, does not contribute to the output power)..

For those who want to see how a similar circuit works, search MC1496 balanced modulator/demodulator data sheet and application notes.

Neil