This is 11m. Anything is possible.
As I told my kids (and grandkids) when they asked how Santa could get all around the world in one night: "By means of magic."
They all still use that phrase!
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As I told my kids (and grandkids) when they asked how Santa could get all around the world in one night: "By means of magic."
They all still use that phrase!
Dxhound, it would appear you know your waveforms and meters fairly well. You are 100% correct in your explanation that a signal driven into flat topping will trick the peak meter into showing more power because the cutoff waveform will hold the peak condition longer. Not that anymore peak power was generated, just distortion.
Some confuse this as being power that is harmonically related to the carrier when more often than not, it's just that flat topping of the envelope. Even the best watt meters render useless readings unless a couple of key parameters are simultaneously observed.
There can be no reflected power while reading forward power. Accuracy is dependant on a perfect 50 ohm load. There can be no flat topping on the positive peaks, they exaggerate PEP readings. Similarly, RF cutoff on the negative peak will reduce the PEP readings. This will require a scope to check. Harmonic content must be below 10%, preferably much lower and there can be no oscillations in any RF stage, including those driving the PA. This will require a spectrum analyzer to confirm.
Antron PSU's
Oddly enough I have a couple of antron 70A supply's that get shut down by a Texas star DX500 running in SSB at about 550w Pep. they can not handle it. Yet 2 of the 70A supplies paralleled have no problem keeping up with a dx1600 at up to 1300Wpep from time to time.
I don't know me...
An RM 50 switch mode supply suffers the same fate with the DX500.
Voltage 14.2
I now use 74A switchmode supplies with the DX500. No problems so far but the fan kicks in a lot and it is noisy.
Oddly enough I have a couple of antron 70A supply's that get shut down by a Texas star DX500 running in SSB at about 550w Pep. they can not handle it. Yet 2 of the 70A supplies paralleled have no problem keeping up with a dx1600 at up to 1300Wpep from time to time.
I don't know me...
An RM 50 switch mode supply suffers the same fate with the DX500.
Voltage 14.2
I now use 74A switchmode supplies with the DX500. No problems so far but the fan kicks in a lot and it is noisy.
A few things you should realize and get used to, doesn't matter if you like it or not.
You can't measure DC power by using Pep unit measures, there are no 'peaks' with DC, so stick with that old 'Avg' measuring units of watts.
Determining RF/AC output power from DC input supplied is never going to be exact, you have to account for efficiency of a number of devices between start and finish. Something on the order of about 50% efficiency is about average.
All power supplies have two ratings ICS and CCS, that deal with how current can be supplied. ICS = surges, short periods of time. CCS = constant current supply, now and forever amen. Use the CCS rating to determine 'how big' a power supply is. The ICS rating is the one you see commonly advertised for power supplies, it's larger than the CCS rating so 'sells' more power supplies.
So how do you do the figuring then?
1. Convert inapplicable measuring units to something useful (Pep->AVG).
2. Measure the output power, voltage of the supply under load, and the amount of current being supplied to the transmitter.
... But I can't measure all those things! Okay, lets do it on paper.
1. Same as before, convert Pep to Avg.
2. Using the nominal figures supplied for your equipment, divide the output power in AVG watts by the voltage of the power supply. That gives you current being drawn.
3. Like it or not, that pig of a transmitter/amplifier/whatever is only about 50% efficient so multiply that current by 2. That's what will be required to get the output you used to start with.
Both ways of determining this, by measuring it, or by figuring it, will come out very close to each other. Try it, see for your self.
All of this assumes you have the equipment to do the measuring and figuring and know how to use it. I think that's a fair assumption, don't you?
- 'Doc
You can't measure DC power by using Pep unit measures, there are no 'peaks' with DC, so stick with that old 'Avg' measuring units of watts.
Determining RF/AC output power from DC input supplied is never going to be exact, you have to account for efficiency of a number of devices between start and finish. Something on the order of about 50% efficiency is about average.
All power supplies have two ratings ICS and CCS, that deal with how current can be supplied. ICS = surges, short periods of time. CCS = constant current supply, now and forever amen. Use the CCS rating to determine 'how big' a power supply is. The ICS rating is the one you see commonly advertised for power supplies, it's larger than the CCS rating so 'sells' more power supplies.
So how do you do the figuring then?
1. Convert inapplicable measuring units to something useful (Pep->AVG).
2. Measure the output power, voltage of the supply under load, and the amount of current being supplied to the transmitter.
... But I can't measure all those things! Okay, lets do it on paper.
1. Same as before, convert Pep to Avg.
2. Using the nominal figures supplied for your equipment, divide the output power in AVG watts by the voltage of the power supply. That gives you current being drawn.
3. Like it or not, that pig of a transmitter/amplifier/whatever is only about 50% efficient so multiply that current by 2. That's what will be required to get the output you used to start with.
Both ways of determining this, by measuring it, or by figuring it, will come out very close to each other. Try it, see for your self.
All of this assumes you have the equipment to do the measuring and figuring and know how to use it. I think that's a fair assumption, don't you?
- 'Doc
To give you a set of real world numbers:
The radio I have, a Kenwood TS-950SDX, is rated for 150 watts PEP. That means that at any given moment, the transmitter can not exceed 150 watts of output. I can hit 150 watts briefly during a voice peak while transmitting, or I can turn the mic gain down and adjust the carrier to 150 watts. Either way, I can't go over that limit (the ALC circuit will ensure this).
The finals in this rig are MRF150s, and are being run at 48.4 volts (according to the service manual). If you select the collector* current meter (Ic) on the front panel, the radio will show you how much current the finals draw when the transmitter is engaged. On a voice peak or maximum carrier power, the Ic reading that I see is about 6.5 amps.
6.5 amps times 48.4 volts is 314.6 watts of DC input power.
So it takes 314.6 watts of DC input to get 150 watts of RF output.
150 divided by 314.6 yields .4770, or about 47.8% efficiency. The datasheet for the MRF150 claims 45% efficiency at 30Mhz so my calculations may be slightly off, but we're definitely in the right ballpark.
What happens to the remaining 164.6 watts of input power? It's dissipated as heat.
For a 50 amp 13.8 volt supply, your maximum DC input power would be 690 watts. If we generously assume an amplifier that's 50% efficient, your maximum RF output power would be about half that, or 345 watts. So to comfortably achieve 750 watts of output, I think you'd need a little over 100 amps of current.
The Toshiba 2SC2879 datasheet claims 35% collector efficiency. And since all the multi-pill amps are made using several push/pull pairs of transistors, there may be additional losses because of all the splitters/combiners needed to tie everything together. So assuming 50% efficiency may be ok as a quick rule of thumb, but in practice it may be overly optimistic.
Now, I also think it's overly optimistic to assume you can get even 750 watts from four 2SC2879s in the first place. (Yeah yeah, I know the old refrain: "Never mind what the manufacturer's documentation says, Bill: you can push a 2879 up to 250 watts! I can swing up to 400 watts on my 2-pill! I've seen it on a Bird! The 'Davemade' label alone gives you 100 extra watts!" Etc, etc..) In my humble opinion, if this "750 watt" amp runs on a 50 amp power supply without blowing any fuses, then I think it's more likely that it's really producing around 400 watts of output instead. And that's with saturating the piss out of things.
But look at it this way: motorcycles are probably better off without doors anyway.
-Bill
* "Collector current" is a misnomer here since the MRF150 is a FET, and FETs have gate/source/drain terminals rather that base/collector/emitter terminals. I think it should be drain current. I suspect Kenwood re-used the same florescent tube display unit from the TS-950S/SD which had BJT finals instead of FETs, so the meter label still says Ic intead of Id.
The radio I have, a Kenwood TS-950SDX, is rated for 150 watts PEP. That means that at any given moment, the transmitter can not exceed 150 watts of output. I can hit 150 watts briefly during a voice peak while transmitting, or I can turn the mic gain down and adjust the carrier to 150 watts. Either way, I can't go over that limit (the ALC circuit will ensure this).
The finals in this rig are MRF150s, and are being run at 48.4 volts (according to the service manual). If you select the collector* current meter (Ic) on the front panel, the radio will show you how much current the finals draw when the transmitter is engaged. On a voice peak or maximum carrier power, the Ic reading that I see is about 6.5 amps.
6.5 amps times 48.4 volts is 314.6 watts of DC input power.
So it takes 314.6 watts of DC input to get 150 watts of RF output.
150 divided by 314.6 yields .4770, or about 47.8% efficiency. The datasheet for the MRF150 claims 45% efficiency at 30Mhz so my calculations may be slightly off, but we're definitely in the right ballpark.
What happens to the remaining 164.6 watts of input power? It's dissipated as heat.
For a 50 amp 13.8 volt supply, your maximum DC input power would be 690 watts. If we generously assume an amplifier that's 50% efficient, your maximum RF output power would be about half that, or 345 watts. So to comfortably achieve 750 watts of output, I think you'd need a little over 100 amps of current.
The Toshiba 2SC2879 datasheet claims 35% collector efficiency. And since all the multi-pill amps are made using several push/pull pairs of transistors, there may be additional losses because of all the splitters/combiners needed to tie everything together. So assuming 50% efficiency may be ok as a quick rule of thumb, but in practice it may be overly optimistic.
Now, I also think it's overly optimistic to assume you can get even 750 watts from four 2SC2879s in the first place. (Yeah yeah, I know the old refrain: "Never mind what the manufacturer's documentation says, Bill: you can push a 2879 up to 250 watts! I can swing up to 400 watts on my 2-pill! I've seen it on a Bird! The 'Davemade' label alone gives you 100 extra watts!" Etc, etc..) In my humble opinion, if this "750 watt" amp runs on a 50 amp power supply without blowing any fuses, then I think it's more likely that it's really producing around 400 watts of output instead. And that's with saturating the piss out of things.
But look at it this way: motorcycles are probably better off without doors anyway.
-Bill
* "Collector current" is a misnomer here since the MRF150 is a FET, and FETs have gate/source/drain terminals rather that base/collector/emitter terminals. I think it should be drain current. I suspect Kenwood re-used the same florescent tube display unit from the TS-950S/SD which had BJT finals instead of FETs, so the meter label still says Ic intead of Id.
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One big factor in all this is the way the ratings are stated in the first place. 350-375watts is quite possible however for some reason people like to think that peak power is twice the rated carrier power. look at most of the CB amps out there. look at the rated power stated. They pretty much all have the peak power rated for twice the carrier power. It may be labeled average and peak but that is what they mean....dead carrier and modulated peak. It has pretty much always been like that and I have never been able to find a sound technical reason for this.
We may be misunderstanding those efficiency ratings on the transistors a little here. The transistor has much less control over the operating efficiency than the circuit around that transistor does and the part with the lower minimum efficiency rating is usually the more powerful part. That minimum efficiency rating is the lowest efficiency the transistor could be operated at while still maintaining full rated power and without exceeding maximum dissipation.
This spec deals with how much headroom the part has in the area of dissipation. It does this by demonstrating a worst case scenario of the part operating at full "rated" power output and maximum dissipation simultaneously. It is rare for a linear RF amplifier to operate below 50% efficiency and most class C junk amps can reach close to 70%.
It mostly comes down to the bias and how long the conduction angle is held on during the sine wave. You could take the same transistor in class E with high level modulation and run the part well over 90% efficiency with no distortion. The circuit controls efficiency, not the active device (tube or transistor).
Side note: Some high end HF rigs are operating the final stages below 50% efficiency. One rig even went so far as to run class A at 1/3 normal output because the finals were working at around 25% efficiency. This practice is done to produce the cleanest output, knowing it is likely that it may be amplified further and fed into a gain antenna. As the power output goes beyond a couple hundred watts, this method becomes less practical due to heat.
This spec deals with how much headroom the part has in the area of dissipation. It does this by demonstrating a worst case scenario of the part operating at full "rated" power output and maximum dissipation simultaneously. It is rare for a linear RF amplifier to operate below 50% efficiency and most class C junk amps can reach close to 70%.
It mostly comes down to the bias and how long the conduction angle is held on during the sine wave. You could take the same transistor in class E with high level modulation and run the part well over 90% efficiency with no distortion. The circuit controls efficiency, not the active device (tube or transistor).
Side note: Some high end HF rigs are operating the final stages below 50% efficiency. One rig even went so far as to run class A at 1/3 normal output because the finals were working at around 25% efficiency. This practice is done to produce the cleanest output, knowing it is likely that it may be amplified further and fed into a gain antenna. As the power output goes beyond a couple hundred watts, this method becomes less practical due to heat.
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how can 4-2sc2879's do 750 watts?
i have a 30 year old maco wattmeter that says my m400 does 850 watts
.....santa claus is coming to town.....
i think you'll find the highly technical term "bullshit" covers why cb amps claim twice ssb output, its the same highly technical rule that covers the A99 9.9dbi gain and certain cb antenna makers audio gain claims amongst many other cb myths.
many manufacturers are only too quick to take advantage of the majority of radio users who have not so good technical knowledge. slowly but surely they are being exposed as rip off merchants, some of these myths are over 4 decades old and have propagated far further than any known radio signal has, it won't disappear overnight,sadly.
some companies have seen the light though, wasn't that long back sirio were claiming outrageous gain figures on their website and their catalogues, nowadays its a tad more honest, maybe due to their increasing involvement in the amateur radio market where pulling technical bullshit isn't as easy but by no means rare, who knows? probably Shockwave, as he deals with them, but even Donald will admit they have changed their advertising/hype policy noticeably in the last few years.
it only takes one major company to realise its better to be honest and hopefully more will follow suit.