[AMRadio] Side band power withincreasedcarrier/was clippertononAM

Thu Apr 14 16:29:34 EDT 2011

What's wrong with just looking at the AM signal with a calibrated scope .

The peak voltage (not peak to peak) voltage gives you all you need assuming 
you know the load resistance.

Bob Macklin
K5MYJ
Seattle, Wa.
"Real Radios Glow In The Dark"
----- Original Message ----- 
From: "Gary Schafer" <garyschafer at comcast.net>
To: "'Discussion of AM Radio in the Amateur Service'" 
<amradio at mailman.qth.net>
Sent: Thursday, April 14, 2011 10:31 AM
Subject: Re: [AMRadio] Side band power withincreasedcarrier/was 
clippertononAM

>
>
> Ok Jeff, let's first define how you find PEP.
> You sum all of the voltages present in the signal, square that and divide 
> by
> the feed line impedance.
> Since there are 3 RF signals present in an AM signal, 2 side bands and a
> carrier, we find out what the voltage is on each and add them together.
>
> With a 25 watt carrier the carrier voltage across 50 ohms will be 35.36
> volts. Where voltage = the square root of (power X impedance). Ohms law.
>
> Since the power in each side band will be 6 db down (1/4) from the carrier
> at 100% modulation that leaves 6.25 watts of RF in each side band. The
> voltage on each side band will be 17.68 volts across 50 ohms.
>
> To find PEP we first add all those voltages up. 17.68 + 17.68 +35.36 = 
> 70.7
> volts. Now we square the total voltage 70.7 x 70.7 = 4998. divide by the
> feed line resistance of 50 ohms and that = 100 watts PEP
> Note that the sum of the voltage in both side bands is equal the carrier
> voltage in this case. Classic 100% modulation.
>
> Let's now increase the carrier. I am going to use 50 watts for ease of
> calculation. We know that if we were to 100% modulate that we would get 
> 200
> watts PEP.
> But since we want to limit our PEP to 100 watts as that is all our solid
> state transceiver will handle.
>
> Again we first find the voltage in the 50 watt carrier. It turns out that 
> it
> is 50 volts.
> Now we know that 100 watts PEP would be 70.7 volts total from above so if 
> we
> subtract our 50 volt carrier from 70.7 volts that leaves 20.7 volts total
> for audio side bands.
> Since that will be divided between the two side bands take half the 20.7
> which = 10.35 volts per side band, the maximum that we can run and have 
> the
> signal stay under our 100 watt PEP radio limit.
> 10.35 squared = 107 divide by the 50 ohm feed line and we have 2.14 watts 
> of
> power in each side band for our now 100 watt PEP transmitter!
>
> Someone else may want to figure what percent modulation this is. Too lazy 
> as
> I don't remember how to do it just now.
>
> This transmitter will be  undistorted but under modulated. But we will 
> have
> reached the 100 watt PEP limit. We will still have lots of room on the
> negative peak side.
>
> As I said earlier, you can judge by looking at the scope of how many
> divisions of audio you have. In this case the carrier will be much greater
> than the audio on the scope.
>
> We could crank up the audio to fill out the negative side so that they 
> just
> hit the 100% point but the positive peaks will be severely clipped due to
> the 100 watt limitation of the transmitter.
>
> 73
> Gary  K4FMX
>
>
>> -----Original Message-----
>> From: amradio-bounces at mailman.qth.net [mailto:amradio-
>> bounces at mailman.qth.net] On Behalf Of Geoff Edmonson
>> Sent: Thursday, April 14, 2011 3:37 AM
>> To: Discussion of AM Radio in the Amateur Service
>> Subject: Re: [AMRadio] Side band power with increasedcarrier/was
>> clippertononAM
>>
>> On 04/13/2011 08:22 PM, Gary Schafer wrote:
>> > Further you can look at the number of divisions on the scope that the
>> audio
>> > occupies. The more divisions the greater the audio power on the
>> signal. If
>> > you raise the carrier above the 375 watt level you will not hit the
>> 100%
>> > mark on negative peaks as soon but you will sooner hit the 1500 watt
>> PEP
>> > mark and you will have less total divisions of audio on the scope.
>> This
>> > means less audio power transmitted.
>>
>> What that means is, you don't have enough audio power available to fully
>> modulate the carrier.
>>
>> As the carrier level increases, the need for enough audio voltage to
>> properly modulate the carrier increases exponentially.
>>
>> 25w out of a typical solid state transceiver, modulated to 100 percent
>> with a sine-wave, produces 100w PEP.
>> If you increase the carrier level to 40w you'll only see 60w PEP and
>> it'll be severely distorted, because there's simply not enough
>> dissipation in the device.  Of course, we're talking low-level
>> modulation levels, but amplify that with a linear on the output of a
>> solid state transceiver, and the problem still exists, only on a much
>> amplified scale.  A linear that only produces 1,000w PEP output, is only
>> good for 250w -max- carrier, and -should- be dropped to 150~200w to
>> ensure enough dissipation available for 100% non-distorted, non
>> flat-topped audio.
>
>
> ______________________________________________________________
> Our Main Website: http://www.amfone.net
> AMRadio mailing list
> Searchable Archives: http://www.mail-archive.com/[email protected]/
> List Rules (must read!): http://w5ami.net/amradiofaq.html
> List Home: http://mailman.qth.net/mailman/listinfo/amradio
> Post: AMRadio at mailman.qth.net
> To unsubscribe, send an email to amradio-request at mailman.qth.net with
> the word unsubscribe in the message body.
>
> This list hosted by: http://www.qsl.net
> Please help support this email list: http://www.qsl.net/donate.html
> 