AM modulation and instantaneous RF power - Page 1
Forum section: electronics / rf-microwave

author: Circlotron
author_rank: Super Contributor
timestamp: May 17, 2018, 06:02:41 am
content: Been reading about old school AM modulators, specifically Radio and Hobbies May 1952 page 56. A little 10 watt setup is described for construction. It makes the statement that the modulation transformer secondary delivers power to the class C final stage that behaves exactly like a resistive load to the modulator. Furthermore, the modulator needs to have at least 50% of the RF power rating so that 100% modulation can be achieved. 10 watts audio will fully modulate 20 watts dc feeding the final stage.Here’s the question - if the final stage appears as a resistive load to the modulator then on 100% positive modulation peaks the instantaneous power to the final stage would reach 80 watts, so how could 10 watts of audio fully modulate 20 watts of RF? The modulated dc feed to the final stage has to swing all the way down to zero so I would expect it would also swing up to double voltage and so quadruple the instantaneous power input.

author: Circlotron
author_rank: Super Contributor
timestamp: May 17, 2018, 06:14:02 am
content: Furthermore, if the dc feed to the final stage swings all the way down to zero volts then it’s power drain will be zero watts at that instant. How is this demanding power from the modulator? That is to say, why does there need to be two modulator tubes in push-pull? If the tube that drives the negative modulation side of things starts to conduct, where is the power demand from it seeing it is reducing the instantaneous dc power input to the final?

author: Circlotron
author_rank: Super Contributor
timestamp: May 21, 2018, 09:57:45 am
content: Bring it to the top for a new set of eyes.

author: PA0PBZ
author_rank: Super Contributor
timestamp: May 21, 2018, 10:21:38 am
content: I think the 'secret' is that the modulator modulates the (constant) DC power supply to the PA, so it modulates (swings) the DC voltage up and down. So for bringing the power up from 50 to 100% you need 50% of the total power, and equally to bring it down from 50% to 0% you need 50% of the power. (Talking about it in a popular non-scientific way)For others here's the article:http://www.americanradiohistory.com/AUSTRALIA/Archive-Radio-and-Hobbies/50s/Radio-and-Hobbies-1952-05.pdf

author: Circlotron
author_rank: Super Contributor
timestamp: May 21, 2018, 11:12:35 am
content: Quote from: PA0PBZ on May 21, 2018, 10:21:38 amThat doesn't sound right. If the no-audio DC input to the final is 50 watts, to get 100 watts input you need another 50 watts. That would mean the audio equals 100% of the DC power, not 50%.to quote the article,"In order to modulate the carrier wave 100 pc, the modulator must be capable of cutting the high tension supply voltage completely on the negative voice peaks and doubling the supply voltage on positive peaks."If the modulator doubles the high tension supply on positive peaks, would this not quadruple the RF output on those positive peaks?Further quote:"For 100 pc modulation, the modulator must deliver audio power equal to half the DC power input to the class C amplifier"That being the case, how would "audio power equal to half the DC power" be enough to quadruple the RF output???

author: PA0PBZ
author_rank: Super Contributor
timestamp: May 21, 2018, 11:31:06 am
content: Quote from: Circlotron on May 21, 2018, 11:12:35 amI think they mean 50% of the total/max DC input (so when the output is at 100%).That way your other doubts also make sense.