Amplitude modulation and amplifier efficiency

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Discussion Overview

The discussion revolves around the calculation of DC input power required for amplitude modulation (AM) using an 80% efficient amplifier combined with a 12kW carrier signal. Participants explore the implications of modulation depth on power requirements and share their solutions to related problems from reference materials.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • Some participants propose that for 100% modulation, the power of the modulating signal must equal half of the carrier power, leading to a calculated output power of 6kW from the amplifier.
  • Participants calculate the DC input power using the formula for efficiency, suggesting that the input power required is 7.5kW.
  • Others share similar problems from reference books, noting discrepancies in the definitions of output power (Po) and carrier power (Pc) across different examples, leading to confusion about the calculations.
  • One participant suggests that the additional power required for modulation should be supplied by the modulator rather than the power amplifier's DC supply, indicating a potential distinction in power supply requirements during modulation.
  • Several participants express uncertainty about the correct interpretation of terms and calculations, asking for further references to clarify the concepts involved.

Areas of Agreement / Disagreement

Participants generally agree on the method for calculating power requirements for 100% modulation but express differing interpretations of the terms and conditions in the reference problems. The discussion remains unresolved regarding the definitions and implications of output and carrier power in various contexts.

Contextual Notes

Limitations include unclear definitions of output power (Po) and carrier power (Pc) in the reference materials, as well as potential misunderstandings about the role of the modulator versus the power amplifier in supplying additional power during modulation.

iampaul
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" A intelligence signal is amplified by a 80% efficient amplifier before being combined with a 12kW carrier to generate the AM signal. if it is desired to operate at 100% modulation, what is the DC input power at the modulator" The choices are a: 12kW b. 7.14kW c.7.5kW d. 15kW

my solution:

For 100% modulation the modulating signal power must be one half of the carrier power.
Pm=0.5Pc=6kW
Pm is the output of the 80% efficient amplifier.
efficiency=Pout/Pin=Pm/Pin
Pin=Pm/efficiency=6kW/0.8=7.5kW
Is this right?

How do I solve problems like this? Sometimes i do the same procedure for a similar question but with different values, but I get a wrong answer. Can you suggest a good reference. I am currently using the electronics communications books by frenzel and blake
 
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iampaul said:
" A intelligence signal is amplified by a 80% efficient amplifier before being combined with a 12kW carrier to generate the AM signal. if it is desired to operate at 100% modulation, what is the DC input power at the modulator" The choices are a: 12kW b. 7.14kW c.7.5kW d. 15kW

my solution:

For 100% modulation the modulating signal power must be one half of the carrier power.
Pm=0.5Pc=6kW
Pm is the output of the 80% efficient amplifier.
efficiency=Pout/Pin=Pm/Pin
Pin=Pm/efficiency=6kW/0.8=7.5kW
Is this right?

How do I solve problems like this? Sometimes i do the same procedure for a similar question but with different values, but I get a wrong answer. Can you suggest a good reference. I am currently using the electronics communications books by frenzel and blake
This seems to be the correct answer for a sinusoidal modulating signal.
 
These are sample problems from Blake's electronic comms book:
1.) A transmitter has a carrier power output of 10W at an efficiency of 70%. How much power must be supplied by the modulating amplifier for 100% modulation
Book Solution: efficiency=Po/Ps (I don't know what Po is exactly, but Ps was defined as the DC input power to the final amplifier)
Ps=Po/efficiency =10W/0.7 = 14.3W
Pa=0.5Ps=7.125W (Pa is the power required from the modulating amplifier)

2.)A collector modulated class C amplifier has a carrier output power of 100W and an efficiency of 70%. Calculate the supply power and the transistor power dissipation of 100% modulation.
Book Solution: For 100% modulation.
Po=1.5Pc=1.5(100W)=150W
Ps=P0/efficiency=150W/0.7=214W
Pdissipated=Ps-Po=64W From Problem 1 it seems that Po is the carrier output power. But in problem 2, Pc is the carrier output power? I really don't get. Can you recommend other references regarding this topic

`
 
iampaul said:
These are sample problems from Blake's electronic comms book:
1.) A transmitter has a carrier power output of 10W at an efficiency of 70%. How much power must be supplied by the modulating amplifier for 100% modulation
Book Solution: efficiency=Po/Ps (I don't know what Po is exactly, but Ps was defined as the DC input power to the final amplifier)
Ps=Po/efficiency =10W/0.7 = 14.3W
Pa=0.5Ps=7.125W (Pa is the power required from the modulating amplifier)

2.)A collector modulated class C amplifier has a carrier output power of 100W and an efficiency of 70%. Calculate the supply power and the transistor power dissipation of 100% modulation.
Book Solution: For 100% modulation.
Po=1.5Pc=1.5(100W)=150W
Ps=P0/efficiency=150W/0.7=214W
Pdissipated=Ps-Po=64WFrom Problem 1 it seems that Po is the carrier output power. But in problem 2, Pc is the carrier output power? I really don't get. Can you recommend other references regarding this topic

`
In problem 2, I think they are asking for the power supply requirements of just the PA. Its power supply furnishes 100W/0.7 = 143W under no modulation. With 100% modulation, the additional power of 71.5W required by the PA is supplied by the modulator rather than the PA DC supply.
I am sorry I cannot recommend books because my own are now 70 years old and not easily available!
 

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