Amplitude modulation and amplifier efficiency

In summary, the book gives a general formula for the power supply requirements of the modulator and the power amplifier, but the specific values may vary depending on the type of amplifier and modulation used. It is important to carefully read the question and understand the given values in order to accurately solve the problem. Other resources such as online tutorials and practice problems can also be helpful in gaining a better understanding of the topic.
  • #1
iampaul
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0
" 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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  • #2
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.
 
  • #3
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

`
 
  • #4
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!
 

Related to Amplitude modulation and amplifier efficiency

1. What is amplitude modulation (AM)?

Amplitude modulation is a method of transmitting information by varying the amplitude of a carrier wave. The amplitude (or height) of the carrier wave is modulated to represent the audio or visual signal being transmitted. This allows for the signal to be received and demodulated at the receiving end to reproduce the original information.

2. How does AM work?

In AM, the audio or visual signal is used to vary the amplitude of the carrier wave. This creates sidebands around the carrier frequency, which contain the information being transmitted. At the receiving end, the carrier wave is removed and the sidebands are demodulated to reproduce the original signal.

3. What is the purpose of an amplifier in AM?

The purpose of an amplifier in AM is to increase the strength of the modulated signal before it is transmitted. This is necessary because the modulated signal is typically weak and needs to be amplified in order to travel long distances and overcome interference.

4. How is amplifier efficiency measured in AM?

Amplifier efficiency in AM is typically measured by the amount of power that is converted from DC power to RF power. This is known as the power efficiency or power gain of the amplifier. The higher the efficiency, the more power is being converted and the less power is being wasted as heat.

5. What factors affect amplifier efficiency in AM?

There are several factors that can affect amplifier efficiency in AM, including the type of amplifier used, the power supply voltage, the input and output impedance, and the modulation depth. Higher efficiency amplifiers are typically more complex and expensive, but they can provide better performance and save energy in the long run.

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