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#### benji_styler

##### New Member
Hi im really struggling with some electronics theory questions which i have to submit 2mrw to complete my work.

Any chance any of you guys could help me out with them? Any help/direction would be much appriciated!!!

Here goes:

1.) An op amp has a gain bandwidth product of 3MHz. What is the maximum gain that can be obtained for an audio amplifier with upper cut-off frequency = 20khz?

2.) The transmitter of a TV broadcast repeater draws a current of 53A from a 450V supply. The output power to the transmitting aeriel is 10 kw. what is the transmitter efficiency?

3.) An audio amplifier delivers 300w maximum continous output power. it draws 2A from a 240V mains supply. How much heat needs to be displaced when the amplifier is running at full power?

4.) A portable radio transmitter has a 50 AH (e.g. produces 50A for 1 hour) 12V battery for the power supply. The transmitter output power into the aeriel is 100W. Its efficiency is 30%. the set-up is required to transmit continously at full power for 90 mins from a fully charged battery. can it be done?

5.) A manufacturer wishes to choose whether to use class A or class B for a new audio power amplifier design with a 100W outpu. The new design has to fit a casing which has sufficient ventilation to be able to dissipate up to 60W of heat continuously. What would you advise and why?

#### audioguru

##### Well-Known Member
To answer those simple questions, you need to know a little about electronics and use a little arithmatic.

#### benji_styler

##### New Member
Could you push me in the direction of the how to do these questions? ie formula need to answe them? i think i can do the arithmatic.

am i up a creek without a paddle? lol

#### dknguyen

##### Well-Known Member
Don't ask people to do your homework for you. These are really basic questions. If you googled you would find the answers. Have you even tried to read your textbook? Have you even read the questions? If you don't know what efficiency, power, current and voltage are, it's pretty hopeless. Sit down and think about what efficiency and power mean. And then think about what voltage and current have to do with power.

Read the word "Gain Bandwidth Product". The word itself is describing to you EXACTLY WHAT IT IS, as is every other definition in the questions.

Perhaps you should actually read the question and think about it? Use common sense. These questions are intuitive enough that if you just guessed what the equations might be, you would probably get it right. If we give you the formula, we basically did the problem for you- that's how simple it is.

Put some effort into your work. If you can figure out F=ma means anything to you, you can solve these problems.

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#### audioguru

##### Well-Known Member
#1- Look at the Gain vs Frequency curve on the datasheet of an opamp that has a gain bandwidth product of 3MHz like the TL072. At 3MHz its gain = 1. The gain increases 10 times for each 1/10th of the frequency. At 300kHz the gain is 10. At 30kHz the gain is 100. At 3kHz the gain is 1000. At 20kHz as in the question, the gain is a little more than 100 but much less than 1000, maybe 200.

#2- The transmitter draws 53A from 450V. What is the power it uses?
It transmits 10kw so its efficiency is the 10kW divided by how many kW it uses.

#3- The amp delivers 300W but uses 2A times 240V= 480W. The difference in the Watts is wasted as heat.

#4- The transmitter output is 100W. Since it is 30% efficient then the power from the battery is 333W. The current from the 12V battery is 333/12= 27.75A so a 50A/hr battery might power it for 50/27.75= 1.8hrs. A little longer than the required 1.5hrs.

#5- Do you know the difference between class-A and class-B? Class-A wastes as much power in heat as is the amp's output. So a 100W class-A amp will heat with 100W which is too high for the case ventillation max of only 60W. The class-B circuit will also deliver 100W but will heat with only about 50W and will be fine.

Where were you when that simple stuff was taught?

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#### dknguyen

##### Well-Known Member
Really, the only excuse is if this is the first time you have ever seen a power or efficiency equation...which really isn't much of an excuse seeing as how you are getting into Gain Bandwidth Products.

#### benji_styler

##### New Member
Hi Thanks for your response. i know it seem like i dont have a clue... I had a major car accident in january in which my head and face wer pretty badly injured. Instead of repeating a year of my course i have tried my best to catch up, however my learning for electronics has seemed to suffer. My memory is a bit patchy due to fracturing my skull and i suffer short temr memory loss. but just trying to get back on with my life

The work im recovering now was taught before my accident.

i really appriciate your help and im sorry if i appear clueless.

for question 2 i worked out the efficiency to be 117% is this correct?

I have a few more questions to complete. i dont mean to be cheeky but would you look at them and help to explain them as you have the other questions. Your well constructed feedback has given me a good understanding of the questions.

Cheers,

ben.

#### audioguru

##### Well-Known Member
Efficiency cannot be more than 100%. Swap your numbers in the division.
Good luck with healing and catching up with electronics therory.
It is late now here in Canada, I'll have a look before bedtime.

#### benji_styler

##### New Member
Heres one more question that i am trying to attempt:

should gain with feedback be used: A/ 1+AB

When a certain amplifier has negative feedback fraction of 0.002 applied, its closed loop gain is found to be 150. What negative feedback fraction should be applied to make it closed-loop gain 100?

thanks for taking the time,

regards.

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#### dknguyen

##### Well-Known Member
G = A/(1+AB) is the equation for gain w/ feedback.

A is the open loop gain at DC/very low frequency (that's why there is no frequency variable. It has been set to and stuff has dissapeared).

B is the feedback gain.

The intial information give you allows you to find A, the open loop gain of the circuit. Once you have A, the amplifier's DC open loop gain (ie. no feedback), you can adjust B until you get the new gain.

It' just throwing around numbers- using info from case 1 to find an unknown required to solve for case 2.

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