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Microphone Amplifier Design

  • Thread starter jegues
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  • #1
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Homework Statement



We are attempting to design a amplifier for a microphone. The end goal is to amplify what is being said through the microphone specified below through a set of speakers or headphones.

See figure attached for the design parameters as well as the parts list we are given.

Homework Equations





The Attempt at a Solution



Note: The output current is limited to a maximum of 20mA.

I'm a little flustered with all the requirements of this design in addition to the limited parts list.
I'm having trouble figuring out how I should start my design or even what direction to move in.

What kind of stages I should be considering with the components I'm provided? Should I focus on one aspect of the design at first (i.e. gain requirement) and sort out the others later on? If so, which design aspect is crucial to achieve before attempting to establish other design parameters?

Can someone give me a push in the right direction so I can get this design started?

I've attached the datasheet specific to the microphone we are provided.

I will be posting whatever design/results/thoughts I come up with for more feedback/improvement.

Thanks again!

EDIT: I went and asked my professor for a good point to start and he had recommened that I start by considering the specific input and outputs we're given, and make an attempt at designing the amplification stage of the amplifier.

He mentioned to me that I need to achieve the required gain as well as the bandwidth, and mentioned that if I could not achieve both in one stage that I should simply use 2 amplification stages to achieve the required bandwidth and gain.
 

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Answers and Replies

  • #2
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Microphones only output a few micro-volts so you will need to amplify this to the desired voltage.

Also, since there is a bandwidth requirement you will need to have a bandpass filter (could be as simple as an RC circuit).

You can accomplish both of these by using an Active Band Pass Filter.

Not sure if this is what your looking for but I hope this helps!
 
  • #3
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Here's one design I tried. (See figure attached)

Note: I assumed 100mV peak input at 1kHz, I'm not sure whether this is actually what my microphone will provide.

Using the values from the scope we can quickly calculate the voltage gain,

[tex]A_{v}|_{dB} = 20log\left( \frac{13.768mV}{999.337uV} \right) = 22.783178dB[/tex]

Which is fairly close to the desired 23dB gain.

Is this indeed correct or am I making a mistake somewhere?

EDIT: I tried my amplifier again attaching the nominal load of 60 ohms and the output and my gain gets completely destroyed. (See 2nd figure attached)

Is there a way to fix this? Or did I do something wrong?
 

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  • #4
NascentOxygen
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The output impedance of the circuit you built is around 10 kohms. Connecting a 60 ohm load will cause considerable loading problems, as you discovered. Your 100 nF coupling capacitor is probably way too small. Consider it a RC high-pass filter. If you are to drive a 60 ohm load, then you'll need to add at least one more stage to transform the impedance. An emitter follower would do it; it has a voltage gain of 1.
 
  • #5
AlephZero
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EDIT: I tried my amplifier again attaching the nominal load of 60 ohms and the output and my gain gets completely destroyed. (See 2nd figure attached)

Is there a way to fix this? Or did I do something wrong?
Did you design the output impedance of your amp? There's a reason your specification says "60R maximum".

Usually, a good way to design a circuit like this is start at the output and work backwards. Most likely, you won't be able to get both hgih gain and low output impedance from a single stage, using the components on your parts list. When you have desgined the output stage to meet the output requirements of the speciifcation, then you know what the preceding stage(s) need to do. You also know the INPUT impedance of your output stage, and you need that to design the stage before it (because, as you just found out, the output impedance that an amplifier is driving affects its gain).
 
  • #6
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The output impedance of the circuit you built is around 10 kohms. Connecting a 60 ohm load will cause considerable loading problems, as you discovered. Your 100 nF coupling capacitor is probably way too small. Consider it a RC high-pass filter. If you are to drive a 60 ohm load, then you'll need to add at least one more stage to transform the impedance. An emitter follower would do it; it has a voltage gain of 1.
So to correct the problems I'm having I could change the coupling capacitor at the output from 100nF to say 100uF?

In addition to this I will pass my output through another emitter follower (i.e. buffer stage) before attaching my 60 ohm load at the output?

Am I understanding this correctly?

EDIT: If I were to simply place an emitter follower stage inbetween the capacitor and the load I would obtain a Class-A power amplifier, no? Aren't Class-A amplifiers highly inefficient? Is there a way I can achieve the same results but with a Class-AB amplifier?
 
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  • #7
AlephZero
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So to correct the problems I'm having I could change the coupling capacitor at the output from 100nF to say 100uF?
You want the impedance of the coupling capacitor to be "small" compared with other impedances, for the frequency range you are interested in. But that won't necessarily change the output impedance of the amp.

In addition to this I will pass my output through another emitter follower (i.e. buffer stage) before attaching my 60 ohm load at the output?
That's a good idea. For the 3 basic configurations of a single-transistor amplifier stage (common emitter, base, and collector) the emitter follower has the lowest output impedance.

You could try an alternative design with a common emitter output stage with a small collector resistor to get a low output impedance. That will mean voltage gain of the stage is also low, but any voltage gain means the previous stage has to do less work. Try it both ways.

Aren't Class-A amplifiers highly inefficient?
Yes, but...

Is there a way I can achieve the same results but with a Class-AB amplifier?
... your requirements specification doesn't say anything about efficiency. The required output is 0.5V into 60 ohms, which is an output power of about 4 mW. Unless you are trying to design an ultra-low-power circuit, efficiency is not very important here. Of course if the required output power was 4kW not 4mW, a Class A amplifier would not be a good design.

EDIT: Re efficiency: http://en.wikipedia.org/wiki/List_of_battery_sizes gives the minimum capacity of any type of "9V battery" as 120mAh (and most types have bigger capacity than that). So for 8h use, you can afford 120/8 = 15mA total current, which is 15x9 = 130mW total power consumption. So compared with the 4mW required output power, you don't need to worry much about power usage.
 
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  • #8
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You want the impedance of the coupling capacitor to be "small" compared with other impedances, for the frequency range you are interested in. But that won't necessarily change the output impedance of the amp.


That's a good idea. For the 3 basic configurations of a single-transistor amplifier stage (common emitter, base, and collector) the emitter follower has the lowest output impedance.

You could try an alternative design with a common emitter output stage with a small collector resistor to get a low output impedance. That will mean voltage gain of the stage is also low, but any voltage gain means the previous stage has to do less work. Try it both ways.


Yes, but...


... your requirements specification doesn't say anything about efficiency. The required output is 0.5V into 60 ohms, which is an output power of about 4 mW. Unless you are trying to design an ultra-low-power circuit, efficiency is not very important here. Of course if the required output power was 4kW not 4mW, a Class A amplifier would not be a good design.

EDIT: Re efficiency: http://en.wikipedia.org/wiki/List_of_battery_sizes gives the minimum capacity of any type of "9V battery" as 120mAh (and most types have bigger capacity than that). So for 8h use, you can afford 120/8 = 15mA total current, which is 15x9 = 130mW total power consumption. So compared with the 4mW required output power, you don't need to worry much about power usage.
Thank you for the very informative reply.

Before I was able to read your response I had already begun to make some reworkings to my amplifier.

See the figure attached for what I've got so far.

For the most part everything from the input to Q1 remained untouched while I replaced the right half of the circuit with a Class-AB buffer.

I was able to tweak the paramters in the amplifier (namely R5) to obtain the desired gain of 23dB's at the output.

This is all good news, but I think I still have one problem remaining.

For a Class-AB amplifier, shouldn't Q3 and Q4 be symmetric with respect to one another? If this is the case, then I should be aiming for half the supply voltage (4.5V) at the junction of R10 and R11.

How can I achieve such a thing? (I tried varying R6 but my gain would go all out of whack if I did so)

Any other comments/suggestions/improvements I can make to my latest design?

Thanks again!
 

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  • #9
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Hello again everyone.

It seems that as though there are a couple of things I had overlooked in the design of my amplifier.

Firstly, on the second page of the data sheet for the microphone (See figure attached, squared in red) there is a resistor that must be hooked to the battery in order to power the circuit for the microphone. This was not included in the circuit I've been working with thus far.

Secondly I did not account for the output impedance of the microphone. (See figure attached, squared in red)

I decided to add these things into my circuit to see how it was going to effect it's preformance.

Before doing so, I had some confusion about whether or not the output impedance of the microphone is in series, or parallel with the small ac voltage obtain from the microphone circuit. For this reason, I divided things into 2 cases, one where it was in series and one where it was on parallel. (Case 1 = Parallel, Case 2 = Series)

As you can see in Case 1, my circuit holds its gain and other various characteristics and works fine.

In Case 2, my circuit loses it's gain, as the voltage at the input is much larger than before. (The order of mV not uV)

First question that needs to be adressed is which case is indeed correct based on the datasheet from the microphone? Case 1 or Case 2?

The second question relates to the answer of the first question.

If it is Case 1: is this output impeadance going to ruin my input impedance that I've established previously? If so, how do I go about fixing that?

If it is Case 2: How do I go about correcting the issues I'm having with my gain? Is there any effect on my input impedance? And if so, how do I correct that?

Thanks again!
 

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  • #10
NascentOxygen
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A mic impedance of 500 ohms doesn't mean you add a resistor of that value. It means the 500 ohms equivalent is already there, as an apparent series resistance, and so you then design the input impedance of your mic pre-amp to be around 500 ohms also, for best signal to noise performance. What is the small signal impedance at the base of your first transistor? (By shunting the mic with 500 ohms you are just wasting valuable audio signal.) You should design your transistor input impedance to be around 500 ohms so the mic is working into 500 ohms and all the current from the microphone is being fed into the base of the transistor.
 

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