# Design of amplifier with NPNs

Discussion in 'Homework Help' started by mihaaa, Jan 16, 2014.

1. ### mihaaa Thread Starter New Member

Jan 16, 2014
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Hi,

I have a question about creating an amplifier using 3 NPN BJTs and resistors.
The problem is that I need to get a wide bandwidth and high voltage gain.
In addition it is required to get the power supply as low as possible, I believe that means to get a low current.

My solution is to use a cascode amplifier(common emitter to common base) but I'm not sure it will be a good solution.

Michael

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2. ### Veracohr Well-Known Member

Jan 3, 2011
600
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Sounds like a good solution to me. Look up "double cascode", to use three transistors.

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Think about the voltage supply requirements for a cascode.

What do you mean by wide bandwidth?

and high voltage gain, how big is Vin?,
so how big will Vout be?

Other solutions to consider might be a long tail pair and buffer.

4. ### mihaaa Thread Starter New Member

Jan 16, 2014
7
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I have a voltage supply of 5V (picture attached in original message)
Vin = 10mV

My goal is to find the bigger M that I can, when M=(voltage gain)*(freq. of smallest pole)/(power of voltage supply)
So I need high voltage gain, wide bandwidth, and low current over the voltage supply.

Thank you for the help!

5. ### studiot AAC Fanatic!

Nov 9, 2007
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So with that in mind, what is the maximum possible gain, given your supply voltage?

6. ### mihaaa Thread Starter New Member

Jan 16, 2014
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5V/10mV = 500, so the maximum possible gain is 500 I believe

7. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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Are you sure?

Your input was 10mV AC was it not?

8. ### studiot AAC Fanatic!

Nov 9, 2007
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10 mV (sinewave) is 10√2 peak = 14.14 mV
And doubled for peak to peak = 28.28 mV - call it 30mV to allow for error margins.

So your max gain is 5000/30 ≈ 170

Do you understand this?

Next can you see any limits on frequency place by the input circuitry (source R and input capacitor and input resistance)?

9. ### mihaaa Thread Starter New Member

Jan 16, 2014
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As far as I know the capacitor create a pole that limits the bandwidth of the circuit

10. ### studiot AAC Fanatic!

Nov 9, 2007
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You have shown Cin as a series blocking capacitor.

This is necessary, do you know why it is there ?

Rs is in series with Cin and the input resistance of your amplifier.
This forms a potential divider chain across the source 10mV sine wave.

11. ### mihaaa Thread Starter New Member

Jan 16, 2014
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The capacitor is blocking the DC and passing the AC to form a high pass filter

12. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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Yes, but which DC?

I suggest you sketch a couple of possible layouts and see where you are going to apply the input (emitter?, base?) and its implications for your design.

In order to have maximum voltage swing at the output, where does the quiescent point have to be ?

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13. ### mihaaa Thread Starter New Member

Jan 16, 2014
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Veracohr, about the "double cascode", I can't find it.
The regular cascode is Vin connected to the base of C-E, and the collector of C-E connected to emitter of C-B.
How the double cascode connected?
Thank you

14. ### Veracohr Well-Known Member

Jan 3, 2011
600
86
It's just another transistor on top, but studiot is right, with only 5V supply it's not a good solution to get high gain.

15. ### mihaaa Thread Starter New Member

Jan 16, 2014
7
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studiot, can you please help me with the type of transistors I should put on the circuit?
Should cascode work?

Thank you

16. ### studiot AAC Fanatic!

Nov 9, 2007
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This is your design not mine.

The point of the exercise is to find out all the things you need to know to do a design.
I assume you would not have been set this exercise unless you had covered enough theory to accomplish it.

You are not yet at the stage to choose transistors (although I am suprised the transistor type is not specified)

I made these points for a reason.

In particular what is Cin? : I don't mean it's value I mean where does it come from and why is it there?

Where did your original diagram come from, did you draw it?

I say this because Cin is usually in parallel with the input. ( I have already observed that you have shown it in series)
I have shown the usual situation with an amplifier when a Cin is specified at the end of my attachment.

However assuming you have been given Cin in series as shown then look at my attachment.

You have Vs across Rs in series with Cin in series with Rin.

You can calculate the impedence of Cin at various frequencies 1k, 10k, 1M, 10M, 100M 500M hz and compare with Rs.

So at what frequency is Cin small compared with Rs?
Why is it important that it is small compared to Rs?
What happens to Vs if Rin is small compared with Rs?

So what does that mean for Rin, what sort of value are you aiming for?

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