Fixed Gain MOSFET Help!

Thread Starter

kauymatty

Joined Apr 28, 2012
5
First off I'm a pretty amateur electronics student and I've been trying to design a light organ circuit.

So, I needed to design a circuit that would amplify the voltage out of a filter and because of what we were taught this year I was advised to use a MOSFET.

Because my light organ must accept AC music signals I thought I should use a fixed gain MOSFET amplifier, the circuit diagram is attached.

However the book doesn't seem to give must information about how to calculate Vin and Vout i.e. the gain and how the MOSFET characteristics really come into it (MOSFET is 2N7000)

Could someone please explain how to work everything out?


Thankyou in advance.
 

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Jony130

Joined Feb 17, 2009
4,980
What is your input voltage range? And how much gain you want?
What frequency range you want to amplify? And what is the load? Your supply voltage? etc...
 
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Thread Starter

kauymatty

Joined Apr 28, 2012
5
Well input voltage range is probably between 0 - 3V, a gain of 2. The entire circuit will have 4 amps, 1 from the audio input, ranges 30-16kHz, then 3 in total after filters e.g. 1 after a bass filter 90-5000, mid-tone 5000-10000, treble 10000-16000.

I want to use a supply voltage of 15V-0V, the output of each of the filter amps will go to probably 2 LEDs, for now, parallel, and a 330ohm resistor
 

Jony130

Joined Feb 17, 2009
4,980
First we need to use data sheet to find K - factor

http://www.onsemi.com/pub_link/Collateral/2N7000-D.PDF



Form figure 1

K = 0.39A/ ( 5V - 1.9V)^2 = 0.04058 = 40.58m

So we need to find Rd and Rs

Id = (Vcc - Vgs) / (Rd + Rs)

\(Vgs=Vt+\sqr{\frac{Id }{K}}\)

so we get

\(Rd + Rs = \frac{(Vcc - (Vt+\sqr{\frac{Id }{K}}) )}{Id} \)

For Id = 5mA

Rd + Rs = 2549.8 ≈ 2.55K


And to get the gain equal to 2

Av = Rd / ( 1/gm + Rs)

gm = 2 * √(Id * K) = 28.5mS

So in our case

Av ≈ Rd/Rs = 2

and Rd+Rs = 2.55K

So we solve for Rd and Rs

Rd = 1.7KΩ ; Rs = 850Ω

But to improve freedom in choosing the operating point it is better to use voltage divider



To archive high output voltage swing Vds must be equal to 0.5Vcc = 7.5V

So we choose Rd = 1K and Rs= 470Ω

Au = 1K/0.47K = 2.1V/V

Now we chose Id current

Vds = 7.5V

Id = ( 15V - 7.5V) / ( 1.47K) = 5.1mA

Now we select R1 and R2

Voltage at the gate is equal to

Vg = Id * Rs + Vgs = 5.1mA * 470Ω + Vgs

\(Vgs=Vt+\sqr{\frac{Id }{K}}=1.9V +\sqr{\frac{5.1mA }{40.50m}} = 2.25V\)

so the gate voltage

Vg = Id * Rs + Vgs = 5.1mA * 470Ω + 2.25 = 4.647V

R2 = Vg/Idz

R1 = ( Vcc - Vg) /Idz

Idz = We can choose any value but we need too remember that R2 and R1 determine the input resistance of our amplifier.

But I can also choose R1 = 470K for example

And then find R2

the new Idz current

Idz = (15V - 4.647V)/ 470k ≈ 22μA

R2 = 4.647V/22uA ≈ 211K ≈ 200k

But in reality you will be forced to pick R2 on the bench.
In order to get Vds voltages equal to 7.5V

As for capacitor use this equation

C = 0.16/(F * R)

R - resistance seen from the capacitor terminals
F - the frequencies you wanted to pass

If you wanted to pass frequencies higher than 30Hz then

C1 > 0.16/( 30Hz * R1||R2) = 39nF
 

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Jony130

Joined Feb 17, 2009
4,980
From the data sheet

http://www.onsemi.com/pub_link/Collateral/2N7000-D.PDF

To find K parameter we need Vt = Vgs(th)
So I pick average Vt value

Vt_max = 3V and Vt_min = 0.8V

Vt_ave = (3V + 0.8V)/2 = 3.8V/2 = 1.9V

I also need Id current for a given Vgs voltage
So I use figure 1 or figure 2 to read the Id for a given Vgs.
I pick randomly Vgs = 5V. And I use figure 1 to find Id current for Vgs = 5V

In real life we a force to forced to measure Vt and Id to find K factor.
Because MOSFET show great process spreader. For example on Vgs(th) = Vt = 0.8V...3V So only real measurement can give you exact K value.

To find K factor for the MOSFET in the lab bench all you need is this simply circuit:



The symbol on the top represents the constant current source.
And for this circuit Vds = Vgs
So now we need to find Vgs for given Id current.
And the only thing that we need to change in this circuit is to increase the value of a constant current source.
For example for Id = 0.1mA we read Vgs = Vt
And for Id = 10mA we read Vgs in linear region.
 

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Thread Starter

kauymatty

Joined Apr 28, 2012
5
Well thankyou for all of your help, this all makes a lot more sense now. Much better than the book I have. Thanks once again.
 

Jony130

Joined Feb 17, 2009
4,980
In reality we don't calculated K factor.
We simply use a voltage divider plus a potentiometer to select the right bias point.
See the example diagram
The voltage gain is set by

Av = ( R3 + R4 + R5)/ R6 = 45/15 = 3V/V


And we use pot PR to set drain voltage equal to 7.5V (0.5Vcc)

And when we pick Rd resistor we use this rule of thumb (but not always).

Rd = 0.1* Rload

and

Id = 0.5Vcc/Rd

Rs = Rd/Av
 

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