Fixed Gain MOSFET Help!

Discussion in 'Homework Help' started by kauymatty, Apr 28, 2012.

  1. kauymatty

    Thread Starter New Member

    Apr 28, 2012
    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.
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    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...
    Last edited: Apr 28, 2012
  3. kauymatty

    Thread Starter New Member

    Apr 28, 2012
    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
  4. Jony130

    AAC Fanatic!

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


    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
    kauymatty likes this.
  5. kauymatty

    Thread Starter New Member

    Apr 28, 2012
    Wow, a lot more than I thought it would be. Thanks a lot for all your help! Hopefully my final project will work now :)
  6. kauymatty

    Thread Starter New Member

    Apr 28, 2012
    Just quickly, where did you get 5V from in the figure 1 calculation and what is Av?
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    From the data sheet

    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.
    kauymatty likes this.
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Oh I forgot about Av

    Av - is a voltage gain

    Av = Vout/Vin = Rd / ( 1/gm + Rs)
  9. kauymatty

    Thread Starter New Member

    Apr 28, 2012
    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.
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
    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


    Id = 0.5Vcc/Rd

    Rs = Rd/Av