Single stage amplifier with current mirror biasing

In summary: So would this be a common source amplifier with source degeneration connected to ground?In summary, we are trying to solve a problem involving finding the current and output resistance of two MOSFETs, M1 and M2. We must first find the value of I_D for M1, then use that to determine I_D for M2, which will be twice the value of I_D for M1. We are also asked to find the small signal parameters for the MOSFETs, but have encountered an issue with the channel length modulation being equal to 0 for both M1 and M2. We must determine what the output resistance would be if the channel length modulation was a small number but not zero. Additionally,
  • #1
CKaiL
38
1

Homework Statement


123.jpg


Homework Equations



ID= kn (Vgs - Vth)^2

The Attempt at a Solution



I asked to find the current I D , M2. [/B]

I think in order to solve this question I need to find Vgs but how do I do that?

RC1 = 2k ohm
R = VDD - Vgs / I D ,M2

VDD = 5V , and R = 2k ohm and I have 2 unknown variable. Vgs and I D ,M2

I think i need to make use of this ID= kn (Vgs - Vth)^2 to solve this question
 
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  • #2
CKaiL said:

Homework Statement


123.jpg


Homework Equations



ID= kn (Vgs - Vth)^2

The Attempt at a Solution



I asked to find the current I D , M2. [/B]

I think in order to solve this question I need to find Vgs but how do I do that?

RC1 = 2k ohm
R = VDD - Vgs / I D ,M2

VDD = 5V , and R = 2k ohm and I have 2 unknown variable. Vgs and I D ,M2

I think i need to make use of this ID= kn (Vgs - Vth)^2 to solve this question
Good idea! But you must first find ID for M1. Combine your last equation with another expression for ID summing currents at the drain.
Then, you assume M1 and M2 have identical I-V characteristics so knowing IDM1 gives you IDM2, and you're done.
 
  • #3
Could I have more information why the characteristics are identical for M1 and M2? They only have the same value for Vth while their Kn is different by a factor of 2?

I forgot to mention that I D,M1 is also given as 2mA for this question so I D, M2 is also 2mA?
 
  • #4
CKaiL said:
Could I have more information why the characteristics are identical for M1 and M2? They only have the same value for Vth while their Kn is different by a factor of 2?
I didn't see that, sorry.
I forgot to mention that I D,M1 is also given as 2mA for this question so I D, M2 is also 2mA?
No. For any MOSFET in the 'saturation' region with λ = 0, which this is since Vgs > Vth & Vds > Vgs - Vth),
I = k(Vgs - Vth)2.
So, since k2 = 2k1, what is I,M2 if I,M1 = 2 mA?
 
  • #5
Hi

So I find the value for Vgs by subbing in Vth and I D,M1. Once I gotten the value of Vgs , i sub this value into the equation to find i D,M2 with a k2? Am I correct?
 
  • #6
CKaiL said:
Hi

So I find the value for Vgs by subbing in Vth and I D,M1. Once I gotten the value of Vgs , i sub this value into the equation to find i D,M2 with a k2? Am I correct?
No point in solving for Vgs. From my last post it should be obvious that I_M2 = 2I_M1. Solve for I_M1 as I discussed earlier.
 
  • #7
Sinc k2 = 2k1 , I2 = 2I1. I2 = 4mA?
 
Last edited:
  • #8
I have another question for this. I am asked to find the small signal parameter of the 2 MOSFET .
The formula are
Output resistance in the small signal equivalent circuit of a MOSFET operating in the saturation region is given as R0 = 1 /(channel length modulation * Current ID)

I have ID value for both 1D M1 and 1D M2 but in this question , channel length modulation for both are equal to 0
when you sub in these value , you will get an error.
 
  • #9
CKaiL said:
Sinc k2 = 2k1 , I2 = 2I1. I2 = 4mA?
Good guess!
 
  • #10
CKaiL said:
I have another question for this. I am asked to find the small signal parameter of the 2 MOSFET .
The formula are
Output resistance in the small signal equivalent circuit of a MOSFET operating in the saturation region is given as R0 = 1 /(channel length modulation * Current ID)

I have ID value for both 1D M1 and 1D M2 but in this question , channel length modulation for both are equal to 0
when you sub in these value , you will get an error.
Suppose the channel modulation was a very small number but not zero - what would be the output resistance then?
 
  • #11
hmm good guess meaning is ID , M2 = 4mA correct or wrong?

For channel length modulation
In the picture it says that the channel length modulation is equal to 0 for channel length modulation M1 and channel length modulation M2.
 
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  • #12
I would also like to check if this is a configuration of a Common source with source degeneration single stage amlifer?
 
  • #13
CKaiL said:
hmm good guess meaning is ID , M2 = 4mA correct or wrong?
Correct.
For channel length modulation
In the picture it says that the channel length modulation is equal to 0 for channel length modulation M1 and channel length modulation M2.
I know. I asked you what kind of number you would get if the channel modulation was a small number but NOT zero.
 
  • #14
CKaiL said:
I would also like to check if this is a configuration of a Common source with source degeneration single stage amlifer?
What terminal is common to the input and output?
What is the definition of "source degeneration"? You want answers on a platter and we're not allowed to give you that - which is to your benefit.
 
  • #15
The source is connected to the gate and Output is connected to the drain
 
  • #16
CKaiL said:
The source is connected to the gate and Output is connected to the drain
 
  • #17
Yes, but what terminal is common to the input and the output? What is the input referred to, and what is the output referred to?
 
  • #18
Input is connected to the gate , Output is connected to the drain. Source is connected to neither input or output
 
  • #19
I am looking at the amplifier at M2
 
  • #20
CKaiL said:
I am looking at the amplifier at M2
Good.
An input is a voltage. A voltage is always between two points.
An output is also a voltage. Same thing.
So, again - what terminal is common to input and output?
 
  • #21
There don't seem to be a common terminal between the input and output if I look at M2 only.The only common terminal I see between the input and output is between M1 and M2 terminal where the body of the M1 (output) is connected to the input of M1
 
  • #22
CKaiL said:
There don't seem to be a common terminal between the input and output if I look at M2 only.The only common terminal I see between the input and output is between M1 and M2 terminal where the body of the M1 (output) is connected to the input of M1
Talking M2 only. Re-read what I said about how voltages always need two points.
The input is between G and S.
The output is between D and S.
See any common terminal yet?
 

1. What is a single stage amplifier with current mirror biasing?

A single stage amplifier with current mirror biasing is a type of amplifier circuit that uses a current mirror to bias the amplifier. This means that the current flowing through the amplifier is controlled by a reference current set by the current mirror. This type of biasing is commonly used in low-voltage and low-power applications.

2. How does a single stage amplifier with current mirror biasing work?

In this type of amplifier, the input signal is amplified by a single transistor, and the bias current for the transistor is set by a current mirror. The current mirror consists of two transistors, with one transistor acting as a reference and the other transistor mirroring its current. The output of the amplifier is taken from the collector of the transistor.

3. What are the advantages of using a single stage amplifier with current mirror biasing?

One of the main advantages of this type of amplifier is its simplicity. It only requires a few components and is easy to design and implement. Additionally, it provides good stability and low distortion, making it suitable for low-power and low-voltage applications. It also has a high input impedance, which reduces loading effects on the input signal.

4. What are the limitations of a single stage amplifier with current mirror biasing?

The main limitation of this type of amplifier is its limited gain. Since it uses a single transistor, the maximum achievable gain is lower compared to multi-stage amplifiers. Additionally, the current mirror can introduce errors and variations in the bias current, which can affect the performance of the amplifier.

5. How is the bias current set in a single stage amplifier with current mirror biasing?

The bias current is set by adjusting the size and characteristics of the transistors in the current mirror. By changing the reference current and the mirroring ratio, the bias current can be adjusted to the desired value. This allows for flexibility in designing the amplifier for different applications.

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