Transcription
This lecture is the first part of MOSFET solved problems. In the last few lectures, we completed everything related to the construction, working, and characteristics of enhancement type MOSFETs and depletion type MOSFETs. Now, in this presentation, we will solve one numerical problem that is based on both enhancement and depletion type MOSFETs.
Let's read the problem first: The current of an enhancement type MOSFET increases at the same rate as a depletion type MOSFET for the conduction region. The current is the drain current \( I_D \) of an enhancement type MOSFET. The current \( I_D \) of an enhancement type MOSFET increases at the same rate as a depletion type MOSFET, so we are comparing the drain current in the case of enhancement type MOSFETs to the drain current in the case of depletion type MOSFETs for the conduction region. We have to tell whether these two currents are increasing at the same rate or not.
The conduction region is the saturation region. In the saturation region, the device is on, and we already know the equation of drain current in both cases: in the case of enhancement type MOSFET and in the case of depletion type MOSFET. We will use these equations to find the solution to this problem.
So, let's move to the solution. I will make two columns: the first column is for enhancement type MOSFET, and the second column is for depletion type MOSFET. The drain current in the case of enhancement type MOSFET is given by:
\[
I_D = K \left( V_{GS} - V_T \right)^2
\]
This is the equation for the drain current in the case of enhancement type MOSFET. In the case of depletion type MOSFET, the equation is the same as the equation in the case of Junction Field Effect Transistor:
\[
I_D = I_{DSS} \left( 1 - \frac{V_{GS}}{V_P} \right)^2
\]
So, this is the equation of drain current in the case of depletion type MOSFET. Now, to find out the rate of change of the drain current, we will differentiate the drain current with respect to \( V_{GS} \). We will perform the differentiation of drain current \( I_D \) with respect to \( V_{GS} \) in both cases: in the case of enhancement type MOSFET as well as in the case of depletion type MOSFET.
But before performing the differentiation, we will first find out what the constant terms are. In the enhancement type MOSFET equation, there are two constant terms: \( K \) and \( V_T \) (the threshold voltage). In the depletion type MOSFET equation, there are also two constant terms: \( I_{DSS} \) (the saturated drain current) and \( V_P \) (the pinch-off voltage).
Now that we have the knowledge of constant terms, we can easily differentiate the two equations. We will first differentiate the equation in the case of enhancement type MOSFET. We will differentiate this equation with respect to \( V_{GS} \). On the left-hand side, we will have \( \frac{d}{dV_{GS}} \), and on the right-hand side, we will have the differentiation of \( K \left( V_{GS} - V_T \right)^2 \) with respect to \( V_{GS} \). Since \( K \) is a constant, we can take it out.
So, we will write the right-hand side as \( K \) times the differentiation of \( \left( V_{GS} - V_T \right)^2 \) with respect to \( V_{GS} \). Let's say \( V_{GS} - V_T = T \). Therefore, we have the differentiation of \( T^2 \) with respect to \( V_{GS} \). If you remember the laws of differentiation, you can easily differentiate \( T^2 \) with respect to \( V_{GS} \):
\[
\frac{d}{dV_{GS}}(T^2) = 2T \frac{dT}{dV_{GS}}
\]
where \( T = V_{GS} - V_T \). So we can write the right-hand side as \( 2K \left( V_{GS} - V_T \right) \frac{d}{dV_{GS}}(V_{GS} - V_T) \). The differentiation of \( V_{GS} - V_T \) with respect to \( V_{GS} \) is equal to 1. Thus, the final result is:
\[
\frac{dI_D}{dV_{GS}} = 2K \left( V_{GS} - V_T \right)
\]
Now, let's begin our differentiation in the case of depletion type MOSFET. We will differentiate this equation with respect to \( V_{GS} \). On the left-hand side, we have \( \frac{d}{dV_{GS}} \), and on the right-hand side, we have \( I_{DSS} \) (which is constant) times the differentiation of \( \left( 1 - \frac{V_{GS}}{V_P} \right)^2 \) with respect to \( V_{GS} \). Let's say \( 1 - \frac{V_{GS}}{V_P} = S \).
So we have \( I_{DSS} \) times the differentiation of \( S^2 \) with respect to \( V_{GS} \), where \( S \) is the function of \( V_{GS} \). We have:
\[
\frac{d}{dV_{GS}}(S^2) = 2I_{DSS} S \frac{dS}{dV_{GS}}
\]
where \( S = 1 - \frac{V_{GS}}{V_P} \). Therefore, we can write:
\[
\frac{dI_D}{dV_{GS}} = 2I_{DSS} \left( 1 - \frac{V_{GS}}{V_P} \right) \left( -\frac{1}{V_P} \right)
\]
The final result is:
\[
\frac{dI_D}{dV_{GS}} = -\frac{2I_{DSS}}{V_P^2} \left( V_{GS} - V_P \right)
\]
Now we will compare the two results to find out the answer to this problem. In the first result, you can clearly see that \( 2K \) is a constant, and \( V_T \) is also a constant. So we have a constant inside the bracket \( (V_{GS} - V_T) \). In the second case, \( -\frac{2I_{DSS}}{V_P^2} \) is a constant, and \( V_P \) is a constant, so we have a constant inside the bracket \( (V_{GS} - V_P) \).
Therefore, for both devices, the drain current will increase at about the same rate. The rate at which the drain current increases in the case of depletion type MOSFET is the same as the rate at which the drain current increases in the case of enhancement type MOSFET.
So this is the answer to this question. I hope the solution is clear to you. If you have any doubts, you may ask in the comment section. The differentiation part you already know how to perform, so there was nothing special to explain in this problem. The important thing was the comparison. In the next lecture, we will solve two more examples.
So this is all. See you in the next one. [Applause] [Music]