Hallo all, I am from electronics background. I have a simple mathematical problem which seems daunting for me.
The general equations for calculating charge is given below. It integrates the capacitance C(v) over the range of voltage 'v'.
q(v) = ∫ C(v)dv. limits are from 0 to v (Charge eqn)
Now to obtain the current from this charge the following equation holds true and i(t) = dq(v(t))/dt . (current eqn).
Now I have a charge which is constant over different range of the voltage. So I dont know how to integrate over the voltage to find the total charge which when differentiated will give me the current.
The following is the value of the capacitance C_GS, C_GD that are dependent on voltages v_GS and v_GD.
C_sov, C_dov and C_ch are constants.
if (v_GS > vth) begin
C_GS = C_sov ;
C_GD = C_dov;
end
else if ((v_DS) > v_GS-vth) begin
C_GS = C_ch/2+C_sov;
C_GD = C_ch/2+C_dov;
end
else begin
C_GS = (2/3)*C_ch+C_sov;
C_GD = C_ch/3+C_dov;
end
Now I need to find the charge
q_g = ∫ C_GS(v_GS)dv in the range 0 to v_GS
q_d = ∫ C_DS(v_DS)dv in the range 0 to v_DS
With these values, I can differntiate to get the respective currents. The problem is with calculating the charges q_d and q_g. Could somone help me in deriving the equations for q_d, q_g for the above case? Thanking you in advance.
EDIT: equations and symbols converted to LaTeX
Hallo all, I am from electronics background. I have a simple mathematical problem which seems daunting for me.
The general equations for calculating charge is given below. It integrates the capacitance $C(v)$ over the range of voltage $v$.
$q(v) = \int_0^v C(v)dv.\qquad\text{(Charge eqn)}$
Now to obtain the current from this charge the following equation holds true and
$i(t) = \dfrac{dq(v(t))}{dt}. \qquad\text{(current eqn)}$
Now I have a charge which is constant over different range of the voltage. So I dont know how to integrate over the voltage to find the total charge which when differentiated will give me the current.
The following is the value of the capacitance $C_{\text{GS}}$, $C_{\text{GD}}$ that are dependent on voltages $v_{\text{GS}}$ and $v_{\text{GD}}$.
$C_{\text{sov}}$, $C_{\text{dov}}$ and $C_{\text{ch}}$ are constants.
if (v_{\text{GS}}>\text{vth}) begin
$C_{\text{GS}}= C_{\text{sov}}$;
$C_{\text{GD}} = C_{\text{dov}}$;
end
else if ((v_{\text{DS}}) > v_{\text{GS}}-\text{vth}) begin
$C_{\text{GS}} = C_{\text{ch}}/2+C_{\text{sov}}$;
$C_{\text{GD}} = C_{\text{ch}}/2+C_{\text{dov}}$;
end
else begin
$C_{\text{GS}} = (2/3)\cdot C_{\text{ch}}+C_{\text{sov}}$;
$C_{\text{GD}} = C_{\text{ch}}/3+C_{\text{dov}}$;
end
Now I need to find the charge
$q_g =\int_0^{v_{\text{GS}}}C_{\text{GS}}(v_{\text{GS}})dv_{\text{GS}}\qquad (\text{note: dv in the original})$
$q_d = \int_0^{v_{\text{DS}}}C_{\text{DS}}(v_{\text{DS}})dv_{\text{DS}}.\qquad (\text{note: dv in the original})$
With these values, I can differntiate to get the respective currents. The problem is with calculating the charges $q_d$ and $q_g$.
Could somone help me in deriving the equations for $q_d$, $q_g$ for the above case? Thanking you in advance.
Edit: The question is reframed. I regret for my confusing statements. The capacitance is piece wise constant over different range of voltage (v_GS,v_DS) for e.g
C_GS(v_GS,v_DS) = C_sov for v_GS > vth;
C_ch/2+C_sov for (v_DS) > v_GS-vth && v_GS
(2/3)*C_ch+C_sov for (v_DS) <= v_GS-vth && v_GS
Now to find the charge q_g, I need to integrate C_GS(v_GS,v_DS) w.r.t V_GS and v_DS in the interval [0,Vg];[0,Vd]. Similarly for C_DS to get q_d.
Edit: I have put it into L$\LaTeX$, but am not sure I got it right. Please check and provide comments The capacitance is piece wise constant over different range of voltage $(v_{GS},v_{DS})$
$C_{GS}(v_{GS},v_{DS}) = \begin {cases}C_{sov} & v_{GS} > vth \\ C_{ch}/2+C_{sov} & v_{DS} > v_{GS}-vth \text{ and } v_{GS}
Now to find the charge $q_g$, I need to integrate $C_{GS}(v_{GS},v_{DS})$ w.r.t $V_{GS}$ and $v_{DS}$ in the interval $[0,V_g]$ and $[0,V_d]$. Similarly for $C_{DS}$ to get $q_d$.