Integration of piece-wise arguments

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 $(v_{GS},v_{DS})$

$$C_{GS}(v_{GS},v_{DS}) = \begin {cases}C_{sov} & v_{GS} > vth \\ C_{ch}/2+C_{sov} & v_{DS} > v_{GS}-vth \text{ and } v_{GS}<vth \\ (2/3)*C_{ch}+C_{sov} & v_{DS} <= v_{GS}-vth \text{ and } v_{GS}<vth \end {cases}$$

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 $q_d$.

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please see if my transcription is correct. – Américo Tavares Jun 24 '11 at 15:09
I didn't find any definition of $C_{DS}$ – Ross Millikan Jun 27 '11 at 13:09

Lots of notation seems only to complicate matters. Let $v_0=0<v_1<v_2<\ldots\$ and $C(v)=c_i$ for $v\in [v_{i-1},v_i]$. Then integral $$q(v)=\int_0^v C(t)\,dt=\sum_{i=1}^{n-1}c_i(v_i-v_{i-1})+(v-v_{n-1})c_n$$ for $v\in [v_{n-1},v_n]$.
I'm not sure the question is well defined. Do you charge up $q_g$ before $q_d$, proportionately, or after? That changes the capacitance values to use. If we charge up $q_g$ first and $vth \gt 0$, we integrate up using to $vth$ using the second set, then use the first, so $$q_g= \begin {cases} (C_{ch}/2+C-{sov})vGS & vGS \lt vth \\ (C_{ch}/2+C-{sov})vth + (2C_{ch}/3+C_{sov})(vGS-th) & vGS \gt vth \end {cases}$$