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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
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2 Answers

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 $v\in [v_{n-1},v_n]$.

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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 $$q_g= \begin {cases} (C_{ch}/2+C-{sov})vGS & vGS \lt vth \\ (C_{ch}/2+C-{sov})vth + (2C_{ch}/3+C_{sov})(vGS-th) & vGS \gt vth \end {cases}$$

Basically you break the integral into pieces where the capacitance changes. Since the capacitance is constant over each piece, the charge is just that capacitance times the voltage of the piece.

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Thank you Millikan . I regret for my confusing and shabbily written question. –  Ramkumar Ganesan Jun 27 '11 at 7:38
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