CMOS (Energy Supply of voltage) [closed]

Question

Can someone please explain why, when \$ln\$ is transitioning from low to high, the energy supplied is \$C_{vdd}\cdot V_{dd}^2\$?

Answer 1

The formula for energy stored in a capacitor is:

$$ E = \frac{1}{2}CV^2 $$

In either state, one of those capacitors is fully charged, with energy \$\frac{1}{2}C{V_{DD}}^2\$ joules, and the other fully discharged, with 0J of energy.

The total stored energy energy in both cases is therefore:

$$ E_{TOTAL} = \frac{1}{2}C{V_{DD}}^2 $$

In other words, the net change in stored energy when state changes from top capacitor charged and bottom discharged, to top discharged and bottom charged is zero.

You might think that the power supply doesn't need to provide any energy at all, since net stored energy hasn't changed, but you'd be wrong. There's a cost to charging a capacitor, and a cost to discharging it, because the charging path has resistance.

The supply has non-zero internal resistance, and the MOSFETs have non-zero on-resistance, and all current that results from the charging and discharging of those capacitors will have to pass via these resistances. Consequently those resistances will dissipate power, and that energy is lost as heat. That energy is what the supply is delivering to the system during each transition between states. I'm going to ignore the supply's internal resistance, and assume that the only resistance in play is the MOSFET's own \$R_{DS(ON)}\$.

Assuming we start with the upper capacitor (\$C_{VDD}\$) discharged (0V across it), and the lower one (\$C_{GND}\$) charged (the full \$V_{DD}\$ across it). When we switch on the lower MOSFET, to discharge \$C_{GND}\$ and charge \$C_{VDD}\$, the system looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

I've renamed stuff to make the algebra easier to write. The voltage across C1 is \$V_{C1} = V_S - V_X\$. The voltage across \$V_{C2} = V_X - 0V = V_X\$. The voltage across R is \$V_R = V_X - 0V = V_X\$. Supply voltage is fixed at \$V_S\$.

I'm not sure that the following is the easiest or quickest proof. There's probably a more intuitive approach that I haven't spotted.

We wish to find the energy delivered to the rest of the circuit by source VS. Energy is the time-integral of power, where power \$P=IV\$, the product of voltage across the recipient and current through it:

$$ \begin{aligned} E &= \int^\infty_{t=0}{P \cdot dt} \\ \\ &= \int^\infty_{t=0}{IV_S \cdot dt} \\ \\ &= V_S\int^\infty_{t=0}{I \cdot dt} \\ \\ \end{aligned} $$

By Ohm's law, current \$I_1\$ through R is:

$$ I_1 = \frac{V_X}{R} $$

Capacitor C1 current is:

$$ \begin{aligned} I &= C_1 \frac{d(V_S-V_X)}{dt} \\ \\ &= C_1 \frac{V_S}{dt} - C_1 \frac{V_X}{dt} \\ \\ \end{aligned} $$

Remembering that \$V_S\$ is constant, so \$\frac{dV_S}{dt}=0\$, that simplifies to:

$$ I = - C_1 \frac{dV_X}{dt} $$

Capacitor C2 current is:

$$ I_2 = C_2\frac{dV_X}{dt} $$

With expressions for all three currents, we can combine them using KCL, which says:

$$ \begin{aligned} I &= I_1 + I_2 \\ \\ - C_1 \frac{V_X}{dt} &= \frac{V_X}{R} + C_2\frac{dV_X}{dt} \\ \\ - \frac{dV_X}{dt}(C_1 + C_2) &= \frac{V_X}{R} \\ \\ \frac{dV_X}{dt} &= -\frac{V_X}{R(C_1+C_2)} \end{aligned} $$

Solving this differential equation for \$V_X\$ is fairly trivial, I won't show any working, just the solution:

$$ V_X(t) = V_X(0)e^{-\frac{t}{R(C_1+C_2)}} $$

\$V_X(0)\$ is the initial potential of node X, at time \$t=0\$, which we know to be \$V_S\$, so this becomes:

$$ V_X(t) = V_Se^{-\frac{t}{R(C_1+C_2)}} $$

You might recognise that last equation as the classic capacitor discharge formula. Interestingly, it has time constant \$R(C_1+C_2)\$, as if C1 and C2 were in parallel. What this tells us is that potential \$V_X\$ starts equal to \$V_S\$, and decays exponentially to zero, with a time constant of \$R(C_1+C_2)\$ seconds.

Plug our expressions for \$V_X\$ and its derivative back into our equation for \$I\$:

$$ \begin{aligned} I &= -C_1 \frac{dV_X}{dt} \\ \\ &= (-C_1)(-\frac{V_X}{R(C_1+C_2)}) \\ \\ &= \frac{C_1}{R(C_1+C_2)}V_X \\ \\ &= \frac{C_1}{R(C_1+C_2)}V_Se^{-\frac{t}{R(C_1+C_2)}} \\ \\ \end{aligned} $$

I'll substitute \$\tau=R(C_1+C_2)\$, just to tidy things up:

$$ I = \frac{C_1}{\tau}V_Se^{-\frac{t}{\tau}} $$

We're ready to calculate energy \$E\$:

$$ \begin{aligned} E &= V_S\int^\infty_{t=0}{I \cdot dt} \\ \\ &= V_S\int^\infty_{t=0}{\frac{C_1}{\tau}V_Se^{-\frac{t}{\tau}} \cdot dt} \\ \\ &= {V_S}^2\frac{C_1}{\tau}\int^\infty_{t=0}{e^{-\frac{t}{\tau}} \cdot dt} \\ \\ &= {V_S}^2\frac{C_1}{\tau}\left[ -\tau e^{-\frac{t}{\tau}} \right]^\infty_{t=0} \\ \\ &= {V_S}^2C_1\left[ -e^{-\frac{t}{\tau}} \right]^\infty_{t=0} \\ \\ &= {V_S}^2C_1\left[\vphantom{\frac{}{}} (0) - (-1) \right] \\ \\ &= {V_S}^2C_1 \end{aligned} $$

That's the energy supplied by voltage source VS (\$V_{DD}\$) to the entire system, during a transition from output high to output low, (or input low to high, since this is an inverter). You can repeat the procedure for a transition from low to high, and I'm sure you'll obtain a similar result, but with C2 (\$C_{GND}\$) in the expression instead.