# How to calculate efficiency of a rectifier with a capacitor filter

I was reading on rectifier circuits and I found out the efficiency of a center tapped transformer is the average power Pdc divided by the rms power Pac and if the resistance of the load is known then it's Vdc^2 divided by Vac^2. This gives around 81.2%. My question is how do you find the efficiency of a capacitor filtered rectifier? I realized I can't use the same formula because I'll get an efficiency of more than 100% which makes no sense. I mean if we found our new average voltage Vdc using the formula Vm (Peak Voltage)- (Vr(Ripple voltage)/2) we can get an average voltage higher than the rms voltage. So is there another way to calculate the efficiency or am I missing something ?

That is my circuit diagram on Simulink

• Can you show us a circuit diagram? Jul 19, 2021 at 17:19
• I'll update the post rn. But the circuit diagram is a center tapped transformer with a filter done in Matlab's SimPower Jul 19, 2021 at 17:23
• It will be the same as always: Pout/Pin, where Pout=Vout*Iout (DC) and Pin=average(Vin*Iin) (AC). Jul 19, 2021 at 17:25
• @aconcernedcitizen How do I find Vout? I mean there are still ripples in my filtered circuit so I don't have a constant output DC voltage. And I thought it was average power divided by rms power? Thank you for your time Jul 19, 2021 at 17:47
• What is RMS power? Jul 19, 2021 at 18:07

There's no analytical way to determine this (as far as I know) because it involves extracting $$\t\$$ from:

$$\mathrm{e}^{-t/\tau}=\sin(2\pi t) \tag{1}$$

Because what happens (in an ideal case) is the capacitor charges to peak voltage, then it discharges according to the exponential law on the resistor until the sine wave catches up, and then it charges again until the peak voltage (and a bit more), then it discharges, etc.

Therefore as an approximation you can consider the ripple voltage to be:

$$V_r=V_p\dfrac{T}{2RC} \tag{2}$$

Half of this value is subtracted from the peak voltage to give the final DC value. This value is then used to calculate the power on the load. Then the current through the capacitor only follows an exponential law during discharge, in rest it's an integral of a sine for that finite period when it's charging. The load sees only the sine voltage for the duration of the charging.

As a short example, consider C to be 1 mF and R to be 100 Ω, with a 100 Vp sine:

As you can see the capacitor discharges until ~3.66 ms, then it charges back up until slightly past the peak, then the cycle repeats. The voltage on the load sees the discharge plus the part of the (absolute valued) sine during the discharge.

The ripple is about 8.4 Vpp, so the DC voltage can be considered $$\100-8.4/2=95.8\;\mathrm{V}\$$ (96.034 V measured), so the power is $$\95.8^2/100\approx 91.8\;\mathrm{W}\$$ (92.224 measured).

The input power is more complicated to calculate, but you can apply the generic mathematical formula:

$$\int{\sin(2\pi t)\mathrm{d}t}=\dfrac{\cos(2\pi t)}{2\pi} \tag{3}$$

This tells you that the shape of the waveform for the current is a cosine, truncated temporally to the time it takes for the capacitor to charge. To this, a sine is added for the same interval it takes for the load to be sustained by the supply. Since outside the interval everything is zero, multiplying the voltage by the current results in a sine multiplied by the sum of a cosine and a sine. In generic terms:

$$\sin(2\pi t)\left[\dfrac{\cos(2\pi t)}{2\pi}+\sin(2\pi t)\right]=\dfrac{\sin(4\pi t)}{4\pi}+\dfrac{1-\cos(4\pi t)}{2} \tag{4}$$

where $$\t\$$ is an interval. In the above example it's from 3.66 ms to 5 ms. The value for the peak current here is ~13.6 A, so the theoretical value for the average power would be the integral of (4):

$$\int_{T_1}^{T_2}{(4)\mathrm{d}t}=\dfrac{2\pi[\sin\left(4\pi T_1\right)-\sin(4\pi T_2)]-\cos(4\pi T_2)+\cos(4\pi T_1)+8\pi^2(T_2-T_1)}{16\pi^2}$$

And if you replace $$\T_1\$$ and $$\T_2\$$ to be 0.366/2 and 0.5/2, respectively (adapt the numbers to a unit period), and then use the values of 100 V and 13.6 A, you get ≈88.8W, which is pretty close to the measured 92.26 W, but less than the load power.

The conclusion is that the efficiency is increased, but so are the harmonics, and about the only ways to determine the efficiency are to either measure it on the breadboard, or simulate it.

• In (4) I considered the terms inside the square brackets to have the same value for the current, but the cosine has the value of 100 V divided by the reactance of the capacitor, ~3.2 Ohms, and the sine only with ~1 A (or 100 V divided by the load). Also, I calculated for a half period, so it gets multiplied by two, so in the end the result is 95.65 W. It's larger, but still not reliable for calculating the efficiency. The conclusion remains. Jul 19, 2021 at 20:29

Well, assuming a diode-voltage drop of $$\\text{V}_\text{D}\$$ we can see that the capacitor will charge to a maximum of $$\\hat{\text{V}}_\text{out}=\hat{\text{V}}_\text{in}-2\text{V}_\text{D}\$$. So the output voltage is given by (ignoring the transient part):

$$\text{V}_\text{out}\left(t\right)= \begin{cases} \left(\hat{\text{V}}_\text{in}-2\text{V}_\text{D}\right)\exp\left(-\frac{t}{\text{CR}}\right)&\text{if}\space\space 0\le t\le\tau\\ \\ \left|\hat{\text{V}}_\text{in}\cos\left(\omega t\right)\right|-2\text{V}_\text{D}&\text{if}\space\space\tau< t\le\frac{\pi}{\omega} \end{cases}\tag1$$

With the fact that the function is periodic, so $$\\text{V}_\text{out}\left(t\right)=\text{V}_\text{out}\left(t+\text{nT}\right)\$$, where $$\\text{T}\$$ is the timeperiod of the function which is equal to $$\\text{T}=\frac{\pi}{\omega}\$$ and $$\\text{n}\in\mathbb{Z}\$$. Where $$\\tau\$$ is defined to be the solution to equation $$\(2)\$$ for $$\\frac{\pi}{2\omega}<\tau<\frac{\pi}{\omega}\$$:

$$\left(\hat{\text{V}}_\text{in}-2\text{V}_\text{D}\right)\exp\left(-\frac{\tau}{\text{CR}}\right)=\left|\hat{\text{V}}_\text{in}\cos\left(\omega\tau\right)\right|-2\text{V}_\text{D}\space\Longrightarrow\space\tau=\dots\tag2$$

So, the output power is defined to be:

$$\text{P}_\text{out}\left(t\right)=\frac{\text{V}_\text{out}^2\left(t\right)}{\text{R}}\tag3$$

So, finding the average gives:

$$\overline{\text{P}}_\text{out}=\frac{1}{\text{T}}\int_0^\text{T}\text{P}_\text{out}\left(t\right)\space\text{d}t=\frac{1}{\text{T}}\int_0^\text{T}\frac{\text{V}_\text{out}^2\left(t\right)}{\text{R}}\space\text{d}t=\frac{1}{\text{RT}}\int_0^\text{T}\text{V}_\text{out}^2\left(t\right)\space\text{d}t=$$ $$\frac{\omega}{\text{R}\pi}\cdot\left\{\int_0^\tau\text{V}_\text{out}^2\left(t\right)\space\text{d}t+\int_\tau^\frac{\pi}{\omega}\text{V}_\text{out}^2\left(t\right)\space\text{d}t\right\}=$$ $$\frac{\omega}{\text{R}\pi}\cdot\left\{\int_0^\tau\left(\left(\hat{\text{V}}_\text{in}-2\text{V}_\text{D}\right)\exp\left(-\frac{t}{\text{CR}}\right)\right)^2\space\text{d}t+\int_\tau^\frac{\pi}{\omega}\left(\left|\hat{\text{V}}_\text{in}\cos\left(\omega t\right)\right|-2\text{V}_\text{D}\right)^2\space\text{d}t\right\}=$$ $$\frac{\omega}{\text{R}\pi}\cdot\left\{\left(\hat{\text{V}}_\text{in}-2\text{V}_\text{D}\right)^2\int_0^\tau\exp\left(-\frac{2t}{\text{CR}}\right)\space\text{d}t+\int_\tau^\frac{\pi}{\omega}\left(\left|\hat{\text{V}}_\text{in}\cos\left(\omega t\right)\right|-2\text{V}_\text{D}\right)^2\space\text{d}t\right\}\tag4$$