# How to find power efficiency in DC-DC converter circuit

We are currently simulating a dcdc converter called the LTC3108.

The input voltage is 500mV and the output is 3.3V through the dcdc converter.

Here we have to calculate the efficiency of the circuit

efficiency = E_out / E_in = I_load * V_out / I_IN * V_IN

The load resistance and the capacitor were placed on the output side, and the output voltage and output current of the resistor were measured.

So, the calculation is that we need to know the voltage and the current. Now the simulation values are as follows.

So, you can see that efficiency calculation is strange ...

How do I calculate it?

E_out / E_in = (500mV * (-267.72mA)) / 3.3V * 33uA = -1225.

What is strange? I want get % value.

• Every time I see something like A*B/C*D I smile. Because it can be interpreted as $\frac{AB}{CD}$ or $\frac{AB}{C}D$. Either way, you got the current backwards, and you calculated $\frac{1}{efficiency}$, and you missed a factor 10 somewhere. Nov 14 '17 at 8:48
• @HarrySvensson even if OP has calculated 1/eff, there's still something wrong. The primary (75u) is switched by the IC and input voltage is DC; so the waveform I_in is incorrect due to $V_L = L \ di/dt$. Nov 14 '17 at 8:52

E_out / E_in = ... But then you fill in the input values for Eout and the output values for Ein???

Why not take it in 2 steps:

Input power= 0.5V*0.268A = 0.134W

Output power: 3.3V * 33uA = 109 uW.

Then Pout/Pin = 109uW/0.134W = 0.000813 = 0.081 %

The 1225 you get is simply 1/0.000813.

Your output power is very low which is unsurprising as your load is a 100 k resistor which means only little current can flow meaning almost no power is dissipated in the load.

It is quite pointless to calculate efficiency with such a light load! In this situation the quiescent current is determining the power consumption.

• How can I improve my efficiency? Do I have to increase the value of the capacitor? Nov 15 '17 at 10:59
• Hmm, I'm sorry if this sounds harsh but I will just say it like it is: If your suggestion is to "change a capacitor" then my conclusion is that you obviously have no idea what you're doing and that you have a lot to learn about DCDC converters. No, changing a capacitor will not help. The reason why the efficiency is low is explained in my answer. I suggest that you study DCDC converters and more specifically DCDC converter efficiency at low currents. Nov 15 '17 at 20:24

Efficiency % = (E_out/E_in) * 100.

Now, your equation is correct but values you have put are not correct. You have swapped values of E_in and E_out.

For E_out = 3.3V * 33uA and for E_in = 500mV * -267.72mA.

Now, we want to calculate the efficiency, so let's take mode of negative current value and put it in to the equation you should get... Efficiency % = (0.0001089 / 0.13386) * 100 = 0.0813 %