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# Derivation of output resistance formula

Category: Datasheet/Specs

The datasheet of ADP5600 says the output resistance is ROUT = 1/(2 × C1 × fOSC) + 4 × RON + 2 × RC1_ESR, but i don't know how to derivate it,. Can someone show me how? Thank you !

• The formula is at 16th page of the datasheet

• Hi,

I think Fig. 41. of the datasheet can help to understand why this is the right formula for estimation of Rout.

If you have output current of Iout, then you have to transfer the following amount of charge in one period:

\Delta Q = I_{out} \cdot T

where T is 1/fosc. However, half of this time is used due to the interleaved operation of two such parts shown in Fig.42.. Therefore there is a voltage change on the capacitor expressed by this formula:

\Delta V = \frac {\Delta Q}{C_1} = \frac {I_{out} T}{2 \cdot C_1 } = \frac {I_{out}}{2 \cdot C_1 \cdot f_{osc}}

So, you can assign an effective resistance to this:

\frac{\Delta V}{I_{out}}= \frac {1}{2 \cdot C_1 \cdot f_{osc}}

During charging the current flows through two switches and the ESR of the capacitor, so they add resistance to the input side:

2 \cdot R_{ON}+R_{C1 \text {_} ESR}

This means additional voltage drop. And the same happens during discharging, so you have to take the double of it:

4 \cdot R_{ON}+2 \cdot R_{C1 \text {_} ESR}

Finally the formula of the datasheet is obtained:

\frac{\Delta V}{I_{out}}= \frac {1}{2 \cdot C_1 \cdot f_{osc}} + 4 \cdot R_{ON}+2 \cdot R_{C1 \text {_} ESR}

I hope it helps.

Zoltan

• Hi,Zoltan

Thank you for your explanation! Now I know the first half about C and T,but I’m still confused about the second half.Charging and discharging are almost the same, so why should the Ron of switch and ESR being doubled?Seeing fig45, Iout is loaded constantly on Rout whenever charging and discharging , so I think they shouldn’t be doubled,could you explain about this ,thank you！

Best Regards

BarBarBar

• Hi,

You have a voltage drop of Iout*(2*RON+ESR) at the input side (C1 is charged to less than Vin), and another drop on the output side (the output voltage is less than the voltage on C1). So, between the input and output voltage you have the sum of these. But I'm not an ADI engineer, they probably know better the structure of this chip and can explain it better.

Zoltan

• Hi Zoltan