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# power consumption

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#### cheniour

##### Member
I have this circuit linear regulator with UVLO so I would like to calculate the power consumption of this block any person can help me, please Vout = 1.5 V (Iss = 15.2 μA, Vout = 1.5 V, switching = 32.768 kHz, δ = 0.809), block function A 1.5-V low dropout (LDO) linear regulator Torex XC6215B with quiescent current consumption 0.8-μA. Below 1.5 V, the input of the linear regulator is disconnected from the supercapacitor by a p-type MOSFET (Si-2302) in order to avoid leakage current. An XC6120C voltage detector turns the low-side switch on when 1.5 V is reached across

What is switching at 32.768khz?
Do you want the power loss when the Si-2305 are open or closed?

Si_2305 closed

The power consumption of the circuit will depend on the input voltage.
Since your schematic shows no load resistance at the output, the power consumption of the LDO will be very low (roughly Vinx0.8uA). However, with an output load resistance added on, the power consumption of the LDO will be: (Input Voltage - 1.5 V) x (1.5V/Rload). In other words, the power consumed by the LDO will be the voltage drop across it multiplied by the current passing through it to the load.

You can model the voltage detector, approximately, as a shunt 7.5Mohm resistor. Put the 2Mohm resistor used to bias your Si2305 in parallel with that resistance and then calculate power consumption as V^2/1.58Mohms.

So, for example, when the input voltage is 3.0V, and if there is a load resistance of 1000 ohms, the LDO power consumption will be 2.25 mW. The detector and switch power consumption will be 5.7uW, which you can choose to ignore as it is much less than the LDO power consumption.

thank youvery much can you explain me how ypu find 5.7µW please ?

In the example, the input voltage is 3V. I estimated that the voltage detector is a shunt resistance of approximately 7.5Mohm based on info on the datasheet. If you want to be more precise, you can look at the datasheet for the XC6120 where you will see a graph of supply current vs input voltage. From the graph, you can find the supply current for an input of 3V, which I find to be 0.65 uA. So, it seems that my estimate of 7.5Mohms was a bit too high. Next, we can calculate the current through the 2M ohm resistor, with the approximation that the voltage across the Si2302 is zero when it is ON, using ohms law. 3V divided by 2Mohms is 1.5uA. For the total current of the voltage detector and the switch I just add the two currents: 1.5uA plus 0.65uA equals 2.15uA. So the total power consumed by that circuit is 3 Volts x 2.15 uA = 6.45uW. This is a little bit higher than my previous estimate of 5.7uW.

As you can see, I have used some approximations to reach a solution. You can calculate this more precisely if the exact power consumption is of interest, but since it is quite low compared to the consumption of the LDO, I assume that approximation was ok.

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