0
\$\begingroup\$

I am trying to understand how much energy a circuit (composed by some passives here called analog, and an MCU) will require. The system is powered through a capacitor, and the "passives" are actually some low power op amps attached directly to the capacitor, like the following schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

The MCU is powered through an LDO, and has a determined active time while otherwise is in sleep (now for simplicity I consider sleep equal to 0mA), but the active part is always on.

I need to understand how much energy the system requires to have a predermined drop in the capacitor voltage. From such results, I will define the next design steps or various tunings.

The approach was to compare the total energy required, given the capacitance in consideration, with $$ E = \frac{1}{2} C (V_{0ms}^2 - V_{10ms}^2) $$

The MCU activates with a given duty cycle, and I need to calculate the energy balance only in this period of time, with a given MCU duty cycle. Example, the MCU wakes for 1ms every 10ms, so duty is 1ms and period is 10ms. Duty cycle is 10%. I only want to know the power balance over 10ms, because afterwards the capacitor will gets fully recharged before the next cycle.

But here the problem:

  • I can calculate how much drop in the capacitor voltage I have with a given energy, but because the energy can be rewritten in $$ E = P t = V I t $$, I can use this form to check, in the duty time activity, how much energy is drawn form the MCU, LDO and active components and check it against the capacitor's energy calculated in the initial formula (related to a given drop).
  • Despite this assumption, because the energy is th eintegration of the power over time, I am not sure if makes sense to calculate the energy of the active part in this way: $$ E = (V_{0ms} - V_{10ms}) I_{active} t_{10ms} + (V_{0ms} - V_{10ms}) I_{MCU} t_{1ms} + (V_{0ms} - V_{10ms}) I_{MCU sleep} t_{9ms}$$ What would be the right approach?
\$\endgroup\$
13
  • 1
    \$\begingroup\$ Right from your first formula you have inbuilt an error. Energy does not equal the change in capacitor voltage squared. You cannot treat energy and voltage this way. \$\endgroup\$
    – Andy aka
    May 23 '20 at 17:14
  • \$\begingroup\$ hi, thanks for pointing out. I copied from my previous question without editing what I meant. maybe is still wrong, but at least if what I thought to use \$\endgroup\$
    – thexeno
    May 23 '20 at 17:46
  • 1
    \$\begingroup\$ If you use my answer in your previous question, the formula is correct and all you need to worry about is current draw, voltage drop and time to calculate a value of capacitance. \$\endgroup\$
    – Andy aka
    May 23 '20 at 18:45
  • \$\begingroup\$ yes, thanks. the thing is that I am not sure how to translate the formula when the voltage is constant (the MCU is powered through an LDO). I guess I need to use the power of the LDO - voltage drop of the capacitor used as input of the LDO and constant current of the MCU when is active, so I have V, I and time to find the energy. I think the assumption makes more sense now.. \$\endgroup\$
    – thexeno
    May 23 '20 at 18:58
  • \$\begingroup\$ @Andyaka I edited the question as now is much less confused. I hope you can check it. otherwise I am willing also to make a new question with this content again. I think my problem is a very naive one though, maybe is not needed to make a new question? \$\endgroup\$
    – thexeno
    May 24 '20 at 9:48
1
\$\begingroup\$

I'm now sure if I understand your question right. I think you want to know (based on the capacitor voltage drop) how much energy is absorbed by the "always on" (static) circuit and how much by the MCU (dynamic) circuit? If you only know the voltage difference of the capacitor (dV) from t=0ms to t=10ms, then this is not possible to calculate, because you have two degrees of freedom in the equations. You would need to measure the capacitor voltage drops of the dynamic or the static circuits each alone.

Also, like Andy aka wrote, your formula is not right: The right one would be $$dE=\frac{1}{2}CV_{0ms}^2-\frac{1}{2}CV_{10ms}^2$$

\$\endgroup\$
16
  • \$\begingroup\$ hi, I corrected the initial formula. my mistake! \$\endgroup\$
    – thexeno
    May 23 '20 at 17:47
  • \$\begingroup\$ can't I consider two theoretical cases one with the static and the other with the dynamic consumption? In both cases knowing an initial voltage and the capacitance. Maybe averaging the MCU peak current over the whole period? In the end it would be the same.... My problem is that becuase there is an LDO, I can't use the assumption of the voltage drop, as it is kept constant. So I guess I should approach in a different way \$\endgroup\$
    – thexeno
    May 23 '20 at 17:51
  • \$\begingroup\$ I do not get what you‘re after. Can you explain more about your problem with the LDO? Which voltage drop is kept constant? \$\endgroup\$ May 23 '20 at 20:34
  • \$\begingroup\$ I just want to calculate the total energy taken from the capacitor to make it drop from an initial voltage to a final one, to maximize quantitatively the duty cycle of the MCU considering the static current consumed by the other components, and not going through trial and error (I would not learn much by doing so, either) \$\endgroup\$
    – thexeno
    May 23 '20 at 20:44
  • 1
    \$\begingroup\$ Ok, i think i got your question now. Let me explain: You can only apply the E=UIt formula when your circuit draws a constant current from the capacitor. But this might not be the case in your circuit. \$\endgroup\$ May 24 '20 at 11:00
0
\$\begingroup\$

So, rephrasing what I found so far, also thanks to discussing in the comments, I think it might help to write a new answer to myself, as I need to put also some graphics in it.

My problem is that I have a theoretical capacitance, and naturally it loses potential as long as the energy is taken from it. This capacitance is used to power some devices, which in the end are not important, it just matters that such devices are taking a known amount of constant current for a known period of time. The capacitor has a known initial voltage, and it is known the minimum allowable voltage we need.

The energy the capacitor can provide, is $$ E_{c} = \frac{1}{2} C (V_{init}^2 - V_{final}^2) $$.

The energy the devices are using can be found through the usage time and their current, but the voltage is dropping. So is NOT possible to assume $$ E_{used} = (V_{init} - V_{final}) I_{active} t $$, because I am interested in finding the total energy used to make such drop in C, not just it's delta:

enter image description here

So in order to find the energy the devices are using (is the first area in the graph), I must simply do some geometry:

$$ E = P_{min} t_{max} + \frac{(P_{max} - P_{min}) t_{max}}{2} $$

And finally I can compare this to the original energy the capacitor can provide to see if the circuit will stay within specifications.

Any thoughts?

\$\endgroup\$
8
  • \$\begingroup\$ Unfortunately I still don't understand what you are trying to find out or why you want to do this. I'm not being awkward (I can be of course!) - I'm just not following what you are trying to evaluate or why. \$\endgroup\$
    – Andy aka
    May 24 '20 at 11:29
  • \$\begingroup\$ There is a device, which harvests energy from a capacitor and I need to adjust the duty cycle of the MCU to use all the energy down to the minimum allowed voltage. I assume the MCU to have constant current (it is powered through a LDO so it sees constant voltage). The meaning of the device is not relevant, as the question is on the mathematical side of it, and not if the product makes sense or not. The only unknown is how to find the energy used by the MCU and to check if will be greater than the energy the capacitor can provide. Because V is not constant, I had some doubts in doing so. \$\endgroup\$
    – thexeno
    May 24 '20 at 11:41
  • \$\begingroup\$ But this takes us back to your earlier question surely. If you know the voltage drop rate and you know how much current the load takes you can calculate the capacitance AND, if you want to know how much energy that capacitor has dispensed between Vupper and Vlower use A - B where A = \$\dfrac{CV_{upper}^2}{2}\$ and B = \$\dfrac{CV_{lower}^2}{2}\$ \$\endgroup\$
    – Andy aka
    May 24 '20 at 11:47
  • \$\begingroup\$ "If you know the voltage drop rate and you know how much current the load takes you can calculate the capacitance" how? from the E = V I t? My approach in the answer doesn't makes sense? Isn't possible to see how much energy the capacitor can provide , and compare it to another energy quantity, which is what the load requires? If is higher, then I need a bigger capacitance. So maybe your approach helps in finding first the capacitance needed, then comparing it with my actual capacitance, and this will give the same answer I am looking for. but then, how? \$\endgroup\$
    – thexeno
    May 24 '20 at 12:32
  • 1
    \$\begingroup\$ \$C = I\cdot \dfrac{dt}{dv}\$. Example: I = 10 mA and dv = 1 volt and dt = 1 second then C = 10,000 uF. \$\endgroup\$
    – Andy aka
    May 24 '20 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.