Energy consumed in a digital circuit

[reddvoid]

I have a digital circuit , to find the energy consumed by the circuit, I am monitoring the current drawn from the supply and integrating it to get the charge pumped into the circuit ... and the total charge*V supposed to give me total energy consumed by the circuit.

But I am getting a small spike of negative current (current going back into the source?)
which gives me negative charge (charge going back into the source?)
which gives me negative energy ( Energy gained by the source?)

Do I have to consider this negative energy while calculating total energy consumed by the circuit or should I neglect it ?

Thanks in advance

 

[anorlunda]

Your units are wrong. Charge is not energy. Current is not power.

$P=VI$

Only in the special case where V is constant, can you say power is proportional to current.

And yes, power can reverse direction transiently, so you need to allow for that in your integration.

 

[reddvoid]

Hi anorlunda,
I did not understand how the units are wrong.
P=VI = V (dQ/dt)
P dt = V dQ = watts second
Watts = Joules/second
so dQ * V should give me Joules ; that is Energy
so, my statement --> " the total charge*V supposed to give me total energy consumed by the circuit " ...seems to be right ,
I don't see where I am wrong.
Can you please point out ?

Thanks

 

[cnh1995]

Hi anorlunda,
I did not understand how the units are wrong.
P=VI = V (dQ/dt)
P dt = V dQ = watts second
Watts = Joules/second
so dQ * V should give me Joules ; that is Energy
so, my statement --> " the total charge*V supposed to give me total energy consumed by the circuit " ...seems to be right ,
I don't see where I am wrong.
Can you please point out ?

Thanks

I don't think it's wrong, basically you're calculating ∫VI⋅dt.

Can you show the circuit diagram and how you are measuring the current? As anorlunda said, power can reverse direction, especially if the circuit contains energy storing element(s).

 

[anorlunda]

I did not understand how the units are wrong.

My bad, I read it wrong.

 

[Baluncore]

But I am getting a small spike of negative current (current going back into the source?)

Move power supply bypass capacitance from the supply to the digital circuit supply rails.
Currents flowing back to the power supply reservoir may be due to the inductive kick when a digital output is turned off.
Is it possible that your supply voltage is dropping suddenly ?
Without a better idea of your circuit, we cannot guess why you have so much energy looking for somewhere to go.

 

[rbelli1]

And yes, power can reverse direction transiently

Currents flowing back to the power supply reservoir may be due to the inductive kick when a digital output is turned off.

I would be very concerned with any circuit that returns current to the source when that was not specifically designed in.

Move power supply bypass capacitance from the supply to the digital circuit supply rails.

I was thinking just this but you beat me to it. Make sure to maintain stability by leaving sufficient capacitance near the power supply too.

Also try to keep the power and ground traces and planes as symmetrical and as short as possible. Modern digital circuits have very high slew rates and even run at low frequency will produce harmonics in very high frequencies. It doesn't take much to produce an RF transmitter (and receiver) with bad power distribution.

BoB

 

[donpacino]

Something to note... You might want to track voltage by time as well. Calculate instantaneous power and then integrate that. Even though your nominal voltage may be at say 5 v. it will vary from 4.8 to 5.2 as your circuit operates. In fact, with those negative current spikes I would not be surprised at all if the voltage changes. Adding bypass caps as others suggested would help this.

Also, if you do not want to actively monitor 2 sources at once, at least measure the voltage with a dmm, don't take the power supplies word for it.

 

[donpacino]

Also how are you monitering current?

A shunt resistor? A hall sensor?

 

[Tom.G]

Here is a schematic for an analog readout Power Meter. The frequency response is limited by the response time of the LDR and the OpAmp. When powered with the same power supply, the OpAmp must accept common mode input voltage down to the negative supply.
Component values are left as an exercise for the reader.

The LED brightness is proportional to load current and the LDR varies its resistance in response. The LDR, being referenced to the Load Voltage, thus acts to multiply the Load Voltage by the Load Current at the circuit output terminal.
The OFFSET control is to bias the LED on at zero Load Current so a negative current can be sensed.