how to measure the power for my design?

[wolfrain]

I have vdd, gnd and vss, and vdd, vss is my dual power supply. At this moment I am doing a filter using several Op Amp in Cadence Virtuoso. What I want to do is to measure the power dissipation for my whole design by measuring the current flowing in vdd in transient analysis and multiplied by the dual voltage (vdd - vss). Actually the current is sin wave and I take the RMS of the current and then multiplied by the dual voltage.

But I think my approach might be wrong, I am not sure.

Any ideas? Or maybe I should do the integral for the product of current and dual voltage in a certain range of time?

Thank you in advance
regards,
wolfrain

 

[radius2]

Hi

Average power is defined as

Pavg = (1/T) * integ(v(t)i(t)) dt (limits: tstart - tstart +T).

Since the supply voltage is constant, aveage power in a ganeral voltage supply
with voltage Vdc is calculated using the definition:

Pavg = Vdc * (1/T) * integ(i(t)) dt (limits: tstart - tstart + T) = Vdc * Iavg

i.e supply voltage times the average current. The rms value of the current
is however defined as:

Irms = sqrt((1/T) * integ(i(t)^2) dt (limits: tstart - tstart +T))

and is not the same as Iavg therefore multiplying by Irms doesn't give correct
result. rms values can however be used if both v(t) and i(t) are the same
type of signals or same types of signals differing by a constant.

Furthermore if you are certain that the same current flows through both
supplies then you can calculate Iavg (I think that there is a function in the
calculator to calculate average values of signals) and then multiply by dual supply.
otherwise the safest is to calculate the power in both supplies separately and add them.

BR

 

[wolfrain]

Hi

Average power is defined as

Pavg = (1/T) * integ(v(t)i(t)) dt (limits: tstart - tstart +T).

Since the supply voltage is constant, aveage power in a ganeral voltage supply
with voltage Vdc is calculated using the definition:

Pavg = Vdc * (1/T) * integ(i(t)) dt (limits: tstart - tstart + T) = Vdc * Iavg

i.e supply voltage times the average current. The rms value of the current
is however defined as:

Irms = sqrt((1/T) * integ(i(t)^2) dt (limits: tstart - tstart +T))

and is not the same as Iavg therefore multiplying by Irms doesn't give correct
result. rms values can however be used if both v(t) and i(t) are the same
type of signals or same types of signals differing by a constant.

Furthermore if you are certain that the same current flows through both
supplies then you can calculate Iavg (I think that there is a function in the
calculator to calculate average values of signals) and then multiply by dual supply.
otherwise the safest is to calculate the power in both supplies separately and add them.

BR

Thank you very much!
I will try it later!


But what do you mean by the same type of signal? If my input is sin wave, then my current is sin wave too. Is it what you mean for the same type of signal? So that in this case can I use RMS ?

 

[radius2]

Hi

Yes exactly, if your voltage and current are both sinusoids then you can show
by using the definition for average power that Pavg = Vrms * Irms * cos(a)
where a is the phase difference between the voltage and current and cos(a) is called the power factor. Then if a=0, the average power is simply Pavg = Vrms * Irms. The 50 Hz signal from the wall outlet is a good example of this.

In your case you want to calculate the power supplied by a a voltage supply which gives constant voltage and I guess a sinusoidal varying current around a DC level (the DC level is the total biasing current for your filter) flows through it. Therefore you need to use the definition and calculate Iavg. Since a sinusoid will have zero in average then your average current will be the DC level i.e. total bias current for the filter if the current sources that give tha bias are ideal. In reality the current sources are implemented with transistors and the bias current will change with the voltage over the current source. Therefore if your signal swings are large, the above mentioned ideal current source approximation might not be accurate.

BR

 

[wolfrain]

Thx a lot!