# Power Calculations

#### garimella

##### Full Member level 4
Elementary general question regarding power budget calculation. In a circuit, Can we algebraically add rms current and dc quiscent current and arrive at the total power? Is this correct approach? or Alternatively should we compute RMS and DC power separately and then finally add them together

#### KlausST

##### Super Moderator
Staff member
Hi,

As long as they don't correlate in frequencies:

V_tot = sqrt(V_RMS^2 + V_DC^2).

Klaus

#### FvM

##### Super Moderator
Staff member
Can we algebraically add rms current and dc quiscent current and arrive at the total power?
Without additional specifications, RMS current already includes the DC component. Respectively P = Irms² * R is the total power. In special cases, one might specify an ACrms current value, measured with the DC component stripped off. We get Ptot = Pac + Pdc = (Iacrms² + Idc²) * R

Alternatively should we compute RMS and DC power separately and then finally add them together
There's no thing called RMS power. We have average or continuous power, peak power. Or may be, AC and DC power.

#### Easy peasy

This bit don't work: " We get Ptot = Pac + Pdc = (Iacrms² + Idc²) * R "

consider the case where the trough of the AC is the same magnitude as the DC

you have to compute the true rms of the current over a period = the cycle of the AC, then P = I^2 R, where I = true RMS.

#### FvM

##### Super Moderator
Staff member

Ptot = Pac + Pdc = (Iacrms² + Idc²) * R is just a special case of Ptot = (Idc² + I1² + I2² + I3² ....)*R. Idc, I1, I2 , I3 being orthogonal components, e.g. harmonic currents. The statement is universally true.
--- Updated ---

Obviously rms quantities have to be measured over full signal periods.

Last edited:

#### c_mitra

consider the case where the trough of the AC is the same magnitude as the DC

Can't consider such cases. It does not make sense.

An arbitrary periodic waveform can be decomposed into an AC and a DC component. The AC component averages zero over a complete cycle but the DC is a constant term (non-zero).

The AC term can be further decomposed into sinusoidal terms but that is beyond the point.

For example, a rectangular pulse train, 0-1V, is not AC; it is AC + DC. Once you remove the DC part, it becomes -0.5 to 0.5V pulse train.

Take an example: a square pulse 0-1V (duty period is 0.5). The RMS value is sqrt(0.5).

It has a DC component 0.5V and an AC component of -0.5 to +0.5V. The RMS value of DC component is 0.5V (no need to calculate).

The AC component has RMS value of sqrt(0.5*0.5)= 0.5V; The rest can be worked out easily.

#### KlausST

##### Super Moderator
Staff member
The AC component has RMS value of sqrt(0.5*0.5)= 0.5V; The rest can be worked out easily.
A sinusoidal waveform with peaks at +0.5V / -0.5V has an RMS value of about 0.35V.

Or did I miss something?

Klaus

Staff member

#### c_mitra

Or did I miss something?

I did mention that we consider a square pulse train (duty cycle 0.5).

Square waveforms are easy to handle.

The average (arithmetic mean) of any periodic waveform over a complete period is equal to the DC component.

This is the constant term that appears in the Fourier transform (corresponding to freq zero = DC).

After we subtract the const term, we are left with AC part...

#### KlausST

##### Super Moderator
Staff member
Hi,

Oh, now I see. Now it makes sense.

Klaus

#### garimella

##### Full Member level 4
consider a simple case, where a current amplifier is feeding 100mA p-p to the load with R=10 ohms @ 1KHz sine. Now the amplifier draws quiscent current say 1mA on +/-15V supply. To calculate power, Pac= irms^2 *10 and Pdc = +15V*1mA (and -15V*1mA). My confusion is that PDC is on +15V line and Pac is related to load. Can we add them to algebraically to get the total power consumed by the amplifier

#### c_mitra

My confusion is that PDC is on +15V line and Pac is related to load. Can we add them to algebraically to get the total power consumed by the amplifier

You have not explained the source of confusion. Yes, power is energy (rate) and is additive. How does the source of voltage matter in the final calculation? You can of course specify, to be explicit, the power consumed by the load and the power consumed by the amplifier.

#### KlausST

##### Super Moderator
Staff member
Hi,

For a DC supply you don't calculate with RMS at all. Just average.
P = V × I, the above is generally true when V or I is fix: power supply, voltage drop on a diode.

For linear loads (ohmic, or complex) where V and I relate on each other one needs to calculate with RMS.

Example:
To do the power loss calculation on an SCR you need both. Because when they are ON, they cause a constant voltage drop combined with an ohmic voltage drop (the one relates on current the other not).
The total power dissipation is the sum of Pc and Pl where Pc = V_const x I_average and Pl = I_RMS ^2 × R

Klaus

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