# Calculating efficiency - voltage drops and AC current

Status
Not open for further replies.

#### ACharnley

##### Member
Perhaps a silly question but...

I have two comparable circuits which take AC in and give DC out. The AC in is a small transformer which gives out 7.9V AC RMS and which drops to 7.30 with a load attached. The circuits give out 5.00V, though one is 4.94 and drops to 4.89 under load.

Should I be measuring the voltage drop values for the efficiency calculation?

#### ACharnley

##### Member
Another question, the AC in, should I be using 1.41/peak or the RMS? If I use peak the efficiency seems way off, however both will be rectifying the voltage so it seems right to give them the DC equivalent?

#### alec_t

##### Well-Known Member
Efficiency would normally be given by (power out)/(power in). This would require a knowledge of both voltage and current for the input and output.

#### ACharnley

##### Member
Yep, but if it's AC in and DC out, should I convert the AC to DC (x 1.41) for the calc?

#### alec_t

##### Well-Known Member
should I convert the AC to DC (x 1.41) for the calc?
That doesn't give you DC; it gives you the peak amplitude derived from the AC RMS value if the AC is truly sinusoidal. For the efficiency calculation you also need to take into account voltage drop in rectifiers and internal resistances. Use the AC RMS input power value.

#### ACharnley

##### Member
Yeah I'm forgetting that the power in is still RMS*I regardless of whether the diodes convert it. Makes sense. Cheers, A.

#### {XtremeDry}

##### New Member
I have a question, Can i determine the input current of AC by using simple ohm's law I=V/Rinternal. so that i can calculate the power of input in order to calculate for efficiency.

#### ChrisP58

##### Well-Known Member
I have a question, Can i determine the input current of AC by using simple ohm's law I=V/Rinternal. so that i can calculate the power of input in order to calculate for efficiency.
Only if the resistance that the AC input sees is the only thing that defines the input power. With a purely resistive load, that would work.

But is you're talking about a power supply, then the input power will NOT be defined by the input resistance. You need to actually measure the RMS values of the both the input voltage and the input current.

#### MrAl

##### Well-Known Member
Hi,

Measurements of rectifier circuits is a little tricky because the diodes only conduct for part of the time, and that could be a very small part of the time. This means a measurement of the voltage alone is not enough.

The exact way is to multiply each voltage point in time with each current point in time, sum the results over one cycle, then divide by the time of one cycle. This can be done using a scope or using an ADC that is fast enough to sample both current and voltage. Because most power supplies operate at 50Hz or 60Hz the ADC really doesnt need to be super fast. 100 points would only require about 6000 samples per second. That's cake in the ADC world.

Alternately you could estimate the current waveshape as a pulse (as viewed on a scope) and go from there.

These kinds of problems are often not very intuitive, in that they are not as simple as they look sometimes. A test tells us for sure if we can use the Vrms*Irms approximation. This test is simply a test to see if the square of the integral of v*i is equal to the product of the integral of v squared times the integral of i squared.

If the following equality is satisfied, then we can use the Vrms*Irms method (Tp is total period):
integrate(i(t)*v(t),t,0,Tp)^2=integrate(i(t)^2,t,0,Tp)*integrate(v(t)^2,t,0,Tp)

If that is not satisfied, then the accuracy will depend on how far off the right hand side is from the left hand side, so if we introduce a factor A^2:
A^2*integrate(i(t)*v(t),t,0,Tp)^2=integrate(i(t)^2,t,0,Tp)*integrate(v(t)^2,t,0,Tp)

If A is between (0.99) and (1.01) for example, then the accuracy is within 1 percent. If A is between (0.98) and (1.02) then the accuracy is within 2 percent, and note we look at A not A^2 which is actually in the formula. You see how this works

In reality though we often deal with waveforms that are only non zero for part of the total time period Tp, so we really have:
A^2*integrate(i(t)*v(t),t,0,Tp)^2=integrate(i(t)^2,t,0,T1)*integrate(v(t)^2,t,0,T2)

where the limits of integration have been changed to show the real limits, and that is assuming the waves both start at t=0 (to make it different than that would require a more complicated expression which would only make it harder to comprehend what the difference is).

Some expressions work out very well because the waveshape is compatible with the Vrms*Irms approach. This includes complete sine waves. For two sine waves in the first equality above the two sides always equal each other exactly so that always works for sine waves as long as they are not interrupted. In the second expression T1=Tp and T2=Tp, and so A=1 which means perfect equality.

Pulse or nearly pulsed waveforms are tricky so it helps to have a test to make sure the Vrms*Irms approach is applicable.

Last edited:

#### KeepItSimpleStupid

##### Well-Known Member
Power in AC circuits get really, really tricky especially if the power supply is switching. Your dealing with non-sinusoidal currents and hopefully power factor correction (PFC). PFC tries to average the current consumption.

For purely resistive loads; P=V*I is supposed to work when Vac is measured in RMS (Root Mean Squared). Most meters ASSUME the input is a sine wave, average it and multiply it by a fudge factor to get RMS. When the waveform changes or the frequency is out of range, all bets are off.

There was a thermal sensor available at one time which used temperature to measure the RMS voltage.

#### Tony Stewart

##### Well-Known Member
Unregulated supplies deliver (quasi-square) pulse currents to caps where the duty cycle is proportional to the % ripple current therefore using peak current you can expect low efficiency as losses are squared with current and not linear.

You can compute load regulation % error and thus from linear load compute effective ESR of source where a pulse ESR impedance is raised by inverse of duty cycle. e.g transistor, diode conductance or ESR or RdsOn is 10x if on 10% of time giving V ripple approximately of 10%. and 1% ripple demands much lower ESR devices and transformer secondary DCR.

This is why SMPS work much better as the duty cycle is high for reasonably low ripple at full load.

Load Regulation error is the effective impedance ratio of source:load. Thus 1% load regulation requires source equiv impedance to be 1% of load.

#### KeepItSimpleStupid

The LT1088 is about $175.00 USD currently, surplus. #### Tony Stewart ##### Well-Known Member Most Helpful Member LT1088 was designed for RF power measurements converting AM power to RMS power by regulating a DC power to match and thus Vout was fairly linear and true RMS by equal heat losses in matched current shunts. Since unregulated power sources only consume power by virtue of the relatively constant fixed ESR when the diode is saturated and the Cap ESR losses, one expects a square edge pulse with slightly sine shaped peaks as with 10% ripple, the peaks only are sine modulated .10%. Since the average voltage using an RC filtered current sense would measure the same as RMS is a pure rectanuglar pulse it will read slightly off by 5% or so. A DVM uses the rectifed average method but assumes a sine wave and thus convert Avg rectified sine to RMS again slightly off. You should assume your transformer W/VA efficiency is only 70% max depending on duty cycle or %ripple because the peak current is inverse with %ripple and transformer inductance loss from saturation during peak currents beyong VA rating result in further coupling losses added to eddy current losses and ohmic loss for a defined hot spot temp rise. Transformer VA ratings only equal W ratings for pure linear resistance loads. after linear regulation from 7.4 to 5V or 63% effic. your net efficiency will be <50% , typical for a poor linear design. The better choice is a SMPS. #### MrAl ##### Well-Known Member Most Helpful Member The LT1088 is about$175.00 USD currently, surplus.
Hi,

Sounds like old world technology before ADC's became so fast and widely available.
The RMS value determines the power heating level, so why not test directly

Heat one resistor with the voltage to be measured, heat another resistor with a known DC current, measure temperature of both resistors. When T1=T2 the DC current and voltage can be used to calculate the power in both resistors and therefore the RMS value.
That's a really old way of doing it, but i dont know if anyone still does it or not due to the ADC usage.

#### Tony Stewart

##### Well-Known Member
HP design their RF power meters this way still sold today for beaucoup bucks.

#### KeepItSimpleStupid

##### Well-Known Member
That chip had a 100 MHz frequency response.

#### Tony Stewart

##### Well-Known Member
Since this a primitive unregulated charger with a linear regulator, accurate prediction is possible and only needs a BW of a few kHz to contain the 1st dozen harmonics which would be adequate using an AC Amp DMM with small calculated correction factors probably beyond the scope of this question.

The more important question is poor DC load regulation for the one with 4.89 V with a drop of 50 mV for some load R and I thus can be attributed to high ESR series regulator in Ohms divided by internal loop gain. Thus the 1% drop from load regulation error implies the series regulator has and equivalent ESR of Rload *1%. Lower is better.

Ground and wire resistance are also factors which affect load regulation outside the loop unless remote sensing is used, which is common even universal laptop chargers using SMPS with at least 4 wires to the DC plug.

Last edited:

#### ACharnley

##### Member
Some interesting replies, but for my own question (before the hijacking!) can I verify that when comparing the efficiency of two devices connected to the same power supply (a transformer), I should be using the power supply voltage with no load on it, and the device voltage with the load connected.

Thanks A.

#### Tony Stewart

##### Well-Known Member
Efficiency is defined as DC output load power(VI)/RMS ac input power (not AC VAR which is higher)

Thus with no load, efficiency is zero, 0, while input AC excitation current creates some losses,

The best case efficiency is at max load, which is also max ripple and your design is <50% efficient best case.

So to minimize ripple to 10% choose Load R*C= 5T where for 50 Hz, T = 10ms and Rmin load= Vout/Iout max, then choose tighter Vac to Vdc values, then 70% efficient is possible or more with huge low ESR cap xx mF.

Your design probably has 30% ripple or RC=T, is my guess, so you needed higher Vac out to avoid dropout on poor 2V + LDO, which is old school design of 50 years ago.

What is your RC value? and dropout value at max I?

Last edited:
Status
Not open for further replies.