Efficiency measurement of step down switching converter

[powersys]

I wish to measure the efficiency of a step-down switching converter. I use digital multimeters (Agilent 33401A) to measure the input voltage (Vin) and current (Iin), and also to measure the output voltage (Vout) and current (Iout). I calculate the input and output power as Pin = Vin*Iin and Pout = Vout*Iout respectively. However I found out Pout > Pin, which gave the efficiency more than 100%.

Could someone pls advise what's wrong in the experiment setup? Can we use digital multimeter here? Please advise...

 

[dfullmer]

Hi,

yeah somethings wrong with your measurement. Either that or our energy problems are over. Can you give me the raw data you collected Voltage and current? Also a description of where you measured them.

Glad to help if I can....

dfullmer

 

[Miguel Gaspar]

For this testing you need two wattmeters, one of pass that goes between line and the converter. And another wattmeter of load.

You can´t use a voltmeter due to the reative voltage and current. That's why you got that results.


regards

 

[Vamsi Mocherla]

Did you measure the RMS values of the output current and voltage? Your digital multimeter could not be giving the accurate RMS values of the voltages and the currents at the output

 

[powersys]

Did you measure the RMS values of the output current and voltage? Your digital multimeter could not be giving the accurate RMS values of the voltages and the currents at the output

I used Agilent 34401A 6½ DMM, set it for ac measurement. Is that OK? Thanks.

Added after 1 minutes:

For this testing you need two wattmeters, one of pass that goes between line and the converter. And another wattmeter of load.

You can´t use a voltmeter due to the reative voltage and current. That's why you got that results.


regards

Do we have digital wattmeter in the market that can measure 1W or 2W, or mili-watt? Thanks.

 

[Beowolf]

Try using an 4-ch osciloscope, you will be able to se the waveforms of current and voltage.


Have you attached some large capacitors on both input and outpot? It will slow the changes on voltage, and some aditional inductors will slow down the current so you coud measure it with DMM.

 

[mentorcrk]

not really

 

[vijayanand_ME]

Hi,
I think ur output voltage is increasing with increase in load. This can be viewed n a digtal meter if ur output voltage waveform contains voltage spikes (noisy)
This spikes can averaged by ur meter and hence a increase in the output voltage. Kindly use CRO to probe the output voltage and reduce the spikes if any.

Vijay anand.G

 

[powersys]

Anyone use Tektronik Current Probe TCP202 and Tektronik Scope TDS3034B to measure the input and output current of a buck converter?

 

[JohnJ]

First of all you are going to have a hard time getting real accurate results. This is because the input and output are going to be measured using different equipment and setups, and are subject to totally different error types.

AC line load measurements can be tricky. Obviously, the best thing for your measurements is a AC line meter of some sort. Fluke and others make some relatively cheap. Last I looked, 5% was real accurate for AC line measurements.

To get that accuracy, you need to compensate for the delay difference between the types of probes (current vs. voltage) and the phase shift caused by the series resistance for the current shunt (using an all pass filter). Then you need to scale and pass the compensated signals to a four-quadrant multiplier. Then run that through a RMS to DC converter. Using off the shelf parts including <1% resistors, you may get 10% accuracy. The DC drift on the multipliers (2 of them) alone will eat you alive.

I only mentioned all that to show you that there is no easy solution. You can’t just get the current and the voltage and calculate with any precision.

Try this: separate the line to DC section from the DC-DC converter. Take your DC current and voltage measurements, (leave the input current shunt in while doing the other measurements). Also put a small RC filter on the output to shut out any residual switching spikes. That will give you the DC-DC efficiency. The efficiency of the AC-DC side of the supply could be calculated as those numbers are relatively well known or could be measured by the heat rise of those components.

I’ve been asked to do what you are doing before, a few months later they bought an AC power test set.

Good luck.

 

[electronrancher]

I think your measurement setup is fine - this type of setup (34401) has been used at the best semiconductor companies to accurately measure efficiency. Don't listen to these guys, they obviously have never done it before.

My guess is that you're measuring VIN before the current meter, and there is some drop across the current meter. You need to measure VIN directly on the board, so any small drop across the current meter will be excluded from your VIN measurement.

On the output side, you need to measure the voltage including the current meter so any drop across the current meter is included.

For the best test, you need 4 meters, but you can do it with 3 if you switch the voltage meter between input and output.

Your two current meters must be connected at the same time during the test - you cannot switch one from the input to the output.

VIN -> Imeter -> Board -> Imeter -> Load

This should show input and output current. Next, you need to measure input and output voltages AT THE BOARD not at power supply and load.

Please re-try the measurement with both current meters connected. When you measure input voltage at the board, you will see if there is any drop across the VIN current meter. If the drop is large you may want to use a higher setting to reduce the internal sense resistance, but in general I think the internal sense resistance on a 34401 is pretty small.

 

[powersys]

First of all you are going to have a hard time getting real accurate results. This is because the input and output are going to be measured using different equipment and setups, and are subject to totally different error types.

AC line load measurements can be tricky. Obviously, the best thing for your measurements is a AC line meter of some sort. Fluke and others make some relatively cheap. Last I looked, 5% was real accurate for AC line measurements.

To get that accuracy, you need to compensate for the delay difference between the types of probes (current vs. voltage) and the phase shift caused by the series resistance for the current shunt (using an all pass filter). Then you need to scale and pass the compensated signals to a four-quadrant multiplier. Then run that through a RMS to DC converter. Using off the shelf parts including <1% resistors, you may get 10% accuracy. The DC drift on the multipliers (2 of them) alone will eat you alive.

I only mentioned all that to show you that there is no easy solution. You can’t just get the current and the voltage and calculate with any precision.

Try this: separate the line to DC section from the DC-DC converter. Take your DC current and voltage measurements, (leave the input current shunt in while doing the other measurements). Also put a small RC filter on the output to shut out any residual switching spikes. That will give you the DC-DC efficiency. The efficiency of the AC-DC side of the supply could be calculated as those numbers are relatively well known or could be measured by the heat rise of those components.

I’ve been asked to do what you are doing before, a few months later they bought an AC power test set.

Yes... I agree with you that probably it's too ambitious to get a real accurate efficiency measurement for dc-dc converter. I'm looking for a method that can give reasonable good result, and the result makes sense. I'm aware that we need some sort of compensating circuits to cancel out the delay (due to probes) between the measured voltage and current waveforms so that an accurate efficiency measurement can be achieved. If I'm not mistaken, Lecroy and Tektronik do have such compensators for the task mentioned above. However, the delay, in my opinion, is critical if we want to measure the efficiency or loss at a particular operating point, e.g. switching loss of a power mosfet.

To measure the efficiency of a dc-dc converter, based on my understanding, my approach is as follows:
1) Measure the TRUE RMS values of input voltage Vin and current Iin.
2) Measure the TRUE RMS values of output voltage Vout and current Iout.
3) Pin = Vin x Iin
4) Pout = Vout x Iout
5) Efficiency = Pout/Pin

Please correct me if my approach is not correct.



Added after 23 seconds:

I think your measurement setup is fine - this type of setup (34401) has been used at the best semiconductor companies to accurately measure efficiency. Don't listen to these guys, they obviously have never done it before.

My guess is that you're measuring VIN before the current meter, and there is some drop across the current meter. You need to measure VIN directly on the board, so any small drop across the current meter will be excluded from your VIN measurement.

On the output side, you need to measure the voltage including the current meter so any drop across the current meter is included.

For the best test, you need 4 meters, but you can do it with 3 if you switch the voltage meter between input and output.

Your two current meters must be connected at the same time during the test - you cannot switch one from the input to the output.

VIN -> Imeter -> Board -> Imeter -> Load

This should show input and output current. Next, you need to measure input and output voltages AT THE BOARD not at power supply and load.

Please re-try the measurement with both current meters connected. When you measure input voltage at the board, you will see if there is any drop across the VIN current meter. If the drop is large you may want to use a higher setting to reduce the internal sense resistance, but in general I think the internal sense resistance on a 34401 is pretty small.

I agree with you that Agilent 34401A is one of the best DMM used in semiconductor industries. Thanks for your suggestions on the arrangement of equipments when measuring voltage and current. I did find out when 34401 was set to measure current, it did give some extra loading to the circuit, which was particularly true when the measured current was small.

The specification manual mentions that the current measurement bandwidth of Agilent 34401A is 5kHz (or 10kHz for Agilent 34410A). However, the switching frequency of the dc-dc converter is 100kHz. Therefore, I think Agilent 34401A is not suitable for measuring the RMS current of the converter. The RMS voltage measurement bandwidth is 300kHz. I'm thinking of measuring the current using shunt resistor or current sensing resistor, which gives voltage output. If following your methods, loading problem can be avoided if the equipments are arranged properly. However, if I wish to measure the inductor current, I'm afraid the additional current sense resistor, though it's small, would affect the characteristics of the entire dc-dc converter. Pls advise.

 

[electronrancher]

you don't want RMS values, you want DC values. the setup using 34401's is fine. trust me, i have used the same setup for production datasheets.

The efficiency calculation you proposed is fine. (VOUT*IOUT)/(VIN*IIN) Try it again, and post the measurements if it still looks wrong.

 

[powersys]

You don't want RMS values, you want DC values. The setup using 34401's is fine. Trust me, I have used the same setup for production datasheets.

The efficiency calculation you proposed is fine. (VOUT*IOUT)/(VIN*IIN) Try it again, and post the measurements if it still looks wrong.

Thanks... Great... Your input, as you have do some serious measurements for production datasheets, is and will be very valuable for us here...

The dc-dc converter I mentioned here is a buck converter using LT1074. Based on the simulation results (as shown in the figure below), the input voltage and current waveforms are not constant dc. Therefore, in my opinion, we should find the RMS values of both waveform is we want to calculate the input power. Pls correct me if I'm wrong. Thanks.

 

[JohnJ]

Ha,
I tapped some serious brain power around here and got a workable answer! (You're lucky we're kinda slow in the engineering department right now). I think this will be accurate enough for your purposes.
:idea:
Simple:
-Voltage probe on input voltage, Current probe on input current (duh).
-Set up scale etc. for maximum accuracy.
-Take the peak voltage. At the SAME TIME POINT, take the current. Multiply, convert peak to RMS.
Done! Input power.
That compensates for any phase shift between the 2 measurements.

:!::!::!::!: CARFULL OF WORING ON LINE DRIVEN CIRCUITS! USE AN ISOLATION TRANSFORMER, NOT A VARIAC(w/o isolation)! Or save up enough to fix the scope or the operator :!::!::!::!:

I talked to 2 power supply design engineers and finally our P.H.D. RF engineer came up with that simplification. The rest of us went right past it! It’s easy to miss the simple stuff. :

 

[powersys]

Ha,
I tapped some serious brain power around here and got a workable answer! (You're lucky we're kinda slow in the engineering department right now). I think this will be accurate enough for your purposes.
:idea:
Simple:
-Voltage probe on input voltage, Current probe on input current (duh).
-Set up scale etc. for maximum accuracy.
-Take the peak voltage. At the SAME TIME POINT, take the current. Multiply, convert peak to RMS.
Done! Input power.
That compensates for any phase shift between the 2 measurements.

:!::!::!::!: CARFULL OF WORING ON LINE DRIVEN CIRCUITS! USE AN ISOLATION TRANSFORMER, NOT A VARIAC(w/o isolation)! Or save up enough to fix the scope or the operator :!::!::!::!:

I talked to 2 power supply design engineers and finally our P.H.D. RF engineer came up with that simplification. The rest of us went right past it! It’s easy to miss the simple stuff. :

Are you referring to use osciloscope or 34401A digital multimeter? Thanks.

 

[max0412]

This discussion has been done before and I’ll post the same app note from National.

**broken link removed**

Read pages 23 (52) on.

I also had this discussion with my third year power electronics instructor (for what its worth he has a master’s and several years experience). He seemed to think a very simplistic setup was adequate pretty much contradicting everything he taught us about waveforms i.e. RMS AVG values and when to use what value. What it will boil down to is how accurate you want to be. All methods will introduce instrument error RC loading. Is your SMPS powering a lifesaving device?

True RMS meters are BW limited and also will typically state what types of waveforms they can measure and at a typical accuracy. This IMHO is moot because you want the AVG values. The output V&I measurements can be done with a simple DVM if you’ve done your design properly it should be DC with a small Fsw ripple (small enough to neglect). For the input assuming your buck is getting its input from an upstream converter (DC input) the input voltage can be measured with a DVM. For offline see the app note. This leaves the input current which is pulsed 100Khz. The way I measured the I/P current in the lab was to use 1% tolerance resistors in the milliohm range (this would have the equivalent RLC of extending your cooper trace a bit).I placed one at the drain of my FET ,the current here is the input current in a Buck. So the voltage across that resistor is proportional to the current through it (I=V/R). You can then calculate the AVG value (DC equivalent) of the current waveform. I used a DSO (digital storage scope and differential probes) to capture the waveform once the converter achieved steady state. A high end scope may even be able to do the calculation for you.

 

[phongphanp]

No good solution, if ur need more precision, attribute standardization.

 

[powersys]

This discussion has been done before and I’ll post the same app note from National.

h**p://www.national.com/appinfo/power/files/f5.pdf

Read pages 23 (52) on.

I also had this discussion with my third year power electronics instructor (for what its worth he has a master’s and several years experience). He seemed to think a very simplistic setup was adequate pretty much contradicting everything he taught us about waveforms i.e. RMS AVG values and when to use what value. What it will boil down to is how accurate you want to be. All methods will introduce instrument error RC loading. Is your SMPS powering a lifesaving device?

True RMS meters are BW limited and also will typically state what types of waveforms they can measure and at a typical accuracy. This IMHO is moot because you want the AVG values. The output V&I measurements can be done with a simple DVM if you’ve done your design properly it should be DC with a small Fsw ripple (small enough to neglect). For the input assuming your buck is getting its input from an upstream converter (DC input) the input voltage can be measured with a DVM. For offline see the app note. This leaves the input current which is pulsed 100Khz. The way I measured the I/P current in the lab was to use 1% tolerance resistors in the milliohm range (this would have the equivalent RLC of extending your cooper trace a bit).I placed one at the drain of my FET ,the current here is the input current in a Buck. So the voltage across that resistor is proportional to the current through it (I=V/R). You can then calculate the AVG value (DC equivalent) of the current waveform. I used a DSO (digital storage scope and differential probes) to capture the waveform once the converter achieved steady state. A high end scope may even be able to do the calculation for you.

As I've never designed and tested a buck converter before, at the beginning, I also thought like your instructor, i.e. "... a very simplistic setup was adequate". When I found out the test results were funny, then I went back to simulation and found out the input voltage and current waveforms are not constant dc.

Few very interesting points you made here. I'm also thinking of your method before, i.e. using a current sensing resistor. But I haven't tried yet. Thanks for your sharing. Guess you're now an expert in power electronic design!

 

[max0412]

Guess you're now an expert in power electronic design!

Hardly.
Sorry if I came off as arrogant. That wasn’t my intent.

I was the same as you my efficiency measurements were way off. For example some people in my class were getting 98% efficiency using my instructor’s procedure. This isn’t likely to happen particularly by a bunch of inexperienced students (Dixon, Middelbrock, Cuk maybe). So I did some more research reviewed my calculation. Noting the input current waveform and came to the conclusion that that is the problem. I want the average value of that wave form. So how do I get it? There is no instrument in the lab that will give it to me directly so I came up with the method above. Using that method brought my efficiency to within 3 to 5 % of calculated. Oddly enough that is the way my instructor trained us to take current measurements rather then use a current probe .Why he felt it wasn’t necessary to use in SMP’s I don’t know. This just makes sense when you calculate the efficiency you use the average value of the input current.

input voltage and current waveforms are not constant dc

Yes the input and output will have ripple you will also have spikes & oscillation on waveforms if you want to consider all these in your calculation you will be doing a lot of extra work for a small increase in accuracy.

No good solution, if ur need more precision, attribute standardization.

Regarding the standards the app-note above is written by application engineers from national so I think it’s safe to say they are outlining the standard procedure. If you do more digging you can find more on the subject.