13 EE362L H Bridge Audio Amp

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EE362L, Power Electronics, H-Bridge as Audio Amplifier

Version November 13, 2008

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The purpose of this experiment is to use your H-bridge inverter as an audio amplifier and analyze

its performance. Step A of the experiment can be performed at any lab bench. Step B requiresyou to use our CD music test station. Sign up for a one-hour time slot for the music test station,

and please be considerate of others who are waiting to use it.

You are welcome to try out the display inverter and music test station anytime. But for purposesof your lab report, you must use your own inverter.

A. Inverter Performance with Input from Sinusoidal Waveform Generator

(perform this step at any lab bench)

1. No DBR power to the inverter yet!2. Make sure that both 10F input and output capacitors are connected to your inverter.3. Make sure that the scope is plugged in through a ground buster.4. Connect a three-headlight load to your inverter output.5. Turn on the benchtop waveform generator. Adjust the settings for 100Hz sinusoidal

output and 5V peak. Note the benchtop waveform generator will display 5Vpp. Itis important that you have no DC offset in the waveform.

6. Either1. Use a BNC-to-wall-wart-plug cable to connect the waveform generator output

to the jack of the inverter ma control potentiometer, or

2. use a BNC-to-alligator cable to connect the waveform generator output across

the outer two terminals of your 500 ma control potentiometer. Note if you

do this at one of the Labview-equipped stations, pay attention to polarity because

the scope and waveform generator chasses may be connected together in back.

7. View the waveform generator output on the scope, confirm 5V peak, and adjust thewaveform generator output voltage if necessary.

8. Plug in the DC wall wart.9. Use the scope to confirm that your triangle wave generator chip is producing a

symmetric triangular wave output. If necessary, adjust the skewness.

10. With ma = 0, use a multimeter to measure the DC value of VGS for each MOSFET.Expect about 4.0Vdc.

11. Use a scope probe to view, one by one, all four VGS waveforms to confirm that yourinverter is firing properly.


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12. With the variac off and its control knob set to zero, connect a variac, 25Vtransformer, and DBR to provide DC power to your inverter. Use the variac to raisethe DC voltage to 35-40V.

13. Connect scope probe #1 and its ground clip to view Vcont, and scope probe #2 and itsground clip to view the inverter output voltage. Use the averaging over one cycle

feature of the scope to denoise the scope trace.14. Adjust the ma potentiometer so that ma is slightly less than 1 (i.e., the fuzzy flat

spot at the maximum and minimum of the sine wave disappears).

15. Measure the rms value of Vcont with a multimeter. Measure Vcont with the samemultimeter. If necessary, adjust the gain of the op amp so that Vcont has the same

rms value as Vcont.

16. Confirm that the H-bridge inverter is working properly and producing a replica ofVcont.

17. Disconnect scope probe #1.18. Using a span of 1kHz, a center frequency of 500Hz, a Hanning window, and an FFT

sampling rate of 10kSa/s, display the FFT of the inverter output voltage (i.e., Channel#2) and save a screen snapshot. For your report, you will estimate the THD of the

voltage waveform using the most significant harmonics (see Appendix).

Waveform

generator

output

Inverter

output

100Hz Test

Dead spots at zero crossings are

characteristic of PWM because of blanking


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19. Raise the benchtop waveform generator frequency to 1kHz, and repeat Step 18. Thistime, for the FFT use a span of 10kHz, a center frequency of 5kHz, and a sampling

rate of 100kSa/s.

1kHz span,

500Hz center

FFT of inverter output with 1kHz input signal

Save screensnapshot #2

10kHz span,

5kHz center

FFT of inverter output with 100Hz input signal

Save screen

snapshot #1


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B. Inverter Performance with Input from CD Player

(sign up for a one-hour slot and perform this portion at the Music Test

Station)

Four 8 speakers in series present 32 to the inverter

495F in series with speakers to block the flow of DC current


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Two small 1:1 audio transformers sum the CD stereo channel outputs to produce a mono

signal. Each transformer is energized by one stereo channel. The two output windings are

connected in series to yield the mono signal.

The CD player is modified to produce a Vcont input signal for our inverters


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You are welcome to use your own music CD if you prefer. The songs on Dr. Gradys CD are

Track 1. The Hollies, Long Cool Woman in a Black Dress

Track 2. Eagles, Peaceful Easy Feeling

Track 3. America, Sister Golden Hair

Track 4. Electric Light Orchestra, Dont Bring Me DownTrack 5. Charlie Dore, Pilot of the Airways

Track 6. The Beach Boys, Help Me, Rhonda

Track 7. The Moody Blues, Your Wildest DreamsTrack 8. Ronnie Milsap, Any Day Now

Track 9. Fleetwood Mac, You Make Loving Fun

Track 10. Steppenwoulf, Magic Carpet RideTrack 11. Roger Miller, Chug-A-Lug

The audio output signal of a portable CD player or radio drives the small speakers inside. Youwill use this signal as the Vcont input for your inverter, drive the four-speaker unit with the

inverter output, and compare inverter input and output waveforms. We use two small audio

transformers to add the two stereo output channels of the CD player to create a mono signal.

The DBR provides the power to drive the speakers. The CD player is used only to provide the

Vcont signal.

20. No DBR power to the inverter yet!21. Connect the mono audio output signal of the CD player to the jack of your ma

potentiometer.

22. Lower the ma control potentiometer to the minimum. Play a song, and observe themono output of the CD player on a scope. Raise the CD volume control to themaximum. There should be no clipping. If clipping occurs, reduce the CD volume

control. See the figure on the next page for an example of clipping.

iPod update

An iPod interface cable is provided so that you can play your iPOD into the CD player,and use the CD player to convert stereo to mono Vcont.

The interface cable connects to the CD players channel A and B auxiliary audio inputs.Use the CD players AUX mode.

Turn Vdc all the way down, turn the CD player volume to max, turn the Vcontpotentiometer to max, and turn the iPod volume to about . View Vcont on the scope.Adjust the iPod volume control until the peaks of Vcont are about 3.5 to 4.0V. Avoid

clipping.


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Audio output of CD player no clipping

With some desktop radios, clipping occurs at about 3V


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23. Connect a variac, transformer, and DBR. Start with zero Vdc.24. Set the variac knob to zero.25. Raise the ma control potentiometer to the maximum. Even with no Vdc, you will

hear weak music because the MOSFET gates are firing.

26. Gradually raise the DBR voltage to about 15V. Simultaneously view the CDplayer output and the inverter output on a scope. You can carefully raise Vdc to40V, but it will be loud! Comment on the audio quality of the amplifier.

27. Borrow the clamp-on ammeter from Dr. Grady or a TA. Select DC milliamps, andpress the meter zero button to zero the reading.

28. Clamp the ammeter around one of the DBR output wires. Measure the current to theinverter while the music is playing.

CD player

output

Inverter

output

Top curve: Audio output of CD player to inverter,

Bottom curve: Output of inverter to speakers

(scope set to average over one cycle)

Save screen

snapshot #3


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29. Multiply your Vdc by the average current to obtain average power to the speakers.Your value should be less than 25W. (Your inverter can easily produce 200W!)

30. Examine the two transfer function graphs near the back of this document. Then,loosen one leg of the inverters 10F output capacitor to electrically remove it,

and listen to the difference. The output capacitor was originally selected for 60Hz

inverter operation and yielded an output filter resonant peak near 5kHz. Without it,the 100H inductor yields a smooth high-frequency roll off, without the resonant

peak.

31. Comment on the difference in sound quality with and without the capacitor. Does thedifference make sense after comparing the two transfer functions?

32. Turn off the equipment and make it ready for the next team.Comments

Your inverter, which is used mostly for motor drives and power-to-grid purposes, is alsoknown as a Class D amplifier. It is very efficient, but the music quality is not as good as

conventional linear amplifiers (which, by the way, have only 50% max efficiency).

Better music quality can be achieved by raising the switching frequency to 100kHz. Ofcourse, higher switching frequency means higher losses.

Class D amplifiers are best suited for powerful bass applications.Extra parts for the student parts bin, screw cabinet, and TA parts bin

Same as for H-bridge labOther items in ENS212

Clamp-on ammeter (Extech #380942, Mouser #685-380942) The audio transformers are Triad Magnetics #TY-145P, 1:1 turns ratio, 600 CT on each

side (Mouser #553-TY145P).


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Appendix. Analysis of Inverter Voltage Transfer Function, With and Without DC

Blocking Capacitor

c program inverter_transfer_function determines the transfer

c function of the inverter with

c Case 1. L,C1 low pass filter and R load

c Case 2. L,C1 low pass filter, C2 dc blocking capacitor, and R load

c

c L in H, C1 and C2 in F, R in ohms

c

real l

complex zc1,zc1_r,zl

complex zc2,zc2_r,zc1_c2_r

pi = 4.0 * atan(1.0)

c

c L is the inverter output series inductor

c C1 is the inverter output shunt capacitor

c C2 is the speaker bank series capacitor

c R is the speaker ohms

c

l = 100e-6

c1 = 10e-6

c2 = 495e-6

r = 32.0

open(unit=1,file="inverter_filter.csv")

write(1,*) "L = ",L * 1e6

write(1,*) "C1 = ",C1 * 1e6

write(1,*) "C2 = ",C2 * 1e6

write(1,*) "R = ",R

do 10 klog = 0,50

freq = 10.0 ** (klog / 10.0)

omega = 2 * pi * freq

c

c transfer function for case 1

c

zc1 = 1.0 / cmplx(0.0,omega * c1)

zc1_r = zc1 * r / (zc1 + r)

zl = cmplx(0.0, omega * l)

h1 = cabs(zc1_r / (zc1_r + zl))

c

c transfer function for case 2

c

zc2 = 1.0 / cmplx(0.0,omega * c2)

zc2_r = r + zc2

zc1_c2_r = zc1 * zc2_r / (zc1 + zc2_r)

h2 = cabs(zc1_c2_r / (zc1_c2_r + zl)) * cabs(r / (r + zc2))

write(*,*) omega,freq,omega,h1,h2

write(1,*) omega,",",freq,",",h1,",",h2

10 continue

stop

end


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Inverter Vout/Vin Transfer Function with 32 Resistive Load

Inverter Filter Performance with 10uF Output Cap

0

2

4

6

8

10

12

1 10 100 1000 10000 100000

Vout/Vin

Inverter Filter Performance without 10uF Output Cap

0

0.2

0.4

0.6

0.8

1

1.2

1 10 100 1000 10000 100000

Vout/Vin


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Appendix. Total Harmonic Distortion (THD)

The THD (in percent) of a waveform is defined as the rms value of the harmonics divided by the

rms value of the fundamental, or

%100(%)1

242322

+++=V

VVVTHD L ,

where 1V is the fundamental frequency component, 2V is the second harmonic, etc. If the

harmonics are given in per unit of the fundamental, a more convenient form of the expression is

%100(%)

2

1

42

1

32

1

2

+

+

+

= L

V

V

V

V

V

VTHD .

When one harmonic, e.g. harmonic k, is much greater than the others, it dominates the sum ofsquares, and the expression becomes

%100(%)1

V

VTHD

k .

Consider the 100Hz scope trace shown in this document as Screen snapshot #1. The 3rd

harmonic (i.e., 300 Hz) dominates the other harmonics and is 36.2dB down from the

fundamental. This yields a ratio 1

5

V

V

of 0.0155. Thus, the (%)THD

is approximately 1.55%.

13 EE362L H Bridge Audio Amp

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