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Class D Switching Power Amplifiers and Power D/A conversion 

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10th January 2013, 07:23 PM  #1 
diyAudio Member
Join Date: Jan 2013

Audio Amplifier, Current Limiting Voltage Supply problem
Hello everyone,
In the analog electronics project, we are required to build an audio amplifier. The restrictions are:  IC's will not be used.  At least 6 watts will be delivered to 8 ohms speaker. I have made 2 designs, one class ab and one class d amplifiers, in simulation everything works fine, i can get pure sinusoids and upto 100 watts on speaker, with +25 double supply rail and, bridge configuration to double the speaker voltage if necessary. However there is one little problem: The power supply that i use is Agilent E3631A, which gives max +  25 volt limited by 1 A current for each of supplies. This limited current here means that (i suppose): The power cannot give more than 1 Amps in any instant of time, which will clip the current seen on power supply, and on the rest of the circuit at 1 Amps, if the project is not designed considering this fact. Therefore, in order not to clip the audio signal, I need to pass at most 1 amps sinusoid across speaker, which will give (1/sqrt(2))^2*8=4 watts. Now since i use +25 supplies the 1 amp dc means 50 watt is driven from the supply. Where the rest of power goes is : 1. the driven current is not dc but ac with lower effective value, which means the driven power is much less. 2. in class ab the practical efficiencies are around 6070% if and the rest is dissipated on the rest of circuit. I think class ab cannot do much about this current limiting. I am not sure about class d. I would be very thankful if you could have some comments. Am i correct on these, and may there be anyway to get through these? In addition, i cannot model the power supply such that it limits currents over 1A, in spice. How can i do that? Thank you for reading 
10th January 2013, 08:08 PM  #2 
R.I.P.
Join Date: Nov 2003
Location: Brighton UK

Hi,
Why use the power supply if its clearly not suitable ? Why use a supply with current limiting when clearly what you want with an audio amplifier is high peak current ? rgds, sreten. (Look up crest factor for music signals.) Last edited by sreten; 10th January 2013 at 08:21 PM. 
10th January 2013, 08:33 PM  #3 
diyAudio Member
Join Date: Jan 2013

it is because there is only this supply in the analog electronics labaratory, we can only use this. Therefore i am wondering if i can give more than 4 watts to the speaker with this supply..

10th January 2013, 08:46 PM  #4 
diyAudio Member
Join Date: Dec 2003
Location: Munich

...you could consider to tease the E3631A with huge rail caps...

10th January 2013, 08:51 PM  #5 
diyAudio Member
Join Date: Jan 2013

sreten,
Because, it is the only available power supply in labaratory. I don't want to buy high watts adaptor, if i can find a solution to the problem. chocoholic, Do you mean bypass capacitors connected between supplies and ground, by "teasing the supply" and "huge rail caps". Can you explain more or can you give a website's link for me to search? 
10th January 2013, 10:29 PM  #6 
diyAudio Member
Join Date: Mar 2005
Location: Quebec

With a class d amplifier, you should reach near your 100W... Output current is always more than supply current.

11th January 2013, 10:53 AM  #7 
diyAudio Member
Join Date: Dec 2003
Location: Munich

Hi efevi,
using large bybass caps between supplies and GND are the most traditional method to keep a voltage stable during high peak currents. This is valid for high charging pulses like behind commonly used transformers+rectifier bridge+caps, but also for high discharging pulses. I have no edu link, but most likely you will find something by searching on own. Basically it all comes back to the fundamental differential equation of a cap. i_cap(t) = C *dv/dt which you can also write as dv/dt = i_cap(t) / C Let's have a look to your application: T=Duration of period ( i.e. 10ms for a 100Hz sinewave) i_load(t) = Imax*sin(2*pi*f*t) i_labsupply(t) = 1A whenever the voltage is lower than the intended supply voltage. i_cap(t) = charging current of the cap For a class ab amp the charging current of the cap is the difference between the current which is being delivered by the lab supply and what is consumed by the amp. i_cap(t) = i_labsupply(t)  i_load(t) (For a class D amp it is more complicated and you would have to calculate the currents according the ratio of supply voltage and output voltage and power conversion efficiency...) If you did not learn to calculate with differential equations up to now, you can simplify it by a stepped approximation. Cut your signal period into 10 time slices. Means 5 slices per half wave. Let's have a closer look to the positive supply. Calculate the load currents for t0= 0 ==> giving you the value for the interval  1/20T < t < + 1/20T t1 = 1/10T ==> giving you the value for the interval 1/20T < t < 3/20T t2 = 2/10T ==> giving you the value for the interval 3/20T < t < 5/20T t3 = 3/10T ==> giving you the value for the interval 5/20T < t < 7/20T t4 = 4/10T ==> giving you the value for the interval 7/20T < t < 9/20T t5 = 5/10T ==> giving you the value for the interval 9/20T < t < 11/20T (will be zero again) During negative half wave the positive supply is not loaded and all supply current is available to recharge the cap. ...same game vice versa for the neg supply.... Now you can calculate with constant currents within each time interval and use the non differential equation below. delta V / delta t = I_cap / C This tells you that the voltage across a cap does not jump, but changes slowly according its capacitance and the applied currents. ...you have a current limited voltage source, which charges the cap. ...and your amp, which discharges the cap. During the times, when the amp draws more current than the supply can deliver, the voltage will decrease. During the times, when the amp draws less current than the supply can deliver, the voltage will increase again. The lower your frequencies are and the less sagging of the supply voltage you want to allow, the larger caps you will need. P.S. Above equations neglect the idle current of the amp, which slightly increases the currents that have to be delivered from the lab supply. Last edited by ChocoHolic; 11th January 2013 at 11:10 AM. Reason: .Correction of typo in : i_load(t) = Imax*sin(2*pi*f*t) 
12th January 2013, 06:54 PM  #8 
diyAudio Member
Join Date: Jan 2013

Thank you very much chocolic,
I was planning to use the large caps for giving the very high shootthrough current impulses reflected to the supply in class d amp design. Since these are impulses(very short time discharge of capacitors), i thought that large enough caps can effectively deliver shoot through currents, and maintain circuit operation (although to say this is odd , since shootthrough currents are unwanted in classd design, and has to be kept small as possible). I did not think that these caps can be used in class ab stage to provide much longer discharge currents to the amp. However considering that supply does not give current for one half cycle(charging the capactior fastly), and since that at the other half cycle the capactior will begin its duty only after source cannot give 1A, this idea is very reasonable. I made the roughest estimates and reasonable sizes of capacitors can do the job. And in the simulation it does not cause problems dor audio frequencies(2020k). However thinking that i am not going to give a sine wave after finishing the design but a complete acoustic signal, i do not know about its effects yet. However i put two sinusoidal sources with differenct frequency and delay in series, and results seems okay. Thank you very much again 
12th January 2013, 10:03 PM  #9 
diyAudio Member
Join Date: Jan 2003
Location: mississauga ontario canada

Based on your amplifier using 100mA for Iq and everything else, and a capacitor big enough to supply 1.8A for the top half of the lowest frequency, and a peak output swing of 21.5V @ 2.7A, you should be able to get 28.9W before supply sag or clipping.
If the cap could only supply 0.5A for that period of time you would still get 8W Your target is 6W No problem
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13th January 2013, 09:18 AM  #10 
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Join Date: Dec 2003
Location: Munich


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