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 60-70% 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
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 60-70% 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
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?
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?
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.
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:
Thank you very much chocolic,
I was planning to use the large caps for giving the very high shoot-through 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 shoot-through currents are unwanted in class-d 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(20-20k).
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
I was planning to use the large caps for giving the very high shoot-through 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 shoot-through currents are unwanted in class-d 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(20-20k).
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
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
If the cap could only supply 0.5A for that period of time you would still get 8W
Your target is 6W
No problem
ChocoHolic, like you said:
C=Q/V
I=Q/t.......Q=I*t
C=(I*t)/V....or...C=(I*dt)/dV
If we assign values we can get a cap value to start and see if it is a practical size.
The Ipk we need is 1.8A...lets make that a constant load.
The deltat (dt) we will use is one half cycle of 20Hz...0.025S
The deltaV (dV) we will assign is 1V.
That gets us C=(1.8*.025)/1
C=45,000uF
So that is a good size cap and we don't need to go larger than that.
It is a good starting point and we can work from there.
Sacrifice the 20Hz to 40Hz and you can get away with 23,000uF
Allow a 2V drop and you can go to 12,000uF.
Not assigning a constant 1.8A and the value can be reduced further.
So somewhere between 10,000uF and 45,000uF.
If the amplifier needs to produce only 6W then even a 5,000uF would suffice.
efevi, I hope that helps
Good luck
C=Q/V
I=Q/t.......Q=I*t
C=(I*t)/V....or...C=(I*dt)/dV
If we assign values we can get a cap value to start and see if it is a practical size.
The Ipk we need is 1.8A...lets make that a constant load.
The deltat (dt) we will use is one half cycle of 20Hz...0.025S
The deltaV (dV) we will assign is 1V.
That gets us C=(1.8*.025)/1
C=45,000uF
So that is a good size cap and we don't need to go larger than that.
It is a good starting point and we can work from there.
Sacrifice the 20Hz to 40Hz and you can get away with 23,000uF
Allow a 2V drop and you can go to 12,000uF.
Not assigning a constant 1.8A and the value can be reduced further.
So somewhere between 10,000uF and 45,000uF.
If the amplifier needs to produce only 6W then even a 5,000uF would suffice.
efevi, I hope that helps
Good luck
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