Ok guys, tell me if I've got it wrong but something just doesn't add up here.
This amp uses current sources for one half of the output.
If the bias current is set to 1.85 Amps, then that is also the peak available output current.
This means the peak power out is IsquaredR. Assuming an 8R load that is 1.85 x 1.85 x 8= 24 Watts peak, =12 W RMS.
So we have an amp that cosumes 1.85 amps with a total supply volts around 80, so total power consumption of around 148 Watts, giving an output of 12 W RMS.
Are we designing room heaters with a byproduct of some nice sound?
This amp uses current sources for one half of the output.
If the bias current is set to 1.85 Amps, then that is also the peak available output current.
This means the peak power out is IsquaredR. Assuming an 8R load that is 1.85 x 1.85 x 8= 24 Watts peak, =12 W RMS.
So we have an amp that cosumes 1.85 amps with a total supply volts around 80, so total power consumption of around 148 Watts, giving an output of 12 W RMS.
Are we designing room heaters with a byproduct of some nice sound?
Hi Carlos,
This is the first e-mail, now that I know it work's, I will send you 3 more
Eric
[image][/image]
The LM I am using are HVK in TO-3 packaging with a 60 volts limit. I am now modifying the amp to create a form of soft start. This will prevent any mishaps in the future. I will keep you guy's updated.
Eric
The LM I am using are HVK in TO-3 packaging with a 60 volts limit. I am now modifying the amp to create a form of soft start. This will prevent any mishaps in the future. I will keep you guy's updated.
Eric
I can not explain it yet, but the math you are using is not applicable. In the article the guy shows a result on the scope of 80W with a voltage rail of + & - 42V and 1.82 amp. Some how it seems to me that about 50% of the energy going trough is transformed in output power. The other articles of similar design shows the same. If you can decipher French I can send you the same e-mail I sent to destroyer.
Regards,
Eric
yes, and it gets better, if you use resistor loads rather than CCS in the tails. This gives twice as much room heating for the same audio output.
Push Pull ClassA, by comparison, only gives half the room heating capability for the same audio output.
Choose you amplifier to suit how far away from the Equator you want to listen.
Andrew T has the right power calculation. It is ‘in theory’ operating about 54W. Eric is running the idle current and voltage rails a bit lower for now until some circuit mods get done to stabilize the startup. Still running in class A on the output stage, as it is a single ended O/P. It can’t go into Class A/B but it could clip the signal at full cutoff or saturation if the signal gets to large.The amp is using the CCS only to sink DC current to establish an operating point. The parallel MOSFETS are modulating/drive the AC current component (along with the DC idle current) The AC component makes it to the speakers…the DC current to the active sink (they do absorb a bit of AC as well, the regs supposed to reject about 70 dB of the signal.. but will need to see how much)
Peak signal current is double the idle current for single end class A: 2x1.85= 3.7A pk
3.7A pk x0.707= 2.61 A rms
For an 8 ohm load max power(rms) is I rms^2 x R gives (2.61^2)x8=54W
Peak signal current is double the idle current for single end class A: 2x1.85= 3.7A pk
3.7A pk x0.707= 2.61 A rms
For an 8 ohm load max power(rms) is I rms^2 x R gives (2.61^2)x8=54W
could you explain this.
I thought for SE ClassA the max output current = bias current and that for Push Pull ClassA the maximum ClassA output current = 2times the bias current.
The only way you can get 2 x idle current in SE mode is if you are coupled to the load via an output transformer.
With a current source peak output current is the same as standing current, so back to 12 W RMS.
With a current source peak output current is the same as standing current, so back to 12 W RMS.
Sorry, my calculator is stupid, designed for VAT calculations so it rounds off small amounts.
Correct recalculation is now 13.7 W RMS.
Correct recalculation is now 13.7 W RMS.
Maximum usable current (for AC Load) in an SE amp stage:
The ''Q'' point on the AC load line for the MOSFETS in this case is 1.85Adc.
(They each take half of that, but lets just treat the paralleled Mosfets as a single element.) That means for a sine wave signal, the ''negative going half cycle'' going from Q point to full cutoff on the Fets is the lower limit before clipping, and on the ''positive cycle'' it is double the quiecsent current, staying in Class A. So from the signal (AC) point of view we look at max AC load power (in this case we are looking at RMS power into 8 ohm) and in terms of Ac rms current it is defind by (I peak x 0.707)squared x 8 ohm
But as you mentioned the real test by measuring the max unclipped peak-peak voltage available to the load: PL max= Vpp squared/8xRL
or using an RMS Voltmeter: PL=(Vrms) squared / RL
( both ref Malvino, Electronic Principles)
As you said with +/-37V rails maybe the maximum signal at 8 ohm load would be 34Vpx2=68 Vpp that could give up to about 72Wrms, right now it is not that high a peak to peak though because of the lower Q current setting decreases the ac output compliance.
The ''Q'' point on the AC load line for the MOSFETS in this case is 1.85Adc.
(They each take half of that, but lets just treat the paralleled Mosfets as a single element.) That means for a sine wave signal, the ''negative going half cycle'' going from Q point to full cutoff on the Fets is the lower limit before clipping, and on the ''positive cycle'' it is double the quiecsent current, staying in Class A. So from the signal (AC) point of view we look at max AC load power (in this case we are looking at RMS power into 8 ohm) and in terms of Ac rms current it is defind by (I peak x 0.707)squared x 8 ohm
But as you mentioned the real test by measuring the max unclipped peak-peak voltage available to the load: PL max= Vpp squared/8xRL
or using an RMS Voltmeter: PL=(Vrms) squared / RL
( both ref Malvino, Electronic Principles)
As you said with +/-37V rails maybe the maximum signal at 8 ohm load would be 34Vpx2=68 Vpp that could give up to about 72Wrms, right now it is not that high a peak to peak though because of the lower Q current setting decreases the ac output compliance.
The circuit is a source follower (or common drain) amplifier, and the maximum AC output compliance (Vpp) is determined on the AC load line by two parameters:
PP=2xIdq x re where re is AC load (8 ohm in this case), and Idq is the quiescent current or the DC drain current
PP= 2xVdsq where Vdsq is the DC drain to source voltage.
So for an AC load on a source follower the peak output current limit is 2 times the idle (quiescent) current. The constant current devices are only used to sink DC current in this case, the AC component is free to go higher or lower.The MOSFET drains are connected directly to the positive rail.
Sorry to disappoint you Shanx, but not in this case. The current source is set to run at 1.85A so that is the peak current available to the load.
we give a strong hint and rather than go and check from another source this one decides to argue.
On what basis does SE ClassA magically produce double the current sunk into the resistor or CCS load?
On what basis does SE ClassA magically produce double the current sunk into the resistor or CCS load?
Actually I double checked that (My bad number) and you are right, toprepairman. The DC current component must be subtracted from the AC Peak current, as that current is what is available to pass to the AC load, it is at half, so Peak current 1.85Apk. (Pk-Pk would have been 3.7A).
Right formulas...my wrong arithmetic!!
So PP=2x1.85Ax8 Ohm=29.6 Vpp so PLmax=(PP)Squared/8x8 Ohm=13.7W That is the power limit on 8 ohms with the present bias.
Based on voltage presently available of about +/-37, the other max compliance term is say 2x34V=68Vp-p for a PLmax=(68)Squared/8*8= 72 W..
Question:to get into that 70W range about 4.2A DC bias required. Is that right?
Right formulas...my wrong arithmetic!!
So PP=2x1.85Ax8 Ohm=29.6 Vpp so PLmax=(PP)Squared/8x8 Ohm=13.7W That is the power limit on 8 ohms with the present bias.
Based on voltage presently available of about +/-37, the other max compliance term is say 2x34V=68Vp-p for a PLmax=(68)Squared/8*8= 72 W..
Question:to get into that 70W range about 4.2A DC bias required. Is that right?
No arguments mate..just trying to figure out where the amp design stands.. and hopefully to help Eric in evaluating/improving the amp build to reach his goal.
yes and to get that current, output Voltage needs to be ~33.5Vpk.
Total stage quiescent dissipation ~310W, very close to the theoretical 25% efficiency when delivering maximum power, but only 0.226% efficiency when operating at -20dB of average power (ref maximum).
From what i recall of the article, the amp is supposed to dissipate around 70W per channel.
Many electronics magazine writers aren't the most tech oriented, those of LED and the German ELRAD rank just about at the bottom of the list. Mind you, i've seen calculation errors by MSc degree 'publishers', one of which is a member of this club, anyone can have a blind spot sometimes.
Many electronics magazine writers aren't the most tech oriented, those of LED and the German ELRAD rank just about at the bottom of the list. Mind you, i've seen calculation errors by MSc degree 'publishers', one of which is a member of this club, anyone can have a blind spot sometimes.
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RMS Rating
Have a question for all of you out there. If an Amplifier is rated at 75Watts at (1kHz) what would this mean as far as RMS ratings? Is it relative and to what?
Thanks, Francis
Have a question for all of you out there. If an Amplifier is rated at 75Watts at (1kHz) what would this mean as far as RMS ratings? Is it relative and to what?
Thanks, Francis
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