I don't have Slone's book nor have I read it.5th element said:
From the excerpts paraphrased in this Forum, I would never trust anything that Slone says you should do.
How do transient peaks pass the protection?
Can transient peaks exceed the DC protection level?
Can you check the 4A current level that Mega is referring to?
If it is the same schematic as I linked to then changing that resistor will only affect current limit at higher output voltages - it's the slope of the locus that is changed. Increasing the resistor will make the available current near rails increase but around 0V out there won't be any change.
A resistive dummy load doesn't demand any current around 0V out, but a speaker does - this is why it can be fine with the dummy load but not with the speaker (see my previous post).
A resistive dummy load doesn't demand any current around 0V out, but a speaker does - this is why it can be fine with the dummy load but not with the speaker (see my previous post).
megajocke said:
It's a pair of Peerless XLS 830452s connected directly to the amps output. They have a DCR of 3.4 ohms each, the impedance being at its lowest of around 5ohms. these are connected in parallel.
In the test above I used a 40 or 50hz sine wave, giving a load impedance of around 6.5 or 4.5 ohms.
It's too late to actually test anything with the driver at this time, simply because it gets very loud. But if I am remembering correctly, I believe it's only the +ve or -ve side that actually causes problems.
Meaning I could disconnect the +ve sides protection circuitry and the entire problem would vanish.
And connecting the +ve side on its own, creates the same problem.
Why would one side be triggered and the other left alone? I've heard about phase angles being positive or negative, would this be a case of that?
AndrewT said:
If by transient you are referring to something like a bass drum, from memory I don't think it makes much difference at all.
I can have the amp at 90% power say and then along comes the 101% thump and there's the sawing. Or I can have it cold and give it the 101% thump and there's the sawing again.
I can test this in the morning though.
What do you mean by this? There is no protection in the amp vs DC offset. Or are you asking if a short burst of power can exceed the clipping level imposed by the protection circuitry? And if so, do you mean when driving a dummy load, or the loudspeaker?
And to do this I measure the RMS output current of the amplifier whilst driving the loudspeaker?
Hi,
transient protection trigger level vs DC or short circuit protection trigger level.
They should be different to get the best from the capability of the output and driver transistors.
Suppose the short circuit current limits at 5A.
The transient current could limit at 10A for 10ms and 15A for 100us.
The transient limits would allow other than test sinewaves to feed test resistor loads.
The medium to long term protection, to prevent the heatsinks and outputs rising in temperature, are the F3.1A fuses in the supply rails for the example above.
Thers'sa bit in PASS and in ESP on test signal for measuring short term current limiting of the output stage, without damaging the devices nor overheating them. I think the test load is a 0r1 resistor.
transient protection trigger level vs DC or short circuit protection trigger level.
They should be different to get the best from the capability of the output and driver transistors.
Suppose the short circuit current limits at 5A.
The transient current could limit at 10A for 10ms and 15A for 100us.
The transient limits would allow other than test sinewaves to feed test resistor loads.
The medium to long term protection, to prevent the heatsinks and outputs rising in temperature, are the F3.1A fuses in the supply rails for the example above.
Thers'sa bit in PASS and in ESP on test signal for measuring short term current limiting of the output stage, without damaging the devices nor overheating them. I think the test load is a 0r1 resistor.
Mega it looks like you were bang on with your assessment. The ugly noises start when the amp is asked to deliver around 4amps RMS into the loudspeaker load.
This happens when either of the N or P channel protection is in place.
Is there a simple way to increase the current this can pass under these operating conditions?
This happens when either of the N or P channel protection is in place.
Is there a simple way to increase the current this can pass under these operating conditions?
Am I right in saying that the amount of current the protection circuitry will allow to pass at zero crossing, is the same the protection circuitry will limit at under short circuit? And that this is hard set by the value of the RE resistors and that lowering their value would increase this?
The DCR resistance of the paralleled drivers is very low at 1.7 ohms.
To calculate this I take it we assume the amplifiers output power is ~220 watts and that the driven load = 1.7 ohms?
Resulting in a peak current of 16 amps. Half of this = 8 amps.
By the looks of things, if I lower the value of the RE resistors down to half their original value, this will allow the 8 amps to pass.
Then I assume that I would just alter the values of the other resistors in the protection circuitry, so that the ratios between the voltage dividers are the same as before?
Something tells me there's more to it then that.
The DCR resistance of the paralleled drivers is very low at 1.7 ohms.
To calculate this I take it we assume the amplifiers output power is ~220 watts and that the driven load = 1.7 ohms?
Resulting in a peak current of 16 amps. Half of this = 8 amps.
By the looks of things, if I lower the value of the RE resistors down to half their original value, this will allow the 8 amps to pass.
Then I assume that I would just alter the values of the other resistors in the protection circuitry, so that the ratios between the voltage dividers are the same as before?
Something tells me there's more to it then that.
Hi,
if you find the existing circuit limits at 4Apk then halving the value of Re will give ~8Apk.
This will not drive a 1r7 resistor to 220W.
It will provide 8*8*1.7/2=54.4W into 1r7 sinewave power with 8Apk.
It will drive a 4r0 load to 8*8*4/2=128W
But first find a method to prove that the protection is limiting @ ~4Apk.
if you find the existing circuit limits at 4Apk then halving the value of Re will give ~8Apk.
This will not drive a 1r7 resistor to 220W.
It will provide 8*8*1.7/2=54.4W into 1r7 sinewave power with 8Apk.
It will drive a 4r0 load to 8*8*4/2=128W
But first find a method to prove that the protection is limiting @ ~4Apk.
AndrewT said:
I was under the impression that the zero crossing was limiting at 4amps RMS.
I measured the RMS current out of the amplifier when driving the loudspeaker and the ugly sawing occurred exactly at 4amps.
Ergo Mega's 70 watts into 4 ohms. 4(4*1.41) = 22.56Volts peak. or ((4^2)^2)/4 = 64 watts.
Then I took the DCR of the loudspeakers - 1.7 ohms and the 220 watts of the amp and calculated the current required.
P=I^2*R. I = Sqrt (P/R) = 220/1.7 = Sqrt of129.4 = 11.38 amps RMS. 11.38 *1.41 = 16 amps Peak. I then halved the peak to = 8. And figured the value of the RE resistors should be calculated to limit at 8 amps RMS.
The book the amp design came from explains the protection circuitry and says it limits at about 2 amps into short circuit, this is for 1 output pair, so I assumed the 4amps was simply doubling this and hence Mega's 4amp figure. 0.7V/0.33 ohm = 2.12*2 = 4.24amps.
Using 4.24 amps we arrive at 72 watts into 4 ohms.
4.24Apk gives 36W into 4r0.
4Aac = 4Arms gives 72W into 4r0. This is 6Apk
If your output stage is required to supply 16Apk to the load then limiting the output to just 8Apk does not make any sense.
That's why post169 was added.
You can achieve short term limiting @ 16Apk to give the output you want, then fit F8A fuses in the supply rails to limit long term currents and power dissipation.
4Aac = 4Arms gives 72W into 4r0. This is 6Apk
If your output stage is required to supply 16Apk to the load then limiting the output to just 8Apk does not make any sense.
That's why post169 was added.
You can achieve short term limiting @ 16Apk to give the output you want, then fit F8A fuses in the supply rails to limit long term currents and power dissipation.
AndrewT said:
Mega originally said
Assuming 220 watts (the max this seems to be able to provide) and a DCR of 1.7 ohms I arrived at a current of 16 amps peak. Half this for 8 amps. This isn't setting the protection circuit at 8 amps peak, but 8 amps RMS. I figured this was a "rule of thumb" kind of way to set the protection circuitry.
The 4 amps mega is talking about matches the description of the amps protection circuitry in the book and is in relation to the output of the amp, not the rail fuses. I assumed this was 4 amps RMS on the output as I confirmed when measured (meter in series with the loudspeaker load), otherwise you cannot get close to 70 watts into 4 ohms.
As mentioned the calculation in the book was 0.7 volts (Activation of the protection transistor) / by the RE resistor. 0.7/0.33 = 2.12 *2 for the output pair = 4.24 amps. Just like mega stated.
Into a 4 ohm load, assuming that's 4.24 amps RMS we arrive at 72 watts.
your lack of clarity between rms current and peak current is very confusing. Be specific.5th element said:
The protection transistor is fully on at 600mVbe. This will be full limiting.
It starts to turn on ~400mVbe. This is when it will start to have an audible effect. This audible effect must not occur with any valid audio signal passing to the speaker.
AndrewT said:
I shall reiterate.
Going from the book the amplifier is taken from and Mega's 4amps. I assumed this was calculated from the voltage taken to fully activate (which the book states as 0.7 volts) the protection transistor and the value of the RE resistors.
0.7 volts/0.33 ohms = 2.12 amps. Two output pairs, double the current = 4.24 amps.
I assumed this was RMS, as Mega stated this 4 amps into a 4 ohm load would deliver 70 watts.
4.24 amps RMS into a 4 ohm load delivers just over 70 watts.
Mega continued
The amplifier can deliver around 220 watts as a maximum.
The DCR of the loudspeaker being driven is 1.7 ohms.
Assuming 220 watts and a DCR of 1.7 ohms I arrived at a current of 16 amps peak.
sqrt of 220/1.7 = 11.37 amps RMS. 11.37 * 1.41 = 16 amps peak.
Mega stated half the peak current required to drive the DCR of the loudspeaker. The Peak current is 16 amps. Half of this is 8 amps.
I took this to mean set
"0.7 volts/0.33 ohms = 2.12 amps. Two output pairs, double the current = 4.24 amps."
to equal 8 instead of 4.24.
Is this clearer?
AndrewT said:
This is getting tedious. But once again.
The amps here I am assuming are RMS. This is the 4 amps in the quote is 4 amps RMS.
Going from the book the amplifier is taken from and Mega's 4amps. I assumed this was calculated from the voltage taken to fully activate (which the book states as 0.7 volts) the protection transistor and the value of the RE resistors.
0.7 volts/0.33 ohms = 2.12 amps. Two output pairs, double the current = 4.24 amps.
I assumed this was RMS, as Mega stated this 4 amps into a 4 ohm load would deliver 70 watts. This part explained I was talking about RMS for all of the above.
4.24 amps RMS into a 4 ohm load delivers just over 70 watts. This again reconfirming its RMS im talking about for all of the above currents.
Mega continued
Here we have the SAME 4 amps RMS under discussion, but also with the additional statement, about calculating the peak current required to drive 220 watts into the DCR of the loudspeaker.
The amplifier can deliver around 220 watts as a maximum.
The DCR of the loudspeaker being driven is 1.7 ohms.
Assuming 220 watts and a DCR of 1.7 ohms I arrived at a current of 16 amps peak.
++sqrt of 220/1.7 = 11.37 amps RMS. 11.37 * 1.41 = 16 amps peak.
Here is that calculation. P=I^2*R. IRMS = the square root of 220/1.7, Ergo 11.37 amps RMS. I multiplied this by root 2 to gain the PEAK current, this was also rather clear.
Mega stated half the peak current required to drive the DCR of the loudspeaker. The Peak current is 16 amps. Half of this is 8 amps.
I took this to mean set
"0.7 volts/0.33 ohms = 2.12 amps. Two output pairs, double the current = 4.24 amps."
to equal 8 instead of 4.24.
What I am meaning here, as this is the ONLY time I interchange the two values. Is that Mega's original statement.
Implies setting the max zero crossing current to be equal to half the peak current required to drive the 220 watts into the DCR of the loudspeaker.
In other words, 4.24 amps RMS is the amount of current the amplifier is presently set to limit at, at zero crossing.
++Half the peak value as calculated above = 8.
Hence set the zero crossing current to 8 amps RMS.
Mega said you'd want the zero crossing limit to be at least half the peak output current into the DCR of the loudspeaker.
In the context of the 4amps RMS at the start of his statement, I took this to mean set the RMS value to that of half the peak value.
4.24 amps RMS = present zero crossing limit.
8 amps peak = half calculated peak value for driving loudspeaker DCR.
Therefore set the new zero crossing limit to = 8 amps RMS.
Now 8 amps peak = 5.7 amps RMS.
If you took Mega's statement to mean, set the new zero crossing limit to 5.7 amps RMS. Then please say so, as this is the only area of confusion, at least on my part.
Now if you think I have erred in calculation somewhere, by accidentally using the wrong unit please say so.
Just to clarify all of this I will explain once again what it is I am trying to do.
The amplifiers original major problem was the false activation of the protection circuitry. I didn't know why, nor what was going on.
With the discussion and the scope I determined that the problem was that the zero crossing current was set too low.
Mega stated the protection circuit will allow 4amps RMS to pass at zero crossing.
I measured the RMS current being delivered to the loudspeaker and the amplifier did indeed deliver 4amps RMS at the onset of the sawing noise.
The output on the scope, also showed a large peak jump up to the rail limit.
This is concordant with Figure 4 on this website.
http://sound.westhost.com/vi.htm
So having determined the problem, now to figure out a cure.
Mega stated that the zero crossing current should be set to
At least half the peak current required to drive the DCR of the loudspeaker
The amplifier can deliver 220 watts. The DCR of the loudspeaker is 1.7 ohms.
The peak current required to deliver this is 16 amps. Half the 16 amps = 8.
Old zero crossing limit = 4. Hence set the new zero crossing limit to 8.
The amplifiers original major problem was the false activation of the protection circuitry. I didn't know why, nor what was going on.
With the discussion and the scope I determined that the problem was that the zero crossing current was set too low.
Mega stated the protection circuit will allow 4amps RMS to pass at zero crossing.
I measured the RMS current being delivered to the loudspeaker and the amplifier did indeed deliver 4amps RMS at the onset of the sawing noise.
The output on the scope, also showed a large peak jump up to the rail limit.
This is concordant with Figure 4 on this website.
http://sound.westhost.com/vi.htm
So having determined the problem, now to figure out a cure.
Mega stated that the zero crossing current should be set to
At least half the peak current required to drive the DCR of the loudspeaker
The amplifier can deliver 220 watts. The DCR of the loudspeaker is 1.7 ohms.
The peak current required to deliver this is 16 amps. Half the 16 amps = 8.
Old zero crossing limit = 4. Hence set the new zero crossing limit to 8.
Hi,
your current limiter will limit at the peak of the current.
If the amplifier delivers 4Aac into 8r0 then the peak current is 5.66Apk and peak voltage is 45.2Vpk.
A VI limiter will change the current limit as voltage across the output transistors (Vce) changes.
A delay in the VI or I limiter will allow a short term current to pass that exceeds the DC current limit.
A 220W into 4r0 amplifier will have an output voltage = 42Vpk and an output current of 5.24Apk when delivering maximum power into the resistive load.
A 4ohm speaker can draw current peaks much in excess of what a 4r0 resistor can on fast starting and fast stopping signals.
A reasonable estimate for the maximum transient current into a reactive speaker load is max current~=Vpk/Rload/0.35
Your 220W amp may be asked to deliver 15Apk into a very reactive speaker.
The limiter should be set to allow this short term transient current to pass without audible degradation.
A reasonable medium term limit would be about half of this i.e. the DC current limit at zero volts crossing could be ~ 7.5Apk.
The fuses in the supply rails can be ~5.24Ak/2, try F2.5A or F3.1A
I notice some of my numbers are similar to what you are quoting, so we seem to be in some agreement.
Finally, the 7.5Apk DC limiter should be set so that a continuous current of 7.5Apk is the limited current value into a short circuit or into a 0r1 dummy load.
The 15Apk transient limiter will need to use capacitors in the protection circuit to allow 15Apk to pass when the Vbe voltage on the protection transistor is about 400mVbe. This implies ~ 22Apk when Vbe=600mV.
As the capacitor/s charge up the Vbe applied to the protection transistor should rise and thus limit medium term transients to a lower current and long term transients to just a few tens of % above the DC current limit.
your current limiter will limit at the peak of the current.
If the amplifier delivers 4Aac into 8r0 then the peak current is 5.66Apk and peak voltage is 45.2Vpk.
A VI limiter will change the current limit as voltage across the output transistors (Vce) changes.
A delay in the VI or I limiter will allow a short term current to pass that exceeds the DC current limit.
A 220W into 4r0 amplifier will have an output voltage = 42Vpk and an output current of 5.24Apk when delivering maximum power into the resistive load.
A 4ohm speaker can draw current peaks much in excess of what a 4r0 resistor can on fast starting and fast stopping signals.
A reasonable estimate for the maximum transient current into a reactive speaker load is max current~=Vpk/Rload/0.35
Your 220W amp may be asked to deliver 15Apk into a very reactive speaker.
The limiter should be set to allow this short term transient current to pass without audible degradation.
A reasonable medium term limit would be about half of this i.e. the DC current limit at zero volts crossing could be ~ 7.5Apk.
The fuses in the supply rails can be ~5.24Ak/2, try F2.5A or F3.1A
I notice some of my numbers are similar to what you are quoting, so we seem to be in some agreement.
Finally, the 7.5Apk DC limiter should be set so that a continuous current of 7.5Apk is the limited current value into a short circuit or into a 0r1 dummy load.
The 15Apk transient limiter will need to use capacitors in the protection circuit to allow 15Apk to pass when the Vbe voltage on the protection transistor is about 400mVbe. This implies ~ 22Apk when Vbe=600mV.
As the capacitor/s charge up the Vbe applied to the protection transistor should rise and thus limit medium term transients to a lower current and long term transients to just a few tens of % above the DC current limit.
Lets just talk figures for a moment to clear up some confusion on my part.
8ohms. 4 amps RMS. 32 Volts RMS.
P=VI = 32 * 4 = 128 watts.
P=I^2 *R = 16*8 =128 watts.
P=V^2/R = 128 watts.
All of this is in agreement.
You then said.
P = VI
(42/1.41)*(5.24/1.41) = 110 watts. /= 220.
P=I^2 * R
(5.24/1.41)^2 * R = 55 watts. /=220.
P = V^2/R
(42/1.41)^2/R = 222 watts ~220.
This is why I am confused.
From my perspective it seems your current figure is too low.
P=VI I = P/Vrms = 7.4amps rms
Using 7.4 amps RMS in
P = I^2 * R = 219 watts ~ 220 watts.
If you'd said
A 220W into 4r0 amplifier will have an output voltage = 42Vpk and an output current of 10.4Apk when delivering maximum power into the resistive load.
I'd understand.
As those figures satisfy
P= VI
P= I^2*R
P= V^2/R
It seems like your peak current is half the value it should be. Am I missing something here? As everywhere I look on the internet with regards to calculating amplifier power, points towards what I am saying as correct.
If Mega's 4 amps was a peak value. How would you end up at 70 watts?
I can get 0.7/0.33 =2.12 *2 = 4.24 amps RMS. Mega also said 24 volts peak or 17volts RMS
P= 17 * 4.24 = 72 watts.
P = 4.24^2 *4 = 72 watts.
P= 17^2 / 4 = 72 watts.
I'm not trying to be argumentative here and you are the one probably with the degree in the correct field, but still I am sure you can appreciate why I'm having trouble with this.
Going by your figures, I do appear to be missing something.
8ohms. 4 amps RMS. 32 Volts RMS.
P=VI = 32 * 4 = 128 watts.
P=I^2 *R = 16*8 =128 watts.
P=V^2/R = 128 watts.
All of this is in agreement.
You then said.
P = VI
(42/1.41)*(5.24/1.41) = 110 watts. /= 220.
P=I^2 * R
(5.24/1.41)^2 * R = 55 watts. /=220.
P = V^2/R
(42/1.41)^2/R = 222 watts ~220.
This is why I am confused.
From my perspective it seems your current figure is too low.
P=VI I = P/Vrms = 7.4amps rms
Using 7.4 amps RMS in
P = I^2 * R = 219 watts ~ 220 watts.
If you'd said
A 220W into 4r0 amplifier will have an output voltage = 42Vpk and an output current of 10.4Apk when delivering maximum power into the resistive load.
I'd understand.
As those figures satisfy
P= VI
P= I^2*R
P= V^2/R
It seems like your peak current is half the value it should be. Am I missing something here? As everywhere I look on the internet with regards to calculating amplifier power, points towards what I am saying as correct.
If Mega's 4 amps was a peak value. How would you end up at 70 watts?
I can get 0.7/0.33 =2.12 *2 = 4.24 amps RMS. Mega also said 24 volts peak or 17volts RMS
P= 17 * 4.24 = 72 watts.
P = 4.24^2 *4 = 72 watts.
P= 17^2 / 4 = 72 watts.
I'm not trying to be argumentative here and you are the one probably with the degree in the correct field, but still I am sure you can appreciate why I'm having trouble with this.
Going by your figures, I do appear to be missing something.
OOPs
Hi,
220W into 4r0 is 42Vpk and 10.5Apk.
the 5.24Apk is the current into a 8r0 load.
I use 8ohm habitually and just hit the wrong key.
Sorry for the confusion that has caused.
All the currents quoted thereafter are applicable for an 8ohm load/speaker driven from a 42Vpk amplifier.
Using 4ohm requires all the currents and fuses to be doubled.
Hi,
220W into 4r0 is 42Vpk and 10.5Apk.
the 5.24Apk is the current into a 8r0 load.
I use 8ohm habitually and just hit the wrong key.
Sorry for the confusion that has caused.
All the currents quoted thereafter are applicable for an 8ohm load/speaker driven from a 42Vpk amplifier.
Using 4ohm requires all the currents and fuses to be doubled.
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