Hi,
had a 3886 in non-inverting mode and doing some measuring.
The bulb tester was still in line.
I accidentally shorted the speaker output leads and as expected the chipamp protected itself and is undamaged.
But, the light bulb flashed during the very short time (tens of mS) that the leads were shorted.
This implies that the mains transformer supplied a power surge during protection mode.
That also implies that a high current passed through the chipamp during the protection mode.
How does the Spike and other protections work in a 3886?
I had imagined that VI limiting and I limiting would limit any excessive current to a very short transient and that this transient would be supplied by the smoothing caps, not a sustained surge that required a big recharge from the transformer.
had a 3886 in non-inverting mode and doing some measuring.
The bulb tester was still in line.
I accidentally shorted the speaker output leads and as expected the chipamp protected itself and is undamaged.
But, the light bulb flashed during the very short time (tens of mS) that the leads were shorted.
This implies that the mains transformer supplied a power surge during protection mode.
That also implies that a high current passed through the chipamp during the protection mode.
How does the Spike and other protections work in a 3886?
I had imagined that VI limiting and I limiting would limit any excessive current to a very short transient and that this transient would be supplied by the smoothing caps, not a sustained surge that required a big recharge from the transformer.
Hi Andrew,
National obviously won't give away too many secrets, but your answer may be in this document
Regards
David
National obviously won't give away too many secrets, but your answer may be in this document
Regards
David
898 does not help.
On the one hand it says Spike monitors junction temperature and goes on to say that when Tj=250degC it cuts off stating a typical pulse width of 200uS.
Then a couple of paragraphs later it describes a completely different VI protection strategy designed to ensure the output transistors never reach max Tj.
Almost as if the document were written by two different people and neither had read the other's contribution.
On the one hand it says Spike monitors junction temperature and goes on to say that when Tj=250degC it cuts off stating a typical pulse width of 200uS.
Then a couple of paragraphs later it describes a completely different VI protection strategy designed to ensure the output transistors never reach max Tj.
Almost as if the document were written by two different people and neither had read the other's contribution.
Andrew, write to Bob Pease at National : rap@galaxy.nsc.com
He a Staff Scientist, book author and has a magazine column.
http://www.national.com/JS/searchDo...SE&categories=Applications:Bob+Pease+Articles
He a Staff Scientist, book author and has a magazine column.
http://www.national.com/JS/searchDo...SE&categories=Applications:Bob+Pease+Articles
It is probably more like different protection circuits working in parallel. One tries to maintain the IC below max Tj, but must not be too fast to prevent convenience tripping. Another one reacts in situations, when the first one is too slow, e. g. during shorts.AndrewT said:
I do not get it.
What is the problem here?
Does not LM3886 protect against damage?
If so, there is a problem 😀
What is the problem here?
Does not LM3886 protect against damage?
If so, there is a problem 😀
The problem is that the Chip passed a large surge through it that made the test bulb flash.
I have just +-22Vdc on the chipamp.
What if I were running +-40Vdc? How big would the current surge be?
Why does there need to be any long term surge if the protection is working like the descriptions?
I have just +-22Vdc on the chipamp.
What if I were running +-40Vdc? How big would the current surge be?
Why does there need to be any long term surge if the protection is working like the descriptions?
The light bulb tester did its job and limited the current. The current that flowed was probably just around or slightly below the threshold for SPiKE. And that is why it took such a long reaction time.
According to what is published in AN-898, one could assume that there is an integrator at work that detects a voltage drop somewhere. The higher the voltage drop as a result of current flow, the faster the reaction time would be.
If you had not used the light bulb tester, the reaction would probably have been much faster. The same will be true for ±40 V power supply, because in both cases a higher current would have flown.
There will certainly be situations, where the reaction time is too slow or the current too high during even the shortest reaction time. Then SPiKe will not be able to protect the IC, as can be seen in this contribution. Which shows again how good it is to use a light bulb tester.
Maybe we should all stock a few light bulbs for the coming years, when the sale of new light bulbs will be prohibited in the EU.
According to what is published in AN-898, one could assume that there is an integrator at work that detects a voltage drop somewhere. The higher the voltage drop as a result of current flow, the faster the reaction time would be.
If you had not used the light bulb tester, the reaction would probably have been much faster. The same will be true for ±40 V power supply, because in both cases a higher current would have flown.
There will certainly be situations, where the reaction time is too slow or the current too high during even the shortest reaction time. Then SPiKe will not be able to protect the IC, as can be seen in this contribution. Which shows again how good it is to use a light bulb tester.
Maybe we should all stock a few light bulbs for the coming years, when the sale of new light bulbs will be prohibited in the EU.
I can't agree with that.
The bulb flash indicated that the transformer tried to supply a demand.
That demand can only be to recharge the smoothing capacitors.
The smoothing caps could only discharge significantly if a high current was drawn from them for a considerable time.
i.e. the bulb flash confirmed that a high current draw happened and that this high current flowed for a long time. Considerably longer than 200uS.
There is no way the bulb in the primary circuit could/should reduce the current through the shorted output. The caps supply the transient currents, the transformer re-charges the caps.
Check this arithmetic:
11A through the shorted output for 200uS pulls down 23mF by ~100mV. Is this correct?
If these numbers are correct then a 2mS current pulse of 11A will reduce the +-22V on the capacitors to ~+-21V. This will still allow a significant shorted output current. If the short lasted 20mS then 11A could discharge the caps by about a half, although the current flowing will rapidly reduce as supply voltage on the caps falls.
Does this sound feasible?
Could Spike have let the output current stay around 11A to 7A for between 2mS and 20mS before shutting down?
The bulb flash indicated that the transformer tried to supply a demand.
That demand can only be to recharge the smoothing capacitors.
The smoothing caps could only discharge significantly if a high current was drawn from them for a considerable time.
i.e. the bulb flash confirmed that a high current draw happened and that this high current flowed for a long time. Considerably longer than 200uS.
There is no way the bulb in the primary circuit could/should reduce the current through the shorted output. The caps supply the transient currents, the transformer re-charges the caps.
Check this arithmetic:
11A through the shorted output for 200uS pulls down 23mF by ~100mV. Is this correct?
If these numbers are correct then a 2mS current pulse of 11A will reduce the +-22V on the capacitors to ~+-21V. This will still allow a significant shorted output current. If the short lasted 20mS then 11A could discharge the caps by about a half, although the current flowing will rapidly reduce as supply voltage on the caps falls.
Does this sound feasible?
Could Spike have let the output current stay around 11A to 7A for between 2mS and 20mS before shutting down?
Depends on the time scale. If the light bulb did not reduce the current, it would be useless. A high current heats the filament. The filament resistance rises and voltage drops across it. Consequently there is less voltage left at the transformer primaries, hence as well at the secondaries. As a result the capacitors would only be recharged to a lower voltage level. After the first surge the short circuit current would be reduced.AndrewT said:
On top of that a shorted transformer also delivers a limited current only.
If the transformer would not recharge the capacitors, yes.AndrewT said:
Guesswork: 11 A for 200 µs, then limited to 7 A until thermal shutdown was reached.AndrewT said:
Could the short impedance have been higher than 2 Ohms, so that 11 A could never have flown? Or higher than 3,2 Ohms, so that even 7 A could never have flown? What was the output signal at that time? Could the output current have been limited by that already? E. g. 0,2 Ohm short impedance would lead to less than 7 A, if the output voltage was below 1,4 V at that time.
And how fast should SPiKe react? 200 µs correspond to a 5 kHz transient, 2 ms to 500 Hz and 20 ms to 50 Hz. All still in the audio range.
Whatever SPiKE did was the right thing, because the IC is still working.
There is no way of knowing how much current was really flowing for how long, unless you want to risk an IC and repeat the situation with adequate measuring and recording equipment in place.
that could be a possible protection scenario.pacificblue said:
But, how long will the device allow 7A to flow with Vce~20V. This is shown on Page9 SOA graph as lasting ~8mS.
That could be just enough to pull down the supply sufficiently.
The 200uS pulse+the 8mS pulse pulls a total of ~2.5V from the smoothing capacitor. This would probably be enough to require a significant re-charge that would flash the bulb.
But, as you say "guesswork".
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