Can you explain how the current limit increases with decreasing voltage across the transistor? I'm with AndrewT on this one.
It work's like this (maybe..)
When the output voltage is low there´s very little current flowing in R10 and R44, becurse of the low voltage drop across these resistors.
As the output voltage increases the voltage across R10 and R44 will increase and they will lead some current away from the base of the protecting transistors (Q12 and Q14). This means that the "trigger-level" og Q12 and Q14 is dependent of the output voltage.
Agree?
TroelsM
When the output voltage is low there´s very little current flowing in R10 and R44, becurse of the low voltage drop across these resistors.
As the output voltage increases the voltage across R10 and R44 will increase and they will lead some current away from the base of the protecting transistors (Q12 and Q14). This means that the "trigger-level" og Q12 and Q14 is dependent of the output voltage.
Agree?
TroelsM
Ahh, it's the simple things you miss 🙂 It's the addition of R10 and R44 that make the design a little different to what I had seen before, which was fixed current limit.
Re: It work's like this (maybe..)
hi again
I guess you are aware that this solution is very sensitive to reactive loads (as a speaker with crossover)
when current and voltage are not in phase this may "overreact" or not protect power transistors from burning...
regards
TroelsM said:
hi again
I guess you are aware that this solution is very sensitive to reactive loads (as a speaker with crossover)
when current and voltage are not in phase this may "overreact" or not protect power transistors from burning...
regards
Real load's and stuff
Hi again.
It's my understanding that if the resistors are the "right" size the TIP's will be kept under the SOA-curves. It's correct that a very reactive load could draw a lot of current with a very small voltage at the output, but it's not very likely with "normal" speakers. but even under those conditions the transistors should be protected.
It would be nice to have a high current limit no matter the output-voltage, but that's not possible with the TIP's.
The curcuit should however allow high, short current peaks, and it should be possible to "slow" down the curcuit with some capacitors. Any Idea's?
TroelsM
Hi again.
It's my understanding that if the resistors are the "right" size the TIP's will be kept under the SOA-curves. It's correct that a very reactive load could draw a lot of current with a very small voltage at the output, but it's not very likely with "normal" speakers. but even under those conditions the transistors should be protected.
It would be nice to have a high current limit no matter the output-voltage, but that's not possible with the TIP's.
The curcuit should however allow high, short current peaks, and it should be possible to "slow" down the curcuit with some capacitors. Any Idea's?
TroelsM
Hi,
High transient current tolerance is achieved by adding caps across the protection transistor bases. C4 & C5 achieve this by producing an RC time constant with R16 & R35.
But why are R16 & R35 & companions of differing values?
Still no reaction to my suggestion to add diodes to collectors of protection transistors.
regards Andrew T.
High transient current tolerance is achieved by adding caps across the protection transistor bases. C4 & C5 achieve this by producing an RC time constant with R16 & R35.
But why are R16 & R35 & companions of differing values?
Still no reaction to my suggestion to add diodes to collectors of protection transistors.
regards Andrew T.
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