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In my opinion,
1) Q1,Q95 are level shifter
2)Q93 is a emitter-follower,
3) C4 & C5 are not compensation cap. because those nodes they connected are all low impedance nodes. so r*c is too small, u can run a AC simulation to check this.
"In my opinion,
1) Q1,Q95 are level shifter
2)Q93 is a emitter-follower,
3) C4 & C5 are not compensation cap. because those nodes they connected are all low impedance nodes. so r*c is too small, u can run a AC simulation to check this.
" -----------SAY AS SAY NOTHING !
C4 & C5 should be compensation cap as it increases output parasitic cap. Same as two stage Op-Amp, with compensation cap connected from 1st stage output to the output stage. Two caps coz' there are two pre-driver stage.
IMO, C4 & C5 are not intended to add capacitive load to the buffers. If you consider Q1 and Q95 are unity gain buffers, C4 and C5 are just connected across the output of two buffers with the same polarity (ie both terminals of C4 and C5 are moving in the same direction). In small signal analysis, there is no current flow through these two cap and they have no effect in small signal domain.
I guess the purpose of C4 and C5 are trying synchronize the driving signal to Q98 and Q9 in large signal transient because the up and down slew-rate of C4 and C5 are unequal (Consider charging and discharging paths).
For Q93, I guess it is used to provide an additional pull-up path to cutoff the power transistor quickly.
Before I can explain in detail, please explain clearly what you don't understand.
If you wish to get the answer you want quickly, you should let us know what you have observed and provide info as much as you know. Otherwise we just guess what you are expecting.
Hi Fanrong
You have here a TransLinear circuit
(Barrie Gilbert)
Please refer to chapter 2 in:
Analogue IC design -The current mode approach-
Toumazou & Lidgey
For very fast signals capacitors acts like dc voltage source. This makes better transient response because all 4 transistors works more symetrically.
Q93 normally should not work?
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