what are the capacitors doing in this circuit

[asimkumar]

this circuit was suggested for a light receiver circuit, which is amplifing pulses(~10mV 20us) from photo diode. this is working good but i want to know what type of amplifier is this and what is the c16 c17 doing

 

[g527727372]

common B amplifier.

 

[chuckey]

C16,17,18 are to tailor the frequency response - reduce the higher frequencies.
Frank

 

[asimkumar]

C16,17,18 are to tailor the frequency response - reduce the higher frequencies.
Frank

it means i will get higher frequency response by removing these capacitors

 

[chuckey]

yes. I have to send at least ten characterers
Frank

 

[asimkumar]

common B amplifier.

can u give me some links about common B design guide

 

[g527727372]

can u give me some links about common B design guide

Q5 supply current for Q6.Q6 works at the common-base amplifier which can be drawed as below:

 

[chuckey]

I redrew the circuit to make the emitters go "downwards" :- , can't see any common base amplifier or even any current source.
Frank

 

[godfreyl]

It's an interesting circuit. The feedback around the two transistors means the input impedance at Q6's emitter is much lower than it would be with a simple common base configuration.

Idling at about 1/4 mA, a common base circuit would have an input impedance of about 100 Ohms. With this circuit the input impedance should be less than one Ohm, giving reasonable accuracy and linearity with the 4.7 Ohm series resistor.

 

[g527727372]

I1 is fixed by 0.8V and R24. so we can call it a current source. the current is flow into Q6 to provide Ie and Ic.
Q6 is act as common base amplifier.
and if you break the line at "X" point. the Ib of Q6 will be decide by 7.2V and R32. this is not good.Because if Ib of Q6 is fixed,Ic of Q6 is fixed too,and Q6 can not amplifies signal.So the line which has "X" point is suggest to provide "changing Ib".when Ic of Q6 changes ,Ib changes too ,so it can get the changing part from Ic of Q5,and the current of R32 is not changed.
All that ,Q5 is suggest to provide static current for Q6,include Ie and Ib. And Q6 is a common base amplifier.

 

[godfreyl]

I redrew the circuit to make the emitters go "downwards" :- , can't see any common base amplifier or even any current source.
Frank

It depends how you look at it. If you change the picture slightly as below, then it looks like a common base circuit because the input is to Q6's emitter and the output is from Q6's collector.

That's a bit simplistic though. It ignores that the input is also connected to Q5's base, and that the base of Q6 is not held at a constant voltage, as it would be for a true common-base circuit.

 

[g527727372]

there is no doube Q6 work at common-base state.
and the base of Q6 is held at a constant voltage.
4.7 Ohm resistor should be consider.and the current change on this resistor,is the reason cause Vo change.（Vi/4.7）*15K=Vo;
I do some simulation to support it ,for better vision I increase 4.7 Ohm to 470 Ohm,or else the Vo is too large:

can see from the figure Vo is amplify,and Vp is held constant.

 

[godfreyl]

can see from the figure Vo is amplify,and Vp is held constant.

Yes, exactly - Vp is held constant. That is the voltage on the emitter, not the voltage on the base.

Compare the signal voltage at the emitter and the base. They are not the same. The voltage swing at the base is much greater.

 

[g527727372]

Vb of Q6 is not held constant.
Ve of Q6 is not held constant too.I simulate again and found that 4.7 Ohm resistor should not be replaced by 470 Ohm.
then I use small signal model to calculate Vb,Ve,Vo,and input impedance.
found the input impedance of Q6 is really small, about 2 Ohm.Just as you said.
But why the small input impedance gets more accuracy and linearity?

 

[chuckey]

Hi here my latest take on this, Break the AC circuit at the emitter of Q6, just a cascaded amplifier, join the loop and and take into account the 4e7 source resistor. :-


Frank

 

[g527727372]

Base on the feedfack,I draw the small signal model.and use two method to calculate :Vo=2400Vi ,Rin=0.69Ohm.

for feedback reason ,Rin is small..