help : i dont know how work one subcircuit in low noise amplifier with bipoar process

[steven_jun]

help i dont know how work one subcircuit in low noise amplifier with bipolar process

in the analysis low noise amplifier ,i dont konw the function of one subcuit ,the process is bipolar

can you tell me the function or some relevant information or document

 

[erikl]

The circuit at the top serves as a bias voltage generator for the actual LNA:

The 4 BJTs (bipolar junction transistors) Q1x & Q2x together with the NMOS NM0 constitute a balanced differential amplifier with a 1:1 (R14:R13) negative feedback, so regenerating a buffered temperature-dependent Vbe (created at the connection between I2 & Q0). The voltage level at the drain of NM0 will be 2*Vbe, at the junction of R14/R13 = input of the AMP this will be 1*Vbe. If the proper LNA BJT DZ17 is thermally coupled to the bias generator BJTs, this guarantees a largely temperature-independent collector current of DZ17 .

 

[lovexxnu]

This part is just to provide a DC operation point, works like current mirror and use the buffer (opamp) to insulate the two BJT. it's not related to the independent temperature, actually there is no independent temperature current generated.

---------- Post added at 18:13 ---------- Previous post was at 18:01 ----------

The circuit at the top serves as a bias voltage generator for the actual LNA:

The 4 BJTs (bipolar junction transistors) Q1x & Q2x together with the NMOS NM0 constitute a balanced differential amplifier with a 1:1 (R14:R13) negative feedback, so regenerating a buffered temperature-dependent Vbe (created at the connection between I2 & Q0). The voltage level at the drain of NM0 will be 2*Vbe, at the junction of R14/R13 = input of the AMP this will be 1*Vbe. If the proper LNA BJT DZ17 is thermally coupled to the bias generator BJTs, this guarantees a largely temperature-independent collector current of DZ17 .

R13 and R14 is to adjust the operation point of the opamp to avoid that the output is limited by the BJT in LNA.

 

[erikl]

... there is no independent temperature current generated.

Seems you don't understand the temperature-dependent Vbe effect on the drain current stabilizing effect of a BJT in case of effective thermal coupling.

 

[lovexxnu]

why the voltage at the drain of NM0 is 2Vbe? I think the current is decided by I0, Q0 and DZ17 are current mirror.
could you say some more about the independent current or provide some reference? does it generated by NM0 and the Vbe?
I did little on this kind of circuit.
thank you

 

[goldsmith]

Dear friend.
Hi
It's work is very simple. I don't know that what is your problem?
Goldsmith

 

[steven_jun]

thank you ,my friends , I have a problem on this circuit.
R13=8*R R14=2R, R0=4R , i dont know how to determine the collector current of QZ17,because i dont know the drain voltage of the nmos NM0

 

[erikl]

why the voltage at the drain of NM0 is 2Vbe? I think the current is decided by I0, Q0 and DZ17 are current mirror.

No and yes: BJT current mirrors need to have the same Vbe potential, i.e. have both their emitters and bases in common, which obviously is not the case in this circuit. It's a buffered current mirror, however:

Consider the circuit on the top including R0 & R13 as an OpAmp, the node I2-Q0-R0 being the non-inverting input (Vbe of Q0), negative feedback from NM0 drain output via resistive divider R14/RZ18 to the inverting input (R13). Then the gain from the non-inverting input to the output (NM0 drain) is vu = 1+(R14/RZ18) ≈ 2 -- with the given values. Of course this is only an approximation, as the input impedance Rbe(DZ17) is in parallel to RZ18; due to the strong negative feedback via RZ8 this input resistance Rbe(DZ17) >> RZ18, however.

could you say some more about the independent current or provide some reference? does it generated by NM0 and the Vbe?

Right: With a temperature-independent current I2, a temperature-dependent Vbe(Q0) is generated. Neglecting DZ17's input resistance, the buffered voltage at DZ17's base is Vbe(DZ17) = (1+(R14/RZ18))*Vbe(Q0)*RZ18/(R14+RZ18) = Vbe(Q0) , so you could consider DZ17 as a "buffered current mirror transistor" of Q0. As DZ17 receives the same (but buffered) temperature-dependent Vbe as that one generated by Q0, it will reproduce the same temperature behavior as I2 -- good thermal coupling between the 2 BJTs implied.

You can find more about BJT current mirrors in any book about (analog) transistor electronics, e.g. in Hans Camenzind's "Designing Analog Chips", chap. 3.

---------- Post added at 13:40 ---------- Previous post was at 13:29 ----------

R13=8*R R14=2R, R0=4R , i dont know how to determine the collector current of QZ17,because i dont know the drain voltage of the nmos NM0

... and we don't know all the other values (other resistors, transistor areas, currents, VDD, VPOW). May be you can make an estimation with my equation given above?

 

[steven_jun]

i think : because R13=2*R0 and the IB(Q16)=IB(Q18)=0.5*I0/(1+beta(Q16),so the VBE(DZ17)=VBE(Q0)+0.5*IB(Q16)*R13.
then the collector current of Q(DZ17) is determinated. do you think ?

thanks !

 

[erikl]

R13=2*R0

Ok.

IB(Q16)=IB(Q18)=0.5*I0/(1+beta(Q16)

No: IB(Q16)≈IB(Q18) ; their collector and base currents ain't necessarily equal, as their collector voltages ain't necessarily equal.

VBE(DZ17)=VBE(Q0)+0.5*IB(Q16)*R13.

?? How do you get this equation?

... then the collector current of Q(DZ17) is determinated. do you think ?

Yes, it is. But not according to your above equations. It's much easier: Ic(DZ17) ≈ Ic(Q0) ≈ I2 , if both BJTs are of equal size and thermally well coupled, see my contribution above. The ratio of R13/R0 shouldn't change a lot.