Sziklai buffer (simple or enchanced)

[antonis]

Sziklai compound is preferred over Darlington mainly for:
Single V_BE drop..
Higher V_BE constancy hence much better linearity..
(Q1 Collector current (the current trough R1) is almost constant – until Q2's current requires a Base current comparable to Q1's one..)⁽¹⁾
Lower output impedance..
(Sziklai's output resistance (r_e) is x times lower than that of a single BJT (0.025/Ic), and  x + a bit more  lower than that of a Darlington, where x is the ratio of Q2 current to Q1 one..)⁽²⁾
Higher input impedance..
(about 4 times with current items values..)⁽³⁾



Q1/Q2 form a classic CFP of almost 100% NFB. Q1 Emitter / Q2 Collector junction is biased at 4.7V (PS at +9.22V) and, disregarding for the time being SW1 & R11, of 1.4mA working current.
Due to Q1's small working current (about 67μA) and moderate (at this level of current) DC gain (about 200), there is an insignificant drop across R5 (about 73mV).
R5 could be of much higher value but it shouldn't add much on input impedance due to already very effective bootstrapping..(see ⁽³⁾)
R3/R4 values could be 10 times lower (for a stiffer divider) but their Thevenin equivalent resistance is effectively set in parallel with R2 (as well as R7//R8) dominating R2's  AC value..
R7+R8 is just a split and bootstrapped pull-down (anti-pop) resistor..
About C1 value: It might seem overlarge for such a high input impedance but you'd never rely on input cap value for HPF for two (mainly) reasons:
1.Noise (no further analysis here..)
2. Low frequency peaking in conjunction with bootstrap cap and bootstrapped resistor
(to make the long story short: The bootstrap loop has a gain which always exceeds unity at a frequency f = 0.159 / √(C1*C2*R5*R3//R4).. The gain at this frequency is given by  √[1 + C2*R5*R3//R4 / C1*(R5 + R3//R4)²]
By making C1 very large we have dual benefit: Very low peaking frequency (well outside audio range) and very low gain at this particular frequency..
(0.4Hz and 1.04 here)
C3 & C4 could be bigger but 1μF value is convenient for using either electro (polarized or not) or film caps..
R9 can be omitted.. It's there to prevent prospective Sallen-Key frequency peaking (like above), formed by R7/R8/C4 and unknown input capacitance..
R5 prevents oscillation and C2 prevents RF rectification..
(both can be omitted albeit strongly recommended ..)
By switching R11 in parallel with R2 you effectively double working current (due to halving Emitter resistor) hence doubling negative swing margin..
e.g. for 3k3 Emitter resistor, an equal value load can't be driven lower than 2.35V (half of Emitter voltage) where it can go down to 1.55V for 1k6 Emitter resistor.. 
(another way to see it is: In the former case, 1.66V RMS can be passed undistorted to the 3k3 load where in the later it can be passed 2.3V RMS..)

By adding another BJT (ignore Q4 for the time being) we can go far better..
We obtain an even higher V_BE constancy (don't forget that Q1 V_BE is the actual configuration V_BE) 'cause R2 now does whatever R1 did on 2 Sziklai pair (see ⁽¹⁾)..
We effectively "isolate" Q1's V_BE from any variation..
Doubling Q3's current results into 280nA variation of Q1's collector current hence 100μV ΔV_BE, which is 25 times lower than that of a 2 BJT Sziklai pair and 180 times lower than that of a single BJT (or a Darlington)..
(not bad in the cost of an extra BJT..)

Q4 and relevant circuitry form a variable current source..
MPSA42 is favored for its high Early voltage (any n-p-n, high voltage, low beta could be substitute) hence high output impedance⁽⁴⁾..
By implementing current source instead of Emitter resistor we effectively raise load drive capability despite load's value 'cause there is no voltage dividing effect when Q3 sinks(*) current.. 
(*) going lower than Emitter biased level..
R13/C6 can be omitted if you're convinced of your power supply cleanness..
R12 value isn't critical.. Any value ensuring current through it lower than 10 times Base current should be fine..
(to put it into equation: R12 < (9V – 1.2V)*h_FE / I_C..)
LOAD pot simply sets Q4 current..


^(citations) (no need to read them.. )

⁽¹⁾ Supposing a typical beta of 270 for Q2, doubling its Collector current results into about 5μA upturn in Q1 Collector current simultaneously with 18mV Q2 V_BE upturn (Ebers-Moll derivation), hence a total of about 7μA for Q1 Collector current.. So, Q1's ΔV_BE (which is pair's actual V_BE) is V_T*log_e(I_C2/I_C1) (Ebers-Moll again) = 0.025*log_e(74μA/67μA) = 2.5mV which is more than 7 times lower than that of a singe BJT (or a Darlington)..

⁽²⁾ When looking back into the emitter, the impedance is just dV_BE / dI_E.. For a single BJT, I_E=I_C so Rout = r_e = 0.025/I_C.. For our Sziklai pair I_E consists of I_E1+I_C2 with a ratio of 0.074/1.35(*) = 0.05 so its output impedance is 18 times lower than that of a single BJT (and somewhat more lower than that of a Darlington). Another way to see it is the greatly reduced V_BE change versus Collector current..
(*) R2 1.42mA current consists of Q1 74μA (67μA through R1 + 7μA Q2 Base current) + Q2 1.35mΑ..

⁽³⁾ More effective bootstrapping due to much lower output impedance..
(r_e < 1 Ohm according to ⁽²⁾, so R5 bootstrapped value is 2,500 times higher than its ohmic value.. A = (R2//R3//R4//R7//R8) / (R2//R3//R4//R7//R8 + re), R5(bstrp) = R5 / (1-A)..)

⁽⁴⁾ Collector output resistance (r_c or r_o)  is calculated by (V_A + V_CE) / I_C, where V_A is particular device Early voltage.. This stands for a grounded Emitter current source.. When there is Emitter degeneration (in the form of feedback, like in CE amps), "effective" Early voltage is raised by [1 + ( I_E * R_E) / V_T ] times or  1 + V_E / 0.025..
(about 25 times for V_E of a single diode voltage drop..)
Of course, higher V_E should result in higher "effective" Early voltage, but this should be in the cost of lowest voltage compliance (V_E + V_CEsat)..



P.S. Sincerely sorry for lengthy/rambling post..

[fryingpan]

Just out of curiosity: base current for the usual Sziklai pair @ 9V has, in my simulations, amounted to 2nA or so. Do you need to go with the trouble of bootstrapping (and adding so many components)? If the voltage divider current is even relevant for a Sziklai pair, that means you could easily use a parallel pair (with respect to the Q1 base) with total resistance = 80Mohm or so... and due to the large amount of feedback (and voltage gain) in the Sziklai pair, I would expect that its inherent impedance should be in the Gohm or at least n*100Mohm range?

[antonis]

base current for the usual Sziklai pair @ 9V has, in my simulations, amounted to 2nA or so.

More close to 370nA, I'd say..
(your simulator overestimates Q1 DC current gain..)

What should be the resistor values of voltage divider for the current across them to be about 4μA (> 10*I_B)..??
(answer: R3 + R4 = 2M2..)
I let you make the calculations for exact R3 & R4 values and their equivalent parallel resistance..

P.S.1
Typical leakage current for BC549 is 15nA (at room temperature and as high as 5μΑ at max working temperature)..
(much larger than that calculated by your simulator..)
You can see how much bias instability could result from the use of big resistors..

P.S.2
You overlook another big value resistors issue, called NOISE..

[fryingpan]

base current for the usual Sziklai pair @ 9V has, in my simulations, amounted to 2nA or so.

More close to 370nA, I'd say..
(your simulator overestimates Q1 DC current gain..)

What should be the resistor values of voltage divider for the current across them to be about 4μA (> 10*I_B)..??
(answer: R3 + R4 = 2M2..)
I let you make the calculations for exact R3 & R4 values and their equivalent parallel resistance..

P.S.1
Typical leakage current for BC549 is 15nA (at room temperature and as high as 5μΑ at max working temperature)..
(much larger than that calculated by your simulator..)
You can see how much bias instability could result from the use of big resistors..

P.S.2
You overlook another big value resistors issue, called NOISE..

I didn't use BC series transistors, but it shouldn't make much difference. 2n390x are jellybean transistors. I'll double check, I might misremember. (FWIW, I use LTSpice).

[antonis]

I didn't use BC series transistors, but it shouldn't make much difference. 2n390x are jellybean transistors. I'll double check, I might misremember. (FWIW, I use LTSpice).

Beware of beta determined for lower than 100μA collector current..

[fryingpan]

That said (I am not doubting you Antonis, quite the opposite, also, simulators should be taken with a grain of salt...) shouldn't the noise contribution of two large resistors to ground (for AC) be lower than smaller resistors in a bootstrap configuration? After all, it is positive feedback and the resistors are "in circuit"... (Of course, even without taking noise into account, the use of large resistors is problematic for many other reasons).

[Rob Strand]

base current for the usual Sziklai pair @ 9V has, in my simulations, amounted to 2nA or so.

More close to 370nA, I'd say..
(your simulator overestimates Q1 DC current gain..)

The base current depends on the the choice of the BE resistor, which is R2 in the second circuit in antonis's first post.  If you have no BE resistor the base current will be lower.  (In antonis's case also R1, depending on which base current you are concerned with.)

It's a design choice.   Typically you want some BE resistor so the circuit operates fast enough but also that it sets reasonable operating currents in the driver transistor.   A remember, transistor hFE drops at low currents.    You will find the open loop gain varies with the BE resistor.

That said (I am not doubting you Antonis, quite the opposite, also, simulators should be taken with a grain of salt...) shouldn't the noise contribution of two large resistors to ground (for AC) be lower than smaller resistors in a bootstrap configuration? After all, it is positive feedback and the resistors are "in circuit"... (Of course, even without taking noise into account, the use of large resistors is problematic for many other reasons).

Simulators do what you tell them and only tell you what *you* interpret from the results.

[antonis]

shouldn't the noise contribution of two large resistors to ground (for AC) be lower than smaller resistors in a bootstrap configuration?

Thermal (Johnson-Nyquist) noise is proportional to actuall (ohmic) resistance..  

[fryingpan]

shouldn't the noise contribution of two large resistors to ground (for AC) be lower than smaller resistors in a bootstrap configuration?

Thermal (Johnson-Nyquist) noise is proportional to actuall (ohmic) resistance..

Yes, but if resistors tied to AC ground should shunt away most of the noise, having a 300kohm resistor "in circuit" means having all of their Johnson noise added to the signal. Academic, I know.

[antonis]

if resistors tied to AC ground should shunt away most of the noise,

Only with a cap across them..

Like the cap set in parallel with op-amp bias resistor..

[fryingpan]

if resistors tied to AC ground should shunt away most of the noise,

Only with a cap across them..

Like the cap set in parallel with op-amp bias resistor..

It really depends on the source impedance... if the source impedance is low, the current noise will not translate to a high voltage noise. Right?

[antonis]

Shall we continue our argument into another thread..??