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Copyright 2010 by Wayne Stegall
Updated September 2, 2010.  See document history at end for details.
Link to old version.

Active Inductor Example

Low-Noise JFET Phono Preamp

This is an example only.  It is not intended to be final or complete design.  It should only serve as an example of equalization and active inductor calculations and not be built.  The SPICE verified two stage design Low Noise 2-Stage All JFET Phono Preamp would better serve building an actual prototype.
This is a redesign using LSK170 JFETs on the input.  You still may want to use the 2N3819 design because of possible difficulty of obtaining LSK170's.

Figure 1:  Schematic

Parts List

Q1-Q4 LSK170C n-ch jfet
Q5,Q6 BS170 n-ch mosfet
56Ω, 5%
56Ω, 5%
56.2kΩ, 1%
8.25kΩ, 1%
39nF, 1% or 5%
C2N,a 13nF, 1% or 5%
360pF, 1% or 5%
R1L 38.3kΩ, 1%
CL 10,000F, 25V electrolytic
1kΩ, 5%
82Ω, 5%
1kΩ, 5%  (or 220Ω, see text)
100F, 25V electrolytic
1kΩ pot (or 2kΩ, see text)
S1,S2 SPST reed relay, normally closed, coil rated 10mA@5V

Remaining values are left for you to calculate.  See text.

Initial Design Decisions

vg,Q6 = (VDD-vgs-Q5-max)+vsg-Q1-max
= 12V

Gain Stage Design

Although I used a convenient fourier series analysis tool to do the small signal analysis, there were some design decisions.  I wanted to choose gate and source resistors small enough to make their noise contribution insignificant compared to the JFET.  This would lower noise and allow the possiblity of noise improvement with better JFETs.
Resistor noise equation (per rtHz, figure of merit, does not account for bandwidth)
vn-resistor-rtHz = sqrt(4kTR)
Rearrange resistor noise equation to calculate equivalent input noise resistance of JFET.
R =
4 x 1.3806504e-23 x 298.15K
 = 60.73254378Ω
I would like to choose RG and RS much lower than 60Ω to make their contribution insignificant.  However, I believe a 6Ω source resistor would greatly raise distortion and bias for a small signal transistor.  Rather, instead I will match the source resistor to the JFET noise resistance.
R1-R8 = 56Ω.
Because the fourier series analysis of this low noise design showed higher distortion than the 2N3819 version (but still very good) and because decreasing the source resistors furthur would increase the distortion, I made no furthur analysis with lower source resistor values.

Fourier Series Analysis of Q1-Q4 (excluding drain load)

Analysis of Common Source Circuit
transistor type = JFET (Depletion mode FET approximation)
IDSS = 12.25mA
MOSFET constant = 25mS
threshold voltage = -700mV
source resistor = 56Ω

Input signal (peak) = 7.071mV
Input bias = 0V
Output signal (peak) = 69.341565A
Output bias = 4.7314585mA
Gain = 9.8064722mS

Total distortion = 57.385979nA = 0.08275841% = -61.6438dB

Breakdown by harmonic

harmonic value percent dB
0 4.7315158mA 6823.49150778% 36.68dB DC Offset
1 69.341565A 100.00000000% 0.00dB Fundamental
2 57.270744nA 0.08259223% -61.66dB
3 115.23468pA 0.00016618% -115.59dB

Distortion figures for first stage are before RIAA equalization.  At the output of the RIAA network, they are lower because of the downward sloping equalization. (A half or a little more for frequencies on or beginning the downward slope; in decibels, 6db lower)

RIAA Analysis

Refer Calculating Passive RIAA Equalization in article Phono Equalization Calculations.

From this fourier analysis we can calculate R1 of Passive RIAA network (aka R2L)
R2L =
(Low frequency RIAA boost relative to 1kHz) x (1kHz output voltage)
(Output signal current) x (#parallel devices)
10 x 1.414213562V
69.341565A x 4
= 56.08594638kΩ
Round to nearest 1% value.  R2L = 56.2kΩ

Given R1:R2 = 6.877358491
R2N = 56.2kΩ
= 8.171742112kΩ
Round to nearest 1% value.  R2N = 8.25kΩ

Given R1C1 = 2187S
C1N =
= 38.91459075nF
Round down to nearest 5% value: 39nF
Choose parallel 5% value to makeup remainder:  0
C1N = 39nF + 0

Given R1C2 = 750S
C2N =
= 13.34519573nF
Round down to nearest 5% value: 13nF
Choose parallel 5% value to makeup remainder:  360pF
C2N = 13nF + 360pF

Active Inductor Analysis

Refer article Active Inductor Load.
R2L chosen already in RIAA analysis.
BS170 specification:  gfs = 320mS @ 200mA.

Calculate MOSFET constant from gfs@iD  specification.
kn = gfs2
= 128mA/V2

Calculate new gfs from operating current and  kn.

iD = 4.7314585mA x 4 - (4V/56.2kΩ)  =   18.925834mA - 71.42857143A = 18.85440543mA
gfs = 2 x sqrt(kniD) = 2 x sqrt(128mA/V2 x 18.85440543mA) = 98.25200038mS

R1L =


left-paren Vbias

left-paren 6V
 = 38.46756952kΩ
Round to nearest 1% value:  R1L = 38.3kΩ
CL =
(98.25200038mS x 38.3kΩ)+1
2π x 5Hz x 38.3kΩ
 = 3.127458306mF
3300F would work here.  I would rather choose a larger common value (CL = 10,000F) and solve for a lower pole.
(98.25200038mS x 38.3kΩ)+1
2π x 10mF x 38.3kΩ
 = 1.563729153Hz
Power on relay added at beginning of design.
Calculate charge time for CL:
i = C δv

Rearrange capacitor equation above and calculate time.  (2V presumes vgs not much more than vT)
Δt =
10mF x 2V
= 1.060760048Sec

Q6 Circuit Design

My design goal here was reasonably low output impedance.  1kΩ output impedance would have been low enough to minimize voltage loss into a typical 47kΩ load.  I choose 500Ω instead on the far chance of driving a 600Ω load with a tolerable voltage loss.

Fourier Series Analysis of Q6 Circuit

Note:  Bias and source resistor represent equivalent circuit simplified for analysis.

Original Design R13 = 1kΩ, R15 = 1kΩ.  Large 2nd harmonic dominates distortion.  Expect to be very tubey

Analysis of Common Source Circuit
transistor type = MOSFET
MOSFET constant = 128mS
threshold voltage = 2V
source resistor = 500Ω

Input signal (peak) = 1.4142136V
Input bias = 7V
Output signal (peak) = 2.7487914mA
Output bias = 9.4563897mA
Gain = 1.943689mS

Total distortion = 3.0913658A = 0.11246273% = -58.9798dB

Breakdown by harmonic

harmonic value percent dB
0 9.4592357mA 344.12344914% 10.73dB DC Offset
1 2.7487914mA 100.00000000% 0.00dB Fundamental
2 2.8643766A 0.10420495% -59.64dB
3 206.32369nA 0.00750598% -82.49dB
4 18.586854nA 0.00067618% -103.40dB
5 1.875836nA 0.00006824% -123.32dB
6 202.86746pA 0.00000738% -142.64dB

For a somewhat more neutral sound, consider that raising the bias relative to the signal will lower the distortion.
Lower R13 and raise R15 to do this.
Alternative design:  R13 = 220Ω, R15 = 2kΩ.

Analysis of Common Source Circuit
transistor type = MOSFET
MOSFET constant = 128mS
threshold voltage = 2V
source resistor = 200Ω

Input signal (peak) = 1.4142136V
Input bias = 11V
Output signal (peak) = 6.8377279mA
Output bias = 42.131413mA
Gain = 4.8350036mS

Total distortion = 4.7943711A = 0.07011644% = -63.0836dB

Breakdown by harmonic

harmonic value percent dB
0 42.136007mA 616.22819497% 15.79dB DC Offset
1 6.8377279mA 100.00000000% 0.00dB Fundamental
2 4.6032189A 0.06732088% -63.44dB
3 181.69099nA 0.00265718% -91.51dB
4 8.965676nA 0.00013112% -117.65dB
5 495.5468pA 0.00000725% -142.80dB

Miscellaneous Calculations

R12 limits charge current through Q6 to C2 to <500mA
R12 >= 20V/500mA = 40Ω
Choose R12 = 82Ω.

Want to set C2 for as low a highpass pole as CL.  First find Requiv of pole.  Output impedance of Q6 is the inverse of its operating ac transconductance gfs.
gfs = 2 x sqrt(kniD) therefore
Rout-Q6 =
2 x sqrt(kniD)
2 x sqrt(128mA/V2 x 42.131413mA)
 = 6.808665637Ω

More tubey version:
Requiv = (Rout-Q6 || R13) + R15 = Rout-Q6 x R13
Rout-Q6 + R13
+ R15 = 1.006762621kΩ

C2 =
2π x 2.245671674Hz x 1.006762621kΩ
 = 70.39580633F
Choose C2 = 100F.

The same calculations for the more neutral version give
Requiv = 2.006604273kΩ and C2 = 35.31930408F
Choose C2 = 47F.

Choose R10 and  R14 to set 10mA through relay coil.
R10 =
 = 1.3kΩ

Calculate load current from power supplies VDD1 and VDD2.
iDD1 = iQ1-BIAS x 4 + iRELAYCOIL = 18.85440543mA + 10mA  = 28.85440543mA
iDD2 = iQ6-BIAS + iRELAYCOIL = 9.4563897mA + 10mA = 19.4563897mA (more tubey version)
iDD2 = iQ6-BIAS + iRELAYCOIL = 42.131413mA + 10mA = 52.131413mA (more neutral version)

Design Decisions Left to You

C1 must be chosen by calculation or experimentation to exceed CL charge time calculated above (432.2244809mSec).  C3 would then be chosen for a somewhat longer delay to prevent a pop from the activation of the other relay.

1There is a seeming discrepancy in the fourier analysis between the cited output bias and the same value given in the analysis.  The first value is the bias without any signal and the second with a signal. This effect is created by the same asymetrical transfer curve that gives us a desireable second order distortion characteristic.  The positive half of the signal is amplified more than the negative half.  As a result the dc bias is modulated by the average signal level. This small component amounts to AM demodulation of the  music signal.  Abstractly, I had already anticipated this effect.  I would be curious what effect this has on the overall sound.  Is is detrimental, or does it add an extra euphonic bass urge?

Document History
January 2, 2010  Created
January 2, 2010  Minor Updates.
January 5, 2010  Recommend split power supply to prevent feedback.  Update schematic to show change.  Recommend specific power supply.
January 5, 2010  Choose relays S1 and S2, calculate their series resistors, and calculate specific load current for each power supply tap.
January 9, 2010  Added footnote explaining how fourier series analysis shows modulation of dc bias by music signal.
January 17, 2010  Added design criteria for class-a biasing decisions.
January 17, 2010  Create new version of document based on LSK170 input JFETs.  Only Gain stage, RIAA, and Active Inductor analyses changed.
January 22, 2010  Minor Corrections.  Modify schematic for clarity.  Capacitor cited wrong in CL charge time calculation.   Calculated result already correct.  Change C2 calculation to match CL pole.  Kept same C2 choice.
January 23, 2010  Correct schematic in vicinity of Q4, R7, and R2N.
January 27, 2010  Minor improvement to writing style.
March 11, 2010  Corrected for improper gfs presumptions.
September 2, 2010  Added recommendation not to build circuit.