banner
 Home  Audio Home Page 


Copyright © 2012 by Wayne Stegall
Updated March 14, 2012.  See Document History at end for details.




One-Bend Amplifier

Part 1:  Compare a current-feedback amplifier allowing a one-bend distortion characteristic to its voltage-feedback equivalent.


Introduction

In the article Transfer Curve Shape and Distortion  I suggested that a circuit topology that alternated device polarity might produce a euphonic distortion result by bending the transfer curve in only one direction.  I decided to compare a useful design using this approach to its similar more common alternative.

The One-bend circuit

The complete schematic may obscure the principle.  Therefore, look at J1, M1, and M3-M6.  Only they affect the transfer curve.  J2 is only a casode, and M2 a current source.  J1 bends the transfer curve, M1 bends it the same way.  Square-law cancellation dictates that the source follower on the output has the bend of one MOSFET in the direction of the bank of FETs M3-M4 or M5-M6 which has the greatest transconductance.  NMOS transistors usually have the greatest transconductance of comparable NMOS and PMOS choices that might be paired up in this way.  On this presumption, the output stage also bends the transfer curve in the same direction as J1 and M1 did.  Then feedback will straighten the one bend to a nice, low, euphonic distortion result.

Figure 1:  Class-a, three-stage, current-feedback FET amplifier.
ampmoscfb


Its common sibling

The comparison circuit is identical except that the two-bend differential input stage will complicate the one-bend tendencies of the remainder of the circuit.

Figure 2:  Class-a, three-stage, voltage-feedback FET amplifier.
ampmosvfb


Common Design Decisions



Power Supply Design

I decided to split the power supplies to allow the first and second stages to have much less ripple than the output stage.  This resulted in the following power supply design.

Figure 3:  Split power supply for both circuits.
pwrsupply

The current out of VDD1 and VSS1 average at most 47mA with peak 94mA from VSS1.  For no more than average 0.1V ripple
C1P and C1N i
fΔv
=
47mA
120Hz × 0.1V
 = 3.917mF

It is a property of enhancement MOSFETs that they will not leave current mode for linear mode under any conditions until VDS drops below VGS-VT.  Since ripple for the first two stages has been specified very low, ripple in the output stage should be filtered by the drain until it exceeds VT a value typically >3V.  Allowing 3V of ripple on dynamic peaks would specify less for continuous operation and be entirely acceptable.  That simulation showed clipping 5V before rail voltage suggests more ripple headroom in the output circuit than this calculation.

The current out of VDD2 and VSS2 peak at most 8A during dynamic peaks.  For no more than average 3V ripple
C2P and C2N i
fΔv
=
8A
120Hz × 3V
 = 22.22mF

Circuit Adjustments

Much of the circuit is adjusted rather than having a calculated bias.
  1. Choose a value for R5 that will bias J5 for 10-12.5mA before assembly,
  2. Set R3, R7, and R13 so that the transistors they bias are initially cutoff.
  3. Connect Node A to ground through a jumper.
  4. Connect Node B to ground through an ammeter.
  5. Adjust R7 until ammeter reads 45mA.
  6. Connect Node B to ground through a jumper.
  7. Connect Node A to ground through an ammeter.
  8. Adjust R3 until ammeter reads 45mA.
  9. Remove jumpers from nodes A and B
  10. Adjust R13 for desired output bias current.
  11. Adjust R3 for 0V DC offset.
  12. Repeat steps 10 and 11 until they both measure correctly.

SPICE Results

At first I intended to alter the open loop gain by trimming R11 and R12 of each circuit until they had comparable distortion levels to base the comparison.  However, I found odd changes in the voltage-feedback transfer curve that called for a wider comparison.

SPICE deck for cfb circuit.
SPICE deck for vfb circuit.
SPICE deck for power supply

Transfer Error Curves

Any localized sharp kinks in these curves are due to localized difficulties in SPICE convergence.  You should imagine them to all be smooth.  All transfer curves were generated using idealized power supplies because DC analysis removes any capacitive filtering of simulated real supplies.

Figure 4
Current feedback
Voltage feedback
Low Feedback
Small Signal
R11=R12=1.5kΩ

trvflfss
Low Feedback
Large Signal
R11=R12=1.5kΩ

trvflfls

Current feedback Voltage feedback
Medium Feedback
Small Signal
R11=R12=1.8kΩ


Medium Feedback
Large Signal
R11=R12=1.8kΩ



Current feedback Voltage feedback
High Feedback
Small Signal
R11=R12=6kΩ


High Feedback
Large Signal
R11=R12=6kΩ




Analysis of Transfer Error Curves


Fourier analysis for current-feedback circuit, gain matched to distortion of following voltage-feedback analysis, 1W into 8Ω.

Fourier analysis for vout:
  No. Harmonics: 16, THD: 0.00623235 %, Gridsize: 200, Interpolation Degree: 3

Harmonic Frequency  
Magnitude  
Norm.Mag  
Percent  
Decibels










1 1000
4.02368
1
100
0
2 2000
0.000250739
6.23159e-05
0.00623159
-84.1080
3 3000
3.9235e-06
9.75101e-07
9.75101e-05
-120.219
4 4000
9.4909e-08
2.35876e-08
2.35876e-06
-152.546
5 5000
1.24382e-08
3.09125e-09
3.09125e-07
-170.197
6 6000
1.7024e-08
4.23094e-09
4.23094e-07
-167.471
7 7000
1.2994e-08
3.22939e-09
3.22939e-07
-169.817
8 8000
1.87858e-08
4.66881e-09
4.66881e-07
-166.615
9 9000
1.22738e-08
3.05039e-09
3.05039e-07
-170.312
10 10000
1.65326e-08
4.10884e-09
4.10884e-07
-167.725
11 11000
8.43652e-09
2.09672e-09
2.09672e-07
-173.569
12 12000
1.52337e-08
3.78601e-09
3.78601e-07
-168.436
13 13000
1.01237e-08
2.51604e-09
2.51604e-07
-171.985
14 14000
1.36079e-08
3.38196e-09
3.38196e-07
-169.416
15 15000
8.07009e-09
2.00565e-09
2.00565e-07
-173.954



Fourier analysis for voltage-feedback circuit, medium open-loop gain, 1W into 8Ω

Fourier analysis for vout:
  No. Harmonics: 16, THD: 0.00667249 %, Gridsize: 200, Interpolation Degree: 3

Harmonic Frequency  
Magnitude  
Norm. Mag  
Percent  
Decibels










1 1000
4.10897
1
100
0
2 2000
0.000274035
6.66919E-05
0.00666919
-83.5185,
3 3000
8.61175E-06
2.09584E-06
0.000209584
-113.572
4 4000
3.15514E-07
7.67865E-08
7.67865E-06
-142.294
5 5000
2.111E-08
5.13754E-09
5.13754E-07
-165.784
6 6000
4.68468E-09
1.14011E-09
1.14011E-07
-178.861
7 7000
1.28597E-08
3.12966E-09
3.12966E-07
-170.090
8 8000
7.86997E-09
1.91531E-09
1.91531E-07
-174.355
9 9000
3.98815E-09
9.70597E-10
9.70597E-08
-180.259
10 10000
2.53434E-09
6.16781E-10
6.16781E-08
-184.197
11 11000
5.24857E-09
1.27734E-09
1.27734E-07
-177.873
12 12000
4.42277E-09
1.07637E-09
1.07637E-07
-179.360
13 13000
3.99287E-09
9.71743E-10
9.71743E-08
-180.248
14 14000
6.1974E-09
1.50826E-09
1.50826E-07
-176.430
15 15000
2.35759E-09
5.73765E-10
5.73765E-08
-184.825


Caveats

Although I did these designs mainly to illustrate a point, a lot care went into their correctness.  In spite of this, some things are or may be incomplete.
I might add these to this article at a later time, or deal with them in new articles.


Correction

By accident these simulations were done with the kp of the PMOS transistors of the output stage 5% greater than that of the kn of the matching NMOS transistors.  This is near to simulating square-law distortion cancellation and seems not to have affected the outcome of the experiment.  In next article, I adjust kn up 5% and kp down by the same amount as would simulate hand choosing transconductance factors favoring an overall NMOS characteristic.

This mistake is due to a shortsighted presumption that the KP SPICE parameter was equivalent to kn and kp.  I had my SPICE book handy and should have known better.  Transconductance factors are calculated properly by the following formula:

kn or kp =
KP × W
2 × Leff


Links

Next article:  One-Bend Amplifier  March 14, 2012.  Part 2:  Improving important details of a current-feedback amplifier.




Document History
January 31, 2012  Created.
January 31, 2012  Added another step to circuit adjustments.
March 14, 2012  Added correction concerning output stage transconductance factor and added a link to next article.