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Copyright 2016 by Wayne Stegall
Created January 16, 2016.  See Document History at end for details.




Dual-slope Integrator

Reduce high frequency amplifier distortion

Introduction

Amplifiers are often diminished in their distortion performance at high frequencies by open loop responses that fall with the slope of a single slope integrator.  This occurs whether imposed by the integrator at the front of a delta-sigma class D amplifier or by dominant-pole compensation of a more conventional class A or B types.  However, if the integrator slope would fall off at a faster rate, it would impose less reduction in gain as the feedback crossover point is reached.  Then lower distortion would prevail to higher frequencies in the closed loop circuit until the approach of the feedback crossover frequency required the loop gain to fall back to unity.  The black line in figure 1 below shows a response that suits this task.  A second-order slope prevails at lower frequencies until a lowpass zero restores a first-order slope for higher frequencies.  This may seem counter intuitive because you would think 180 phase shift at the lower frequencies would create oscillations.  However, if the closed loop creates a feedback crossover frequency higher than the zero frequency the global feedback loop will override any such tendency.

Figure 1:  Dual-slope integrator response superimposed on related single-slope integrator response.
integrator
Legend:
green:  first order integration.
black:  dual-slope integration.

Circuit and Calculations

To get a basis to use this integrator it is necessary to calculate an analysis of the circuit.  The op-amp circuit of figure 2 below was chosen for simplicity.  Although this circuit could appear at the front of a delta-sigma amplifier, the discrete transistor stage in a class A or B amplifier would calculate similarly.

Figure 2:  Dual-slope integrator implemented with operational amplifier.
integrator

(1)
va
vin
 =   1
sR1C1
, iC1 =
vin
R1

(2)
iC2 = iC1 va
R2
 = 
vin
R1
+
vin
sR1R2C1

(3)
iC2 = vin(sR2C1 + 1)
sR1R2C1

(4)
vout = va   iC2
sC2

(5)
vout = vin
sR1C1
   vin(sR2C1+1)
s2R1R2C1C2

(6)
vout = vin(s(R2C2 +R2C1)+1)
s2R1R2C1C2

(7)
H(s) =
vout
vin
 =   sR2(C1+C2)+1
s2R1R2C1C2

(8)
fzero =
1
2πR2(C1+C2)

Above fzero:

(9)
H(s) = C1+C2
sR1C1C2
 = 
1
sR1(C1 <series> C2)

 

Closed Loop

Now lets close the global feedback loop as in figure 3 below and analyze again.

Figure 3:  Dual-slope integrator placed into context with additional open-loop gain and the application of global feedback.
closedloopgen

Text
(10)
H(s) = 
A
1+Aβ

(11)
H(s) = 
sA2R2(C1+C2) + A2
s2R1R2C1C2

1+β
sA2R2(C1+C2) + A2
s2R1R2C1C2


(12)
H(s) =  sA2R2(C1+C2) + A2
s2R1R2C1C2 + sβA2R2(C1+C2) + βA2

(13)
H(s) =  sR2(C1+C2) + 1
β s2R1R2C1C2
βA2
 + sR2(C1+C2) + 1

In the form
s2
ω02
 + 
s
ω0Q
 + 1

(14)
ω0 =
     
βA2
R1R2C1C2

(15)
Q =
     
R1R2C1C2
βA2

R2(C1+C2)
 =
1
C1+C2
     
R1C1C2
βA2R2

(16)
Q = ωzero
ωpole


Application

Although full calculations have been made, real applications would fall into a more complex stability analysis with other poles adding phase shift.  So its seems that equations 8 and 9 become most useful:  equation 9 to place the first-order integration in its proper place then equation 8 to set the zero below the feedback crossover frequency by an amount deemed proper for the overall effect in the SPICE analysis and the hardware prototype.

SPICE Verification

For SPICE verification the feedback summing node of figure 3 was moved to the inverting terminal of the integrator's op amp.  The result is shown in figure 4 below.

Figure 4:  Circuit used for SPICE verification 
closedloop-spice

The SPICE amplifier simulated is specified for:

The first simulations puts fzero at the feedback crossover frequency to see how bad the peak is there.

Figure 5:  SPICE deck configured for fzero = ffeedback-crossover. 
* dual slope integrator example
* Spice Opus 2.31
v1 vin 0 dc 0 ac 1 sin 0 0.1V 1kHz
r1 vin vn 1.62k
c1 vn vc1c2 6.19n
r2 vc1c2 0 130
c2 vint vc1c2 6.19n
eopa vint 0 0 vn 100k
ea2 vout 0 vint 0 100
rf vout vn 51.1k
.end
.control
set units=degrees
ac dec 20 1k 1meg
plot db(vout)
.endc


Figure 6:  ≈ 3.5dB peak configured for fzero = ffeedback-crossover.
bode-spice1

The second simulations puts fzero at the half of the feedback crossover frequency to see how much the peak is reduced.

Figure 7:  SPICE deck configured for fzero = ffeedback-crossover.
* dual slope integrator example
* Spice Opus 2.31
v1 vin 0 dc 0 ac 1 sin 0 0.1V 1kHz
r1 vin vn 1.62k
c1 vn vc1c2 6.19n
r2 vc1c2 0 255
c2 vint vc1c2 6.19n
eopa vint 0 0 vn 100k
ea2 vout 0 vint 0 100
rf vout vn 51.1k
.end
.control
set units=degrees
ac dec 20 1k 1meg
plot db(vout)
.endc

Figure 8:  ≈ 2dB peak configured for fzero = ffeedback-crossover.
bode-spice2

Additional experimentation reveal some additional lowering of peak at the cost of a much lower zero frequency.  This is undesirable because much of the gain in lower distortion is lost as the the zero frequency is lowered.  It is perhaps desirable to accept the +0.5dB rise at 20kHz for a distortion improvement.  At this point, I did additional experimentation with adding lead compensation to the global feedback loop:  i.e. feedback capacitance CF in parallel with the RF already specified.  Setting fzero back to the feedback crossover frequency produced excellent results with an experimentally chosen value of lead compensation.

 

Figure 9:  SPICE deck configured back to fzero = ffeedback-crossover with lead compensation added at an experimentally chosen value.
* dual slope integrator example
* Spice Opus 2.31
v1 vin 0 dc 0 ac 1 sin 0 0.1V 1kHz
r1 vin vn 1.62k
c1 vn vc1c2 6.19n
r2 vc1c2 0 130
c2 vint vc1c2 6.19n
eopa vint 0 0 vn 100k
ea2 vout 0 vint 0 100
rf vout vn 51.1k
cf vout vn 43p
.end
.control
set units=degrees
ac dec 20 1k 1meg
plot db(vout)
.endc


Figure 10:  No peak when configured back to fzero = ffeedback-crossover with lead compensation added at an experimentally chosen value.  
bode-spice3




Document History
January 16, 2016  Created.