How to Bias Op-Amps Correctly
op-amps, operational amplifiers, bias, biasing op-amps
Renesas Electronics Corporation
REN r13an0003eu0100-biasing-op-amps APN 20191206 Application Note
Operational Amplifiers
How to Bias Op-Amps Correctly
Abstract
The inputs of an operational amplifier (op-amp) must be DC-biased to ensure proper device operation. A basic requirement that many textbooks neglect to discuss in detail. Consequently, engineers new to op-amps might overlook this important requirement, which can lead to malfunctioning circuits.
This application note tries to rectify this shortcoming by explaining the need for input biasing and suggesting practical solutions that ensure proper circuit operation.
Contents
1. The Need for Input Biasing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2. AC-Coupled Amplifiers with Dual Supplies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3. AC-Coupled Amplifiers with Single Supply. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3.1 Design Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 4. Appendix. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4.1 Deriving the -3dB Frequency, fCT, of the Overall Circuit Gain, GT. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 5. Revision History. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
List of Figures
Figure 1. Figure 2. Figure 3. Figure 4. Figure 5. Figure 6. Figure 7. Figure 8. Figure 9. Figure 10. Figure 11. Figure 12. Figure 13. Figure 14. Figure 15. Figure 16. Figure 17.
Input Biasing Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 DC-Coupled Buffer and Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 A Malfunctional AC-Coupled Amplifier with Missing DC Bias Path. . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Correctly AC-Coupled Amplifier for Dual-Supply Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Examples of Non-Functional AC-Coupled Instrumentation Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . 3 Correctly Biased INA Inputs use High-Value Input Resistors Between Each Input and Ground . . . . 3 Input Biasing with Voltage Reference for Highest PSRR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Input Biasing with Buffered Voltage Divider to Maintain High PSRR . . . . . . . . . . . . . . . . . . . . . . . . . 4 Transfer Function and Frequency Response of the Supply-Noise Gain of the Voltage Divider. . . . . 4 Transfer Function and Frequency Response of the AC-Coupled Input Stage . . . . . . . . . . . . . . . . . . 5 Transfer Function and Frequency Response of the Closed-Loop Gain of the Op-Amp, ACL. . . . . . . 5 Frequency Responses of the Various R-C Units of the Circuit in Figure 7 on page 4 . . . . . . . . . . . . 6 AC-Coupled Noninverting Amplifier with ISL28134 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 AC-Coupled Inverting Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 AC-Coupled Inverting Amplifier with ISL28134 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 AC-Signal Equivalent Circuit of Figure 8 and its Frequency Responses . . . . . . . . . . . . . . . . . . . . . . 8 FCT/FpG Ratio vs Passband Gain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
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Operational Amplifiers
How to Bias Op-Amps Correctly
1. The Need for Input Biasing
Figure 1 shows the differential input stage of an op-amp. The base terminals of transistors Q1 and Q2 form the non-inverting and inverting op-amp inputs, IN+ and IN-, respectively. For the op-amp to operate correctly, these inputs must be DC biased. That is, the DC bias currents (IB+ and IB-), must be able to flow into or out of the input terminals. The direction of the bias currents depends on the type of transistors. For NPN transistors and N-channel enhancement MOSFETs, the currents flow into the inputs, for PNPs and P-channel enhancement MOSFETs, the currents flow out of the inputs.
DC biasing is achieved by connecting the inputs via resistors to a reference potential, Vmid, which is the mid potential of the positive and negative supply voltages (VS+ and VS-) (Figure 1).
The biasing resistors (RB+ and RB-), represent equivalent resistances, because they can consist of one or more resistors as depicted in the DC-coupled amplifier circuits of Figure 2.
VS+ IB+
VS+
IB+
IN+
Q1
RB+
Vmid
RB-
IB-
IN-
next stage
Q2
VS-
Figure 1. Input Biasing Principle
Buffer VIN
IB- OPA
GND
VS-
VOU T GND
VS+ RIN IB+
VIN
IB- OPA
GND
RF
Amplifier
RG
VS-
VOUT GND
GND
Figure 2. DC-Coupled Buffer and Amplifier
In the DC-coupled buffer of Figure 2, IB+ flows through the signal source and its low source impedance into the noninverting input, while IB- flows from the output of the op-amp into the inverting input.
In the DC-coupled amplifier, IB- flows through RF and RG; therefore changing the DC voltage at the inverting input by IB- · (RF||RG). This introduces a DC offset voltage between the inputs. To minimize this offset, the DC potential at the non-inverting input must be adjusted by about the same amount. This is achieved by inserting a resistor (RIN), in series whose value matches the parallel combination of RF and RG, RIN = RF||RG.
2. AC-Coupled Amplifiers with Dual Supplies
While DC-coupled op-amp circuits receive their biasing through the signal source impedance, AC-coupled circuits have this bias path blocked by the input coupling capacitor (CIN). Figure 3 shows an AC-coupled amplifier without a path for the DC bias current to flow. In this case, IB+ charges the coupling capacitor until the common-mode voltage rating of the input circuit is exceeded, or its output is driven into saturation. Depending on the polarity of the bias current, the capacitor charges towards the positive or negative supply rail, with the resulting bias voltage being amplified by the closed-loop DC gain of the amplifier, 1 + RF/RG.
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Operational Amplifiers
How to Bias Op-Amps Correctly
+VS
CIN
0.1
VIN
OPA
0.1
VOU T
+VS
CIN
0.1
VIN
RIN
OPA
0.1
VOU T
-VS
RG
RF
Figure 3. A Malfunctional AC-Coupled Amplifier with Missing DC Bias Path
-VS
RG
RF
-3dB Input Bandwidth
fC-in =
1 2 CIN RIN
with RIN = RF||RG
Figure 4. Correctly AC-Coupled Amplifier for Dual-Supply Operation
For a small bias current, such as 1pA, this process can take hours. Therefore, a casual lab test with an ACcoupled scope is more than likely to miss this problem, and the circuit does not fail until much later. Obviously, this problem must be avoided.
Figure 4 shows a simple solution by connecting the noninverting input through RIN to ground. This forms a new DC path for the bias current. Like in the DC-coupled case before, minimizing the offset due to bias currents requires RIN = RFRG.
Note: AC-coupling forms a high-pass filter with a cutoff at fC = 1/(2 CIN RIN), which sets the minimum input bandwidth of the amplifier. With the circuit gain being defined by the application, RF and RG can be determined. Their parallel value defines RIN, which then allows the calculation of the input capacitor: CIN = 1/(2 fC RIN).
Similar input biasing methods must be applied to the differential input stages of three-amp Instrumentation Amplifiers (INAs). Figure 5 shows INA circuits that are AC-coupled using either two capacitors or a transformer, without providing a DC bias path.
+VS 0.1
+VS 0.1
VIN
INA
VIN
VOU T
VR EF
0.1
INA
VR EF 0.1
VOU T
-VS
-VS
Figure 5. Examples of Non-Functional AC-Coupled Instrumentation Amplifiers
Correct biasing solutions for these circuits are shown in Figure 6, where a high-value resistor (RIN) is added between each input and ground.
+VS
CIN
0.1
+VS 0.1
VIN
RIN 1/10
RIN
INA
RIN
VIN VOU T VR EF
CIN
0.1
RIN 1/10
RIN RIN
INA
VR EF 0.1
VOU T
-VS
-VS
Figure 6. Correctly Biased INA Inputs use High-Value Input Resistors Between Each Input and Ground
There is a small offset-voltage error due to mismatches between the input resistors and the input bias currents. To minimize this error, a third resistor, about 1/10th their value (yet still large compared to the differential source resistance), can be connected between the two in-amp inputs, therefore bridging both resistors.
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Operational Amplifiers
How to Bias Op-Amps Correctly
3. AC-Coupled Amplifiers with Single Supply
The DC biasing of AC-coupled single-supply amplifiers also requires an input resistor connecting the noninverting input to the reference potential, Vmid. In single supply circuits however, Vmid is derived from VS as VS/2 to ensure a symmetrical, maximum dynamic range for both, input and output signals. This can make VS/2 susceptible to supply noise.
The ideal design approach is to use an integrated voltage-reference chip (VREF) with high power-supply rejection ratio (PSRR). A VREF also provides low output impedance, allowing to provide VS/2 potential to multiple reference locations within a larger circuit design (Figure 7).
CIN VIN
VS
VREF CI
VS 1
VS/2 RIN VS/2 OPA
0.1
VS/2 CO
RF RG
CG
VOU T
CIN VIN
VS RB
RB
VS 1
VS/2 RIN VS/2 OPA
0.1
RF
VS/2
RG
CB
CG
VOU T
Figure 7. Input Biasing with Voltage Reference for Highest PSRR
Figure 8. Input Biasing with Buffered Voltage Divider to Maintain High PSRR
For biasing a single-supply op-amp, a more convenient low-cost method exists in form of a voltage divider, whose output must be buffered with a large capacitor (CB), to make VS/2 less sensitive to supply noise, thus maintaining a high PSRR (Figure 8).
AC-coupled single-supply amplifiers also require the capacitive decoupling of the feedback path to ensure 0dB gain at DC. This prevents the VS/2 potential at the noninverting input from being amplified by the passband gain, therefore saturating the output.
To calculate the component values, it is important to understand the interactions of the various time constants: RB-CB, RIN-CIN, and RG-CG. For clarity, their frequency responses are depicted in Figure 9, Figure 10, and Figure 11.
To the VS supply, the biasing voltage divider presents a low-pass filter whose frequency response is:
GSN
VCC VCC
2
1 2
1
1 jCBRB
, with a cutoff frequency of 2
fCB
1 2 CB RB
. 2
Here,
GSN
denotes
the
supply-noise gain.
A (dB)
VS
(Input Noise)
RB RB
VS/2
(Output
CB Noise)
GSN =
VS/2 VS
=
1 2
1 1+jCBRB/2
0
-6
GSN
fCB =
1 2CBRB/2
f (Hz)
Figure 9. Transfer Function and Frequency Response of the Supply-Noise Gain of the Voltage Divider
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Operational Amplifiers
How to Bias Op-Amps Correctly
The RIN-CIN circuit has a high-pass response
GIN
jCINRIN 1 jCINRIN
with
pole
at
fCI
1 2CIN RIN
. This pole and the
pole of the closed-loop gain, fpG, determine the minimum input bandwidth of the amplifier circuit, fCT.
CIN VIN+
A (dB) 0
f (Hz)
VIN
RIN
GIN
fCI =
1 2RINCIN
GIN =
VIN+ VIN
=
jRINCIN 1+jRINCIN
Figure 10. Transfer Function and Frequency Response of the AC-Coupled Input Stage
The
closed-loop
gain
of
the
op-amp
is
given
with
ACL
1
jCG RG RF .
1 jCG RG
It has a pole and a zero-location due to the decoupling effect of CG. The zero
frequency,
fzG
1
2CG RG
RF ,
is the +3dB corner frequency above unity gain. This zero ensures that DC voltages at the noninverting input, such
as
VS/2
and
any
offset
errors,
are
amplified
at
a
gain
of
1V/V
(0dB).
The
pole
frequency, fpG
1 2CG RG
,
defines the -3dB cutoff of ACL. This pole and the pole of the input stage, fCI, determine the minimum input bandwidth of the amplifier circuit, fCT.
A (dB)
VIN
OPA
VOUT
RF
RG CG
ACL
=
VOU T VIN
=
1+j(RG+RF)CG 1+jRGCG
0
ACL
fpG
=
1 2RGCG
f (Hz)
fzG
=
1 2(RG+RF)CG
Figure 11. Transfer Function and Frequency Response of the Closed-Loop Gain of the Op-Amp, ACL
Depicting all discussed frequency responses, Figure 12 shows how GIN and ACL define the total gain (GT), of the AC-coupled amplifier using GT = GIN · ACL, therefore moving the new -3dB cutoff, fCT, to a higher frequency. To ensure enough supply-noise suppression at this new cut-off frequency, practical experience suggests that the corner frequency of the voltage divider should be at least 1/10th of fCT: fCB = 0.1 · fCT.
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Operational Amplifiers
How to Bias Op-Amps Correctly
40 30 20
fzG 10
ACL 0
fpG
GT
fCT
fCB ~ 0.1fCT
GIN
Gain (dB)
-10
GSN
-20
-30
fCB
fCI
GSN= -26dB
-40
-50
-60 1m
10m
100m
1
10
100
1k
10k
Frequency (Hz)
Figure 12. Frequency Responses of the Various R-C Units of the Circuit in Figure 7 on page 4
3.1 Design Example
This section discusses the practical design of an actual amplifier circuit. As previously mentioned, with regard to component selection there are always compromises to be made between power consumption, resistor noise contribution, and supply noise rejection.
The design goal is to build an AC-coupled noninverting amplifier, operating from a single 5V supply, with an overall passband gain of 100V/V (40dB) and a minimum bandwidth of fCT = 16Hz. This requires the corner frequency of the voltage divider to be fCB 1/10th fCT = 1.6Hz. To preserve the power in the voltage divider, a DC quiescent current of 50µA is allowed. This makes the value of RB.
(EQ. 1)
RB
VS 2Iq
5V 2 50A
50 k
The buffer capacitor, CB, can therefore be calculated with:
(EQ. 2)
CB
1 2 fCB RB
2
1 2 1.6 Hz
25 k
4F
To ensure an earlier rather than later supply-noise roll-off, chose the next higher standard value with CB = 4.7µF. Aiming for equal corner frequencies for the input RIN-CIN stage and the closed-loop gain of the op-amp (ACL), demands that fCI ~ fpG. These frequencies are derived from fCT using:
(EQ. 3)
fCI fpG
fCT 1
2
16 Hz 1.554
10.3Hz
10Hz
To see the detailed derivation of Equation 3, see "Appendix" on page 8.
For the noninverting amplifier, a passband gain of 100V/V requires the ratio RF/RG = 99. Here, the compromise between the feedback path loading, noise contribution from RF, and the capacitor size of CG lead to resistor values of RF = 100k and RG = 1.01k. The capacitance (CG), required to place the ACL pole at 10.3Hz can now be calculated with:
(EQ. 4)
CG
1 2 fpG RG
1 2 10.3 Hz 1.01k
15F
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Operational Amplifiers
How to Bias Op-Amps Correctly
To calculate RIN, the offset due to bias current flow must be considered. While IB- flows through RF only, as the DC path through RG is blocked by CG, IB+ flows through RIN and the parallel circuit of the two RB resistors. To minimize this offset, the sum of RIN and RB/2 must equal RF. RIN is the difference between RF and RB/2:
(EQ. 5)
RIN RF RB 2 100k 25k 75k
Then, the input capacitance that produces the corner frequency derived in Equation 3 on page 6 is calculated with:
(EQ. 6)
CIN
1 2 fCI RIN
1 210.3Hz 75k
206 nF
Again, to make the roll-off occur slightly earlier, chose the next higher standard value with CIN = 220nF. Figure 13 shows the final amplifier circuit and its corresponding frequency responses.
220n VIN
5V
75k
49.9k
4.7µ
49.9k
5V
40
1
30
20
0.1
10
A1
VOUT
100k
1.01k 15µ
A1 = ISL28134
Gain (dB)
0 -10 -20 -30 -40 -50 -60
10m 100m 1
ACL GIN GSN GT
10 100 1k 10k 100k 1M 10M Frequency (Hz)
Figure 13. AC-Coupled Noninverting Amplifier with ISL28134
In the case of the inverting amplifier in Figure 14, the biasing voltage divider remains the same as in the noninverting case. The VS/2 potential at the noninverting input also appears at the inverting input due to feedback action.
VS
RB VS/2
VS 1
0.1
CB
RB VS/2 OPA
VOU T
CIN RIN
RF
VIN
Figure 14. AC-Coupled Inverting Amplifier
Note: The gain setting resistor (RG), also represents the input resistance (RIN). With that, the closed-loop gain of the amplifier becomes the overall circuit gain. Its pole frequency is the cutoff frequency of the input stage and therefore, the minimum input bandwidth of the circuit. This simplifies the calculation of component value enormously.
The main difference is, IB+ only flows through the parallel circuit of the two bias resistors. Therefore, reducing the offset due to bias current forces RF to drop in value to match RB/2. For high gain applications with low input bandwidth, this necessitates a much smaller RG value, which in turn requires the increase of CIN. The simple design procedure would be to:
· Make RF = RB/2 for offset reduction, · Then deriving the gain resistor with RG = RF/G · And finally, calculating the input capacitor with CIN = 1/(2 fIN RG)
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Operational Amplifiers
How to Bias Op-Amps Correctly
Applying the above equations to an AC-coupled inverting amplifier with passband gain of G = 100V/V and minimum input bandwidth of fIN = 16Hz, while maintaining RB = 49.9k, results in RF = 24.9k, RG = 249, and CIN = 47µF. Setting the supply-noise roll-off of the voltage divider to 1/10th of fIN makes fCB = 1.6Hz, which yields a capacitor value of CB = 1/(2 fCB RB/2) = 4µF. Again, chose the next higher standard value of 4.7µF.
Figure 15 shows the actual circuit with its corresponding frequency responses.
5V 49.9k
4.7µ
49.9k
47µ 249 VIN
40
5V
30
1
20
10
Gain (dB)
0.1
0
GSN GT
A1
VOU T
-10
-20
A1 = ISL28134
-30
24.9k
-40
-50
-60 10m 100m 1
10 100 1k 10k 100k 1M 10M Frequency (Hz)
Figure 15. AC-Coupled Inverting Amplifier with ISL28134
4. Appendix
4.1 Deriving the -3dB Frequency (fCT), of the Overall Circuit Gain (GT)
Figure 16 shows the AC-signal equivalent schematic of the noninverting amplifier in Figure 8 on page 4 and its individual and overall gain responses.
CIN VIN
RIN
OPA RF
RG CG
VOUT
Gain (dB)
40
fzG
fpG
40
A0
30
fCT
20
A0
10
ACL
0
GT
GIN
-10 fCI
-20
1m
10m
100m
1
10
100
1k
10k
Frequency (Hz)
Figure 16. AC-Signal Equivalent Circuit of Figure 8 and its Frequency Responses
The gain responses for GIN and ACL are:
GIN
jCINRIN 1 jCINRIN
with pole frequency
fCI
1 2CIN RIN
.
ACL
1
jCG RG RF
1 jCG RG
with pole frequency
fpG
1 2CG RG
, zero frequency
fzG
2 CG
1
RG
RF
,
and
passband gain
A0
RG RF RG
.
Because the ratio of pole to zero frequency is:
fpG fzG
RG RF RG
A0 , it follows that
fpG
fzG A0
.
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Operational Amplifiers
How to Bias Op-Amps Correctly
Expressing the transfer functions in their generic forms gives:
GIN
j CI 1 j CI
jf 1 j
fCI f fCI
and
ACL
1 1
j j
zG pG
1 1
jf jf
fzG fpG
This
makes the overall gain:
GT
GIN ACL
jf 1 j
fCI f fCI
1 1
jf jf
fzG fpG
And its magnitude function: GT
f fCI 2 1 f fCI 2
1 f
1 f
fzG 2
fpG 2
By making fCI = fpG and replacing fzG with fpG/A0 the magnitude function becomes:
GT
f 1
fpG 2 f fpG 2
1 f 1
A0 fpG f fpG 2
2
.
To find fCT, |GT| is set A0/2 (the -3dB magnitude of A0) and the generic frequency, f, is replaced with fCT:
A0 2
fCT fpG
2
1
fCT A0 fpG
2
1
fCT fpG
2
1
fCT fpG
2
A02 2
fCT2 fpG2
1
fCT2A02 fpG2
1
2
fCT2 fpG2
fCT4 fpG4
Then, solving for fCT results in:
fCT fpG 11 A02 1 11 A02 2 Therefore, for A0 > 10, fCT fpG 1 2 1.55 fpG .
Using 1.55 for A0 2V/V, yields less than 9% deviation from the theoretical fCT value. Plotting the factor of fpG over A0, one might simply use an average factor of 1.5.
fCT / fpG Ratio (n)
1.56 1.55 1.54 1.53 1.52 1.51 1.50 1.49 1.48 1.47 1.46 1.45 1.44 1.43 1.42 1.41
2
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Passband Gain, A0 (V/V)
Figure 17. FCT/FpG Ratio vs Passband Gain
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Operational Amplifiers
5. Revision History
Rev. 0.00
Date Dec.13.19
Initial release
How to Bias Op-Amps Correctly
Description
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