Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
1.Noise behaviour of Amplifiers.
1. Noise model of an Amplifier.
1.1. An amplifier is made with active and passive components. All these component
produce noise. To find the noise on the output you calculate for every noise
source the transfer function to the output. This will end up into a noise signal on
the output of the amplifier, which is the statistical sum of all noise sources times
there transfer function to the output.
2
V out
A⋅Uin
U in
Uout
Figure 1: A noisy amplifier.
In figure 1 the equivalent schematic of a noisy amplifier is drawn. In this
schematic all the noise sources in the amplifier are concentrated in the signal
2
source Vout
.
1.2.
For noise calculation on a noisy amplifier an equivalent schematic is
drawn in figure 2.This schematic has in it a noiseless amplifier and two noise
sources, a voltage source Vs2 and current source i 2p . The noise sources are partly,
or even completely correlated.
Vs
2
A ⋅ U in'
Uin
i 2p
U in'
Uout
Figure 2: The equivalent schematic of a noisy amplifier.
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
The value of the voltage source Vs2 can be determined by shorting the input to
the ground. Assumed hereby is, the amplifier still works with the input shorted.
The noise source i 2p is shorted now; the noise apparent at the output is due to Vs2
alone.
Wen the input is left floating, the noise due to the current source i 2p only will
appear on the output.
2. The Voltage Amplifier
2.1. Resistive source impedance.
In practice voltage amplifier with a resistive source are commonly used (figure
3). The signal source Vin with his source resistor Rin makes the input voltage for
the amplifier.
Voltage amplifier
Source
2
i in
Vs
2
A
R in
2
Vin
ip
Figure 3: Voltage amplifier with a resistive source.
The output signal of this circuit has two components in it. First the input signal
amplified with the factor A (under condition that the input impedance of the
amplifier is very high). Second the noise signal.
In formula these signals are:
Vout = A ⋅ Vin
Vout
Vin
A
Equation 1.
Output voltage.
Input signal.
Amplification.
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
2
vineq
= Rin2 ⋅ (iin2 + i p2 ) + v s2 = 4 ⋅ k ⋅ T ⋅ Rin + Rin2 ⋅ i 2p + v s2 [In V2/Hz] Equation 2.
vineq
equivalent noise voltage (in V/√Hz)
vs
ip
input noise voltage (in V/√Hz)
input noise current (in A/√Hz)
Rin
k
T
signal source impedance.
constant of Boltzmann, 1.38 ⋅ 1023 J/K
absolute temperature.
The first term in equation 2 describes the noise contribution of the source
impedance. Normally you don't have any influence on the first term, because the
impedance is part of the voltage source and therefor given,
The other two terms are due to the amplifiers own two noise sources. Choosing
the amplifier properly can influence them. This is realised by selecting the right
opamp or by designing an appropriate amplifier.
It is normal to choose vs2 = Rin2 ⋅ i p2 . The voltage due to the current source over Rin
equals the value of the voltage source. The result is a lower noise level compared
to just one of the noise levels.
It is obvious that the source impedance Rin has a high influence on the noise of
an amplifier, partly due to it's own thermal noise and indirect by the voltage drop
of i p .
2.2. In figure 4 a practical schematic is drawn of an opamp voltage amplifier. We
assume that the opamp has a gain of ∞ , so the negative input of the opamp is
virtually ground.
Opamp
Voltage
amplifier
2
i2
Source
R
2
i in
i
2
2
1
Vs
R1
2
-
R in
A0 = ∞
i
Vin
2
p
+
Figure 4: A practical schematic of an opamp voltage amplifier.
The amplifier circuit loads the source with the resistor R1 . This makes the input
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
voltage and the therefor also the gain of the amplifier depending on the value of
the source impedance.
R2
Vout = −
⋅ Vin
Equation 3.
Rin + R1
Due to this dependency usually the non-inverting configuration is preferred
above the amplifier in figure 4.
The spectral equivalent of the noise is:
( Rin + R1 ) 2
R + R1 2
2
vineq = 4 ⋅ k ⋅ T ⋅ ( Rin + R1 +
) + ( Rin + R1 ) 2 ⋅ i p2 + v s2 ⋅ (1 + in
)
R2
R2
Equation 4.
In equation 4 the influence of the new components in the circuit have been added
compared to equation 2. In the first term R1 and R2 are added. In the second
term the current i p also runs through the resistor R1 . In the third term the
contribution of vs due to the feedback have been added.
If the opamp in figure 4 has a differential input stage, the source v s+2 must be
added in equation 4. Due to this a differential input pair is not advisable, when
we are designing an amplifier in a noise critical environment. A single transistor
input stage in front of the opamp, with enough pre amplification, will make the
noise contribution of the opamp negligible.
2.3. Resistive and capacitive source impedance.
Voltage amplifier
Source
2
i in
V s2
A
R in
2
Vin
ip
C in
Figure 5: Voltage amplifier with resistive and capacitive source.
In figure 5 a voltage amplifier is drawn with a resistive and capacitive source.
This resister and capacitor form a low pass filter in the source. The voltage gain,
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
under the condition that the amplifier is ideal, is:
A
A
Vout =
⋅ Vin =
⋅ Vin
1 + l ⋅ ω ⋅ Rin ⋅ Cin
1 + j ⋅ ω ⋅ τ RC
Equation 5.
The low pass RC filter shows clearly in this formula. For noise the formula is:
2
= 4 ⋅ k ⋅ T ⋅ Rin + Rin2 ⋅ i 2p + v s2⋅ ⋅ 1 + j ⋅ ω ⋅ Rin ⋅ Cin
vineq
2
Equation 6.
The difference between equation 6 and equation 2 is that the contribution of the
source vs is frequency depended. The higher the frequency the bigger the
contribution of vs gets. This effect is called "gain peaking". It can be explained
by looking at the differences in bandwidth for Vin and vs . Vin Is connected to the
amplifier through a low pass filter, while vs is directly coupled to the input. The
last one sees the full bandwidth of the amplifier for not to low frequencies.
The conclusion of this part is noise reduction by bandwidth limitation should be
done on the output of the first amplifier.
2.4. Charge measurement with a voltage amplifier.
Most times, in a detector when an event has occurred, a charge is produced. This
charge can be measured with a voltage amplifier, by converting it into a voltage
in a capacitor. Often is it possible to use the detectors internal capacity. The
schematic of this set-up is drawn in figure 6.
Voltage amplifier
Source
Vs
2
A
Iin =
Vin
Rin
2
R in
i
2
in
C in
ip
Figure 6: Current measurements with a voltage amplifier.
The difference between figure 5 and 6 is the set-up of the source. In figure 5 it a
voltage source and in figure 6 the Thevenin equivalent of this source, a current
source with the internal resistor Rin parallel.
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
The resistor Rin is also used to discharge the capacitor Cin . The time constant of
this circuit can be too high; so pulse pile up will occur. To solve this a reset
switch or a smaller resistor can be placed parallel with Cin .
The transfer function of the schematic, when we neglect the discharge, is:
A ⋅ ∫ I in dt
A ⋅ Qin
Vout = −
=−
Equation 7.
Cin
C in
The formula for the noise, in unit of charge, is:
i 2p
1
2
+
+ vs2 ⋅ Cin2
qineq = 4 ⋅ k ⋅ T ⋅
2
2
Rin ⋅ jω
jω
[in C2/Hz]
Equation 8.
The first terms in equation 8 are depended of 1 jω , and are referred to as
2
'parallel noise'. In the last term, 'serial noise', the noise is proportional with Cin2 .
The voltage amplifier has an advantage over the charge amplifier as long as the
input capacitor is small.
3. Current Amplifier.
3.1. Resistive and capacitive source impedance.
One of the ways to measure a current is measuring the voltage drop over a
resister (shunt). The problem with this method is, when we look at the source in
figure 7 and place the shunt between the output points of the source, the current
will be divided between the shunt and the resistor Rin . There for the voltage is
not the result of Rshunt ⋅ i but of (Rshunt // Rin ) ⋅ i .
The conclusion from this is the shunt must be relatively small compared with
Rin .
To avoid this problem for small currents a current amplifier or transconductance
amplifier is used. In figure 7 the schematic of a current amplifier is drawn.
The input of the amplifier is controlled by the feedback loop and kept at 0 Volts,
on condition of a high gain. The input impedance of the amplifier is very high, so
the complete input current will run into the feedback resistor. The result is:
Vout = − R fb ⋅ I in
Equation 9.
The input equivalent current noise is:
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
2
iineq
= 4 ⋅ k ⋅T ⋅
1
+ i 2p +
R fb // Rin
v s2
R fb ⋅ Rin ⋅
1
j ⋅ ω ⋅ Cin
2
[in A2/Hz]
Equation 10.
Current
amplifier
Source
2
i fb
R
Vs
Iin =
fb
2
−A
2
Vin
2
R in
Rin
i in
ip
C in
Figure 7: A current amplifier with a capacitive and resistive source.
The first term in equation 10 describes the thermal noise due to the resisters. In
the last term it is clear that the contribution of vs increases with the frequency.
3.2. Charge measurement with a current amplifier.
To measure charge with a current amplifier it is necessary to add an integrator
circuit on the output of the current amplifier, see figure 8.
Current
amplifier
Source
2
i fb
Integrator
C
R fb
2
Vs
Iin =
Vin
Rin
−A
R
−A
2
R in
2
i in
ip
C in
Figure 8: charge measurement with a current amplifier.
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
The formula for the output is:
R fb
R fb
Vout = −
⋅ ∫ I in dt = −
⋅ Qin
Equation 11.
R ⋅C
R ⋅C
The output voltage is independent of the input capacity.
When we assume that the integration stage is noise less, the equivalent input
noise is:
i 2p
v s2
1
2
qineq
= 4 ⋅ k ⋅T ⋅
+
+
+ v s2 ⋅ Cin2
2
2
2
2
jω
R fb ⋅ jω
( Rin // R fb ) ⋅ jω
[in C2/Hz]
Equation 12
4. Charge Amplifier.
4.1. Parallel resistive and capacitance source impedance.
The easiest way of measuring charge is collecting it in a capacitor and measures
the voltage over the capacitor. This method has one major drawback; the voltage
over the capacitor depends also on the signal capacitance Cin . The charge will be
divided between the two capacitors, what will cause an error.
This error can be avoided by putting the capacity in the feedback loop of an
amplifier. The input side will be virtually grounded, with a high gain. All the
charge will flow into the feedback capacitor, under condition that the input
impedance of the amplifier is high.
Charge
amplifier
2
i fb
R fb
Source
C fb
Vs
Iin =
Vin
Rin
2
−A
2
R in
2
i in
ip
C in
Figure 9: A charge amplifier with capacitive a resistive source.
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NIKHEF, Amsterdam.
Noise behaviour of amplifiers.23-10-98, JDS
Preliminary.
The schematic of a charge amplifier is drawn in figure 9. The size of Cin is for
the output voltage of no interest any more, because the voltage over it is always
0. In the noise calculation it still plays a role.
The transfer function is:
∫ iin dt = − Qin
Vout = −
C fb
C fb
Equation 13.
The circuit in figure 9 is not stable. The capacitor must be discharged to make it
stable. There are two options to do this, a reset switch and a feedback resister.
The output of the amplifier would just make a voltage step every time a charge
appears on the input. The resister or the switch will discharge Cin to 0 V.
The disadvantage of the feedback resister R fb is the time constant of the circuit,
pulse pile up is possible and some extra noise. A reset switch has also a
disadvantage; he introduces dead time in the circuit.
The formula for noise is without the resistor R fb :
2
ineq
q
= 4 ⋅ k ⋅T ⋅
1
Rin ⋅ jω
2
+
i 2p
jω
+ v s2 ⋅ (Cin + C fb )
2
2
Equation 14.
The first two terms are called parallel noise. They strongly increase at low
frequencies. The last term is due to the serial noise, and increases with the square
of the input capacitance.
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