Bioelectric Amplifiers
Essential Electronics Formula
Ohm's Law
Ohm's Law: A voltage of 1V across a resistance
of 1W will cause a current of 1 A to flow. The
formula is
R = V / I
(where R = resistance in W, V = Voltage in V,
and I = current in A)
V = R * I
I = V / R
Reactance
The impedance (reactance) of a capacitor,
which varies inversely with frequency (as
frequency is increased, the reactance falls and
vice versa).
 XC = 1 / (2 Π f C)
 where Xc is capacitive reactance in Ohms, (Π pi) is 3.14159,
f is frequency in Hz, and C is capacitance in Farads.
Inductive reactance, being the reactance of an
inductor. This is proportional to frequency.
 XL = 2 Π f L
 where XL is inductive reactance in Ohms, and L is inductance
in Henrys
Frequency Frequency
There are many different ways for the
calculations of the frequency, depending on
the combination of components.
The -3dB frequency for resistance and
capacitance (the most common in amplifier
design) is determined by
fo = 1 / (2 Π R C)
where fo is the -3dB frequency
When resistance and inductance are combined,
the formula is
fo = R / (2 Π L)
Power
The power in any form can be calculated by
many means:
P=V I
P = V2 / R
P = I2 R
Where:
P
is the power in [W]
V
is the voltage in [V]
I
is the current in [A]
Amplifiers
Amplification Basics
The term "amplify" basically means to make stronger.
The strength of a signal (in terms of voltage) is
referred to as amplitude
Types of amplification
There are three kinds of amplifications: Two major
types, and the third type is derived from the another
two :
Voltage Amplifier - an amp that boosts the
voltage of an input signal
Current Amplifier - an amp that boosts the
current of a signal
Power Amplifier - the combination of the above
two amplifiers
Voltage
and
current
amplifier
Voltage amplifier:
In the case of a voltage amplifier, a small input
voltage will be increased
 so that for example a 10mV (0.01V) input signal might
be amplified so that the output is 1 Volt.
 This represents a "gain" of 100 - the output voltage is
100 times as great as the input voltage. This is called
the voltage gain of the amplifier.
Current amplifier:
In the case of a current amplifier, a small input
current will be increased.
 an input current of 10mA (0.01A) might be amplified
to give an output of 1A
 Again, this is a gain of 100, and is the current gain of
the amplifier.
Power gain Power Amplifier
If we now combine the two amplifiers, then
calculate the input power and the output power, we
will measure the power gain:
P=VxI
(where I = current, note that the symbol changes in a
formula)
The input and output power can be now calculated:
Pin = 0.01 x 0.01
(0.01V and 0.01A, or 10mV and 10mA)
Pin = 100 mW
Pout = 1 x 1
(1V and 1A)
Pout = 1W
The power gain is therefore 10,000, which is the
voltage gain multiplied by the current gain.
Types of Amplifiers
1. Vacuum Valve
2. Transistor
3. Operational amplifier
1. Vacuum Valve
In electronics, a vacuum
tube or (outside North
America) thermionic
valve or just valve, is a
device generally used to
amplify, switch or
otherwise modify, a signal
by controlling the
movement of electrons in
an evacuated space.
2. Transistor
Bipolar junction transistor
(BJT) are two diodes
joined with a very thin
common region
A small electrical input can
be amplified by transistor
Vout
Av 
Vin
,
ic

iE
RL
RL
Av   

re  RE
rM
A simple one-transistor
amplifier with positive
and negative supplies
Bioelectric
Amplifier
Is the amplifier
that used to
process bio-potentials
The gain may be low, medium or high (X10, X100,
X10000)
It is usually ac coupled.
DC-coupling is required where the input signal are
clearly dc or change very slowly (0.05 Hz)
Exceptional for EX.: ECG signal should be AC
coupled despite of the component as low as 0.05
Hz to overcome electrode offset potential from
electrode-skin connection
The high-frequency response is the frequency at
which the gain drops 3dB below its midfrequency
value (for ECG form 0.05 to 100 Hz)
Bioelectric Amplifier
Low gain amp: gain factor bw X1 and X10
Unity gain (X1) used for isolation, buffering and
possibly impedance transformation bw signal
source and readout device.
Used for relatively high-amplitude bioelectric
events (EX: action potential)
Medium gain amp: gain factor bw X10 and X1000
(EX: ECG, Muscles potentials, …)
High gain or low-level signal amp: gain factor over
X10000 to as high as X1000000 (EX: EEG)
Bioelectric
Amplifier
important parameters in Bio amp:
Noise: normally is the thermal noise
generated in resistances and semiconductors
devices.
Drift: change in output signal voltage caused
by change in operating temperature.
High input impedance: 107 to 1012 Ω and it
should be at least an order of magnitude high
than the source impedance.
Integrated circuit (IC) operational amplifier is
well suited as bioelectric amp because of its
properties.
Operational amplifiers
Operational Amplifier (Op-amp)
The op-amp is a device for increasing the power
of a signal. It does this by taking power from a
power supply and controlling the output to match
the input signal shape but with a larger
amplitude (Amplification).
The op-amp is used also to perform arithmetic
operations (addition, subtraction, multiplication)
with signals.
The properties of the negative feedback loop
determine the properties of the circuit containing
an op-amp.
Op-amp
It has two inputs: the inverting input (-) and the
non-inverting input (+), and one output.
It has usually two supplies (±Vss) but it can work
with one.
-Vss
Inverting
input
Non-inverting
Output
+
input
+Vss
Symbol of op-amplifier
What
is
inside
the
Op-amp?
The Op Amp is basically three amplifiers or stages.
The input differential stage; the gain stage, and the output
stage.
Real vs. Ideal Op-amp
Parameter
Ideal Op Amp Typical Op Amp
Open-loop voltage gain A
∞
105 – 109
Common mode voltage gain
0
10-5
Frequency response f
∞
1- 20 MHz
Input impedance Zin
∞
106 Ω (bipolar)
109–1012 Ω (FET)
Output impedance Zout
0
100 – 1000 Ω
Important
Supply Voltage
(±Vss): Parameters
 The maximum voltage (positive
and negative) that can be safely
used to feed the op-amp.
Differential Input Voltage:
 This is the maximum voltage that
can be applied across the + and
– inputs.
Input Voltage:
 The maximum input voltage that
can be simultaneously applied
between both input and ground
also referred to as the commonmode voltage.
 In general, the maximum voltage
is equal to the supply voltage.
Important
Parameters
Input Offset
Voltage (Voff
):
 This is the voltage that must be applied to one of the input
pins to give a zero output voltage. Remember, for an ideal
op-amp, output offset voltage is zero!
Input Bias Current (IB):
 This is the average of the currents flowing into both inputs.
Ideally, the two input bias currents are equal.
Open-Loop Voltage Gain (Ao):
 The output to input voltage ratio of the op-amp without
external feedback.
Common-Mode Rejection Ratio (CMRR):
 A measure of the ability of the op-amp' to reject signals
that are simultaneously present at both inputs.
 It is the ratio of the common-mode input voltage to the
generated output voltage, usually expressed in (dB).
Important
Parameters
Slew Rate (SR):
 Is the time rate of change of the
output voltage with the op-amp circuit
having a voltage gain of unity (1.0).
 SR = max rate at which amplifier
output can change in V/µs
 SR defines the Op-amps ability to
handle varying signals.
 SR defines how fast the amplifier is.
If op-amp is driven at rates > SR
(given in the spec. sheet) 
signal clipping & distortion.
VO
SR 
V/ ms
t
with t in ms
Examples
Input offset voltage
adjustment:
Supply Voltage:
Output signal shape of Op-amp
 Inverting mode:
 Invert the input
signal.
 Vo= -AVin.
 Non-inverting mode
(follower):
 Input signal dose
not change.
 Vo=+1Vin
Vin
Vo
Vin
Vo
Output signal shape of Op-amp
Differential mode:
V1
 The out put voltage is
proportional to the gain stage
and the difference bw V2 and
V1 as long as V1≠V2
 Vo=A (V2-V1)
Vo
V2
 Common mode:
Signal voltages are
common to both inputs
such V3 or where V1=V2
V0=A (V2-V1)=0
V1
Vo
V3
V2
Amplifiers Configurations
Inverting
It consists 1of an
op-amp, anAmplifier
A
input resistor (R1), and a
I2
I
1
feedback resistor (Rf)
The input voltage is applied
to the inverting input
The non-inverting input is
grounded
At the point A:
Vin
Vout
 I1  I 2 , I1 
, I2 
R1
Rf
Rf
Vin Vout
Vout


A  

R1 R f
Vin
R1
Vout
Vout
 Rf
  Vin 
 R1
  AVin



Example
of
an
inverting
amp
Calculate the gain and the output voltage of an
inverting op-amp if the feedback resistor (i,.e., Rf) is
120 kΩ, and the input resistor (R1) is 5.6 kΩ and the
input signal is 50 mV.
Solution:
A 
Rf
R1
120 kW
A
 21
5.6kW
Vout   AVin
Vout  21  50  10 3  1.05 V
Not: For equal resistors (R1=Rf),
it has a gain of -1, and is used in
digital circuits as an inverting
buffer
2-ofNon-inverting
Amplifier
It consists
an op-amp, an input
resistor (R1), and a
feedback resistor (Rf)
The input voltage is applied to the non-inverting input
The inverting input is connected to the summing point
of R1 and Rf and the other end of R1 is grounded
At the point A:
A
Vin
Vout  Vin
I1  I 2 , I1 
, I2 
R1
Rf
Rf
Vout
Vin Vout  Vin
1 


A
Vin
R1
R1
Rf
Vout  AVin or
Vout
 Rf 

 Vin 1 
R1 

I1
I2
Example of an non-inverting amp
Calculate the gain and the output voltage of a noninverting op-amp if the feedback resistor (i,.e., Rf) is
10 kΩ, and the input resistor (R1) is 2.2 kΩ and the
input signal is 100 mV.
A 1 
A  1
Rf
R1
10 kW
 4.6  1  5.6
2.2 kW
Vout  AVin
Vout  5.6  100  10 3  0.56 V
3- Unity gain non-inverting amplifier
A special case of the
non-inverting amplifier
(follower)
The resistor network is not used in this circuit
The output is connected directly to the inverting
input
Used in output buffering and impedance matching
bw. a high source impedance and low-impedance
input circuit
Vout
A
 1 
Vin
Vout=Vin
amplifier
Is used for Differential
finding the difference
of two voltages
each multiplied by some constant (determined by
the resistors)
The output voltage is proprtional to the differnce
between the voltage applied to the two input
terminals
Rf
Vin = V2-V1
R1
Whenever R1 = R2
V1
and Rf = Rg→
Vout
V2
R2
Rg
Comparator
a comparator is a device which
compares two voltages or
currents and switches its output
to indicate which is larger (one is
reference)
Very useful for comparing signals
and working with sensors
Comparator circuits can be built
with op-amps, but there are also
comparator ICs with large slew
rates and short propagation
delays - good for high speed
switching
Where Vs is the
supply voltage
Vout≤Vs+ when (VComparator
+>V-)
Vout≤Vs- when (V+<V-)
Output voltage will
"switch" whenever the
input voltage (at the
inverting input) reaches
the reference voltage
Vref (at the non-inverting
input)
Note that R2 and R1
form a voltage divider.
Use a potentiometer in
place of R2 for an
adjustable reference
voltage
Multiple-input circuits (summing
Many input signal can amp)
be
used at the same time that
represent many voltage
sources (adding of many
signals)
The non-inverting input is
grounded
Output is inverted
Applications of Operation Amplifier
pH
probe
amplifier
14
High impedance amp 10 W
Bias current 0.5 pA
Probe output is 50 mV
Rs is source resistance 500 W
Gain factor is 20
Amplifier output is 1 V
The 100 kW trim
potentiometer will zero out
the op-amp’s voltage (real
≈250 mV)
Gain errors are caused
mostly by inexact feedback
resistance
Instrumentation Amplifier
 It is the solution for the
following problems:
 High gain
 High input impedance
 It consist of three opamp (A1, A2 and A3)
 A1 and A2 are the two
input amp’s and
connected in the noninverting follower
configuration
 A3 is connected as
differential amplifier (dc
coupled)
2 R2
A  1
 The gain is
→
R1
pH probe electrometer instrumentation
amplifier
The two input stage amplifiers are connected to the
high impedance reference and sample electrodes
The circuit is isolated from the earth ground to
prevent noise pickup
The integrator is a integrator
circuit that
produce a voltage output
proportional to the area under a
curve defined by a time
depended function (time average
of the input signal)
The output is:
1
where:
 Vout
 Vin
R
C
t
Vout  
V

RC
in
dt
output potential in [V]
input signal potential in [V]
input resistance in [W]
feedback capacitance in [F]
the time in [sec]
• The integrator functions as lowpass filter
The differentiatorDifferentiator
circuit produce a
voltage output proportional to the time
rate change of the input signal
voltage.
The output is:
dVin
where:
Vout   RC
 Vout
 Vin
R
C
t
output potential in [V]dt
input signal potential in [V]
feedback resistance in [W]
input capacitance in [F]
the time in [sec]
• RC is time constant and must be very
short compared to time constant or
period of the input signal
• The differentiator functions as a highpass filter