The Physics of Implantable
Devices
HRC 2010
Celia Jeffery 2010
Electrical Terms
Voltage
• Electromotive Force or Electrical
pressure that causes current to flow
• Higher voltage indicates stronger
force “pushing” electrons
• Unit of measure : Volts
• Electrical symbol : V
Current
• Number of electrons flowing in a wire
per second
•
•
•
•
Unit of measure : Amps
Or milliamp (mA)
1 Amp = 1000 mA)
Electrical symbol : I
Resistance
• The sum of all the resistances to
the flow of current
• Z (total) = Zc + Z e + Z p ( c = conductor, e = electrode
resistance, p = polarization impedance)
• Unit of measure : Ohm (Ω)
• Electrical symbol : R
Relationships
Assumes constant water pressure (voltage)
Ohms Law
V=IxR
V
I
R
I = V/R
V = voltage (Volts)
I = current (Amps)
R = resistance (Ohms)
R = V/I
Ohm’s Law Applied to Pacing
5V  10mA x 500
Unit Conversions
1.0 mA (milli-Amps) = .001 A or 10-3 A
1.0 uA (micro-Amps ) = 0.000001 A or 10-6 A
1.0 k (kilo-Ohms) = 1000  or 103 
Energy
• Energy expended by applying a force of
one Newton through one metre
• Or 1 Watt second
• Unit of measure : Joule
• Electrical symbol : J
Use of energy in a pacing circuit
Energy required by circuitry
and lost due to resistance
Resistance of pacing lead
and patient interface, plus
the energy required for
depolarisation
Calculating Energy Usage
V = 5V
E = energy used in circuit
V = voltage
I = current
t = time
E=VxIxt
E = 5 x 0.001 x 1
E = 0.005J
E = 5mJ
Calculating Energy
V = 5V
E = energy used in a circuit
V = voltage
E=VxIxt
E = 5 x 0.01 x 0.0005
I = current
t = time
E = 0.000025J
E = 25J per pulse
Calculating Energy
Combine the two equations
1. V = I x R
2. E = V x I x t
E = V2 x t
R
Calculating Energy
When the settings are 5V @ 0.5ms
(taking resistance as 500 )
25
E
0.0005
500
E25μJ
Calculating Energy
When the settings are 5V @ 0.2ms
(taking resistance as 500 )
25
E
0.0002
500
E10μJ
Calculating Energy
When the settings are 2V @ 0.5ms
(taking resistance as 500 )
4
E
0.0005
500
En 4μJ
The impact of technology
When the settings are 2V @ 0.5ms
(taking resistance as 1000 )
4
E
0.0005
1000
En 2μJ
Pulse Amplitude and Pulse Width
6.4 uJ
3.7 uJ
Strength
Duration
Curve
Pulse Amplitude (V)
Voltage
2.5
2.25
2
1.75
1.5
1.25
1
0.75
0.5
0.25
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 Pulse Width
Pulse Width (ms)
Strength
Duration
Curve
Pulse Amplitude (V)
Voltage
2.5
2.25
2
1.75
1.5
1.25
1
0.75
0.5
Rheobase x2
0.25
Rheobase
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 Pulse Width
Pulse Width (ms)
Chronaxie
Application
Decrease the voltage to below the normal
battery voltage if possible
Pulse widths between 0.3 and 0.6 ms seem to
be the most efficient
Calculating Longevity (hard way!)
Need:
1. Battery capacity in Amp-hours
2. Pacemaker total current drain in microAmps,
Then:
1. Convert the battery capacity from Amp-hours to microAmp-hours
by multiplying by one million (1000000)
2. Divide the microAmp-hour battery capacity by the microAmp
current drain and your answer will be in hours.
3. Divide the number of hours by 24 hours in one day and then divide
by 365 days in one year to get to years
Longevity Calculations (hard way!)
Longevity =
1.5 AH x 1000000 µA/Amp
20 µA x 24 hrs/day x 356 days/year
Longevity Calculations (easy way!)
Longevity =
1.5 AH x 1.000.000 uA / Amp
20 µA x 24 hrs/day x 356 days/year
Longevity = 1.5 AH x 114 µA Years/AH
20 µA
The number 114 becomes a conversion constant which
converts the battery from Amp-hours to microAmp-years.
Longevity Calculations (easy way!)
Example
Battery capacity is 1.1 Amp-hours and current
drain is 20 microAmps.
1.1 AH x 114 = 6.3 years
20µA
Capacitor
• Stores a lot of energy in the form of
electrical charge
• Releases energy in one go
• Ability to store charge is Capacitance
(C), measured in Farads usually µF
• Capacitance (C) = Charge (Q)
Voltage(V)
• C = Q/V
Electrical Charge
• 6.24 x 1018 electrons transported in one
second by a current of one amp
• Unit of measure : Coulomb
• Electrical symbol : Q
Power
• The work done by a circuit equal to to 1
Joule per second
• Unit of measure : Watt
• Electrical symbol : W
• W=VxA
A bit of fun!
• We have a device with 150µF Capacitor,
which is charged to 750V.
• What is the capacity or the device and
the power output when the shock is
delivered?
• Can we work it out?
Charge Stored (Q)
= CV
= 150 x10-6 x 750
= 0.113 Coulombs
Energy Stored
= ½ CV2
=½ 150 x10-6 x 750x 750
Device Capacity = 42.19J
Assuming the capacitor discharges in 10ms
Average Current (=q/t) = 0.113
10x10-3
=11.3A
Average power demand = IxV
=11.3x 750
= 8475W
= 8.5 kW!!!