E&i web cast

Webcast Agenda

What you can expect to learn:

• How to select the appropriate solution for your application

• Defining your specifications to yield the most efficient and cost-effective results

• Identifying the impact of linearity on specific medical applications

Specifying The Product

• Understanding the power requirements• Impedance matching• Defining frequency range• A question of linearity• Applying a DC bias• E&I’s solutions

Understanding the Power Requirements

• Begin by defining the peak-to-peak voltage required by your transducer.

• Then calculate the power from this.– Enter Vp-p or Vrms into highlighted cells– Read off required output power

Amplifier GainV p-p V rms V p-p V rms

Watts dBm Volts Volts dB dBm Watts Volts Volts

0.0000 -46.02 0.00 0.00 50 3.98 0.0025 1.00 0.350.0003 -6.02 0.32 0.11 50 43.98 25 100.00 35.360.0010 0.00 0.63 0.22 50 50.00 100 200.00 70.710.0063 7.96 1.58 0.56 50 57.96 625 500.00 176.78

Input OutputInput Power Output Power

Impedance Matching

• In addition to power, the impedance of the transducer will affect Vp-p

• To calculate the Vp-p we have to consider the power actually transmitted to the load

• Consider a 100-Watt power level case below

Source Load r Return Loss Power Forward Power to load Vp-p Irms

50 12 0.613 -4.25 100 62.43 77.419 2.2850 25 0.333 -9.54 100 88.89 133.333 1.8950 50 0.000 100 100.00 200.000 1.4150 75 0.200 -13.98 100 96.00 240.000 1.1350 100 0.333 -9.54 100 88.89 266.667 0.9450 200 0.600 -4.44 100 64.00 320.000 0.5750 400 0.778 -2.18 100 39.51 355.556 0.3150 800 0.882 -1.09 100 22.15 376.471 0.17

Impedance Matching

• One can place a transformer between the amplifier and the transducer

• In this case we increase Vp-p for the same power level

Source Load r Return Loss Power Forward Power to load Vp-p Irms

50 12 0.613 -4.25 100 62.43 77.419 2.2812.5 12 0.020 -33.80 100 99.96 97.959 2.89

50 50 0.000 100 100.00 200.000 1.4150 100 0.333 -9.54 100 88.89 266.667 0.94

100 100 0.000 100 100.00 282.843 1.0050 200 0.600 -4.44 100 64.00 320.000 0.57

200 200 0.000 100 100.00 400.000 0.7150 800 0.882 -1.09 100 22.15 376.471 0.17

Transformers• Looking at a 4:1 transformer explains how impedance

transformation occurs

RL V

V

+

+

+

+

2V

-

-

-

- Secondary

Primary

V

4RL

If V is present across RL then V must also be present across the secondary transformer.If the turns of the primary and secondary transformers are the same then V must also be present across the primary.Now if the Current flow through RL is I then the current flow through each winding transformer must be I/2So the input impedance is 2V/I/2 = 4RL

I

Defining Frequency Range

• There are many factors that need to be considered when determining the frequency range required.

• For example, ablatement for the treatment of liver and kidney cancer occurs at 0.8–1.6 MHz and for biliary duct cancer at 10 MHz.

• It is important to choose a top end frequency range at the lowest for your requirements to avoid unnecessary cost.

A Question of Linearity

• Class A vs. Class AB– The difference between a class A and a class AB amplifier is

simply the point at which the transistors are biased. In the case of class A, the transistor is biased so that over the entire cycle of the RF input, the transistor is operating within its linear portion. In the case of class AB, part of the cycle of the input is actually turning the transistor off.

A Question of Linearity

• Class A vs. Class AB– This means that in the case of a class A amplifier, the output

is a faithful reproduction of the input signal whereas, in the case of class AB, some distortion is inevitable.

• We should always chose class AB if possible, as again, for a given power it is less costly– For most heating applications class AB is sufficient.– However, for some cavitation applications where it is crucial

to know exactly on which cycle of RF cavitation will occur and when it will cease, class A is necessary.

– Also, for most diagnostic applications class A is necessary.

Applying a DC Bias• Activation of a piezoelectric crystal can

sometimes be enhanced by applying a DC bias to the signal.

• In the diagram below, a positive bias is applied to the signal so that the RF is centered on a 55 mV point as opposed to 0 Volts.

Bias

Applying a DC Bias• Below is a simple circuit that allows us to achieve

this

The value of the capacitor and inductor are chosen depending upon the operating frequency.

If you have an application and would like us to provide you with the appropriate values so you can build one, we can do this.

OEM Solutions

50-Watt Module150 KHz – 150 MHz

100 Watt 10KHz – 12 MHz

Matching Transformers

• Custom and standard transformers• Power levels from of 1 W to 1 KW• Frequencies from 10 KHz – 10 MHz• Single or multiple taps• Switchable taps

– Examples:-

Part Number Output Impedance Max Power Level FrequencyOhms Watts KHz

JT-3 3.125 50 10 - 1,000JT-6 6.25 100 10 - 1,000

JT-12 12.5 300 10 - 10,000JT-25 25 500 10 - 10,000JT-100 100 500 10 - 10,000JT-200 200 500 10 - 10,000

Transformers

Custom Amplifiers

E&I offers custom solutions to satisfy your phased array requirements.

We have 10 and 50 watt standard modules and can offer other power levels.

50AB6 10AB61

RF Amplifiers

Our Customers

“Excellent! Thanks for exceeding our expectations.”

- Kona Medical

“Thank you for the amazingly fast delivery time”

- University of Oxford

“First class service!”- Aeroflex International

Next Steps• Would you like a more in depth

discussion on any of the points raised?– Schedule a conference call or visit to your facility?

• Would you like any of the spread sheets e-mailed to you?

• Would you like to provide us with your product specifications?

Tony [email protected] x 202

Jen [email protected] x 205