E&i web cast
- 2. 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
- 3. Specifying The Product • Understanding the power requirements • Impedance matching • Defining frequency range • A question of linearity • Applying a DC bias • E&I’s solutions
- 4. 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 Gain V 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 0.35 0.0003 -6.02 0.32 0.11 50 43.98 25 100.00 35.36 0.0010 0.00 0.63 0.22 50 50.00 100 200.00 70.71 0.0063 7.96 1.58 0.56 50 57.96 625 500.00 176.78 Input Output Input Power Output Power
- 5. 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 2.28 50 25 0.333 -9.54 100 88.89 133.333 1.89 50 50 0.000 100 100.00 200.000 1.41 50 75 0.200 -13.98 100 96.00 240.000 1.13 50 100 0.333 -9.54 100 88.89 266.667 0.94 50 200 0.600 -4.44 100 64.00 320.000 0.57 50 400 0.778 -2.18 100 39.51 355.556 0.31 50 800 0.882 -1.09 100 22.15 376.471 0.17
- 6. 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 2.28 12.5 12 0.020 -33.80 100 99.96 97.959 2.89 50 50 0.000 100 100.00 200.000 1.41 50 100 0.333 -9.54 100 88.89 266.667 0.94 100 100 0.000 100 100.00 282.843 1.00 50 200 0.600 -4.44 100 64.00 320.000 0.57 200 200 0.000 100 100.00 400.000 0.71 50 800 0.882 -1.09 100 22.15 376.471 0.17
- 7. 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/2 So the input impedance is 2V/I/2 = 4RL I
- 8. 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.
- 9. 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.
- 10. 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.
- 11. 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
- 12. 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.
- 13. OEM Solutions 50-Watt Module 150 KHz – 150 MHz 100 Watt 10KHz – 12 MHz
- 14. 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 Frequency Ohms Watts KHz JT-3 3.125 50 10 - 1,000 JT-6 6.25 100 10 - 1,000 JT-12 12.5 300 10 - 10,000 JT-25 25 500 10 - 10,000 JT-100 100 500 10 - 10,000 JT-200 200 500 10 - 10,000 Transformers
- 15. 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
- 16. RF Amplifiers
- 17. 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
- 18. 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 Harris President tonyharris@eandiltd.com 585-214-0598 x 202 Jen Elkins Sales jenelkins@eandiltd.com 585-214-0598 x 205