GE Corometrics 126, Руководство по обслуживанию

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4-48 120 Series Maternal/Fetal Monitor Revision B
2015590-001
Theory of Operation: Dual Ultrasound Board
The actual switching of the ultrasound input/output to the two transducers is
accomplished using PIN diodes D1–D4. The signals PIN1* and PIN2* control the
switching of quad analog switch U25 which 6 V and –6 V to resistors R1
and R2 and consequently bias the PIN diodes on or off. The ±6 V are derived from
±12V regulators VR3 and VR4 through a series of R/C filters which removes any
noise present on these supply lines.
Channel A Pin Diode Circuitry
If Channel A (US1) is selected, R1 is biased to –6 V and R2 is biased to +6 V. The
anode of D1 is held at ground potential by L5. The cathode of D1 will therefore be
at approximately –600 mV since it is forward biased by the current through R1. The
anode of D2 must then be at ground potential due to the –600 mV at its cathode.
The current through D2 will be approximately 15 mA since the +6V applied to the
anode of D2 is through a 402
Ω
resistor R3 (6V/402¾ = 14.9 mA). D3 and D4 will
be off at this time since their cathodes are at +6 V and their anodes are at 0 V. The
15 mA flowing through D1 and D2 stores charge in them which is an order of
magnitude greater than will be depleted when the transmitter drives the transducer.
Thus, even though the transmitter has large voltage excursions, D1 and D2 will
remain on and provide a low impedance path between the transmitter and the
transducer. During reception of signals, these diodes provide a low impedance path
to the pre-amplifier.
Channel B Pin Diode Circuitry
If Channel B (US2) is selected, the process works the same except R2 is biased to –
6 V and R1 is biased to +6 V, turning on diodes D3 and D4 and turning off diodes
D1 and D2.