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High Speed IBM 370 Channel/Control Unit Coaxial Line Driver with a 5-Volt +/- 5% Power Supply

IP.com Disclosure Number: IPCOM000107655D
Original Publication Date: 1992-Mar-01
Included in the Prior Art Database: 2005-Mar-22
Document File: 2 page(s) / 67K

Publishing Venue

IBM

Related People

Braden, JJ: AUTHOR

Abstract

Disclosed is a line driver circuit shown in Figs. 1 and 2 for driving a double-end terminated coax up to 400 ft. long using a single 5-volt power supply level. It can be used for driving single or multiple line receivers.

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High Speed IBM 370 Channel/Control Unit Coaxial Line Driver with a 5-Volt  5% Power Supply

       Disclosed is a line driver circuit shown in Figs. 1 and 2
for driving a double-end terminated coax up to 400 ft. long using a
single 5-volt power supply level.  It can be used for driving single
or multiple line receivers.

      The following circuit operation refers to Fig. 2.  SW1 is
actually a Schottky type open collector inverter.  When open, its
output appears to be high impedance; when closed the output is about
.4 V.  To start the explanation of this circuit, SW1 is closed.  The
Qout is off (output is down = 0 V).

      Because 0.5 ma is flowing through diode D1, node 1 is 0.8 V
below Vcc.  Also, there is no current flowing through the collector
of Qout which makes node 5 = Vcc.  This condition diverts all of the
current form I source 2 through Q2 and into Q3.  The current through
Q3 is mirrored and doubled by Q4 and Q5.  This current, which is
about 2 ma, is going through closed SW1.

      When SW1 is opened, the 2 ma is driven into the base of Qout
which causes it to turn on very fast and produce a positive voltage
at its emitter.  Because Qout is turned on, its collector current
causes a voltage drop across resistor RC.  When the current through
RC is 80 ma, the voltage across it is 0.8 V.  Because diode D1 also
has .8 V across it, the currents through Q1 and Q2 are equal and
about 0.5 ma worst case.  If the beta of Qout is high, the...