Browse Prior Art Database

Word-Line Driver for Bipolar Static Ram Having Low Standby Power

IP.com Disclosure Number: IPCOM000119657D
Original Publication Date: 1991-Feb-01
Included in the Prior Art Database: 2005-Apr-02
Document File: 2 page(s) / 70K

Publishing Venue

IBM

Related People

Chuang, CTK: AUTHOR [+3]

Abstract

A new word-line driver circuit for bipolar static RAMs that minimizes the standby power consumption is presented. The essence of the new circuit is the use of p-n-p transistor circuitry for pulling up and a resistor for pulling down the decoder output. The new driver allows complete turn-off of current in the circuit, while maintaining the word line at a standby low level. The p-n-p transistors in the new circuit need to be fast and have enough current gain.

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Word-Line Driver for Bipolar Static Ram Having Low Standby Power

      A new word-line driver circuit for bipolar static RAMs
that minimizes the standby power consumption is presented.  The
essence of the new circuit is the use of p-n-p transistor circuitry
for pulling up and a resistor for pulling down the decoder output.
The new driver allows complete turn-off of current in the circuit,
while maintaining the word line at a standby low level.  The p-n-p
transistors in the new circuit need to be fast and have enough
current gain.

      Fig. 1 shows the basic schematics of the new word-line driver
circuit.  Here, the QN5 circuitry is a switchable current source, QN6
is the emitter-follower driver, and QN1-QN4 form a current-switch
decoder.  The QP1-QP2 combination is a current mirror with a
mirroring factor (or gain) M.  When the current source is ON and all
the decoder inputs are low for the selected word line, the current I
flows through QN1 and QP1. Due to the mirroring, a current MxI flows
from QP2 into R1, raising the voltage at the node X to (VL+R1xMxI)
which is the desired high level.  This high level must guarantee that
QP2 does not saturate.

      When the current source is ON but at least one of the inputs is
high for the unselected word lines, I flows through the QN2-QN4
circuit and R1.  Because no current flows in QN1 and QP1, the QP2
does not provide a current into R1.  As a result, the voltage at the
node X stays at (VL-R1xI).

      I...