Browse Prior Art Database

Voltage Gate Driver

IP.com Disclosure Number: IPCOM000075256D
Original Publication Date: 1971-Aug-01
Included in the Prior Art Database: 2005-Feb-24
Document File: 2 page(s) / 33K

Publishing Venue

IBM

Related People

Batheret, PH: AUTHOR [+2]

Abstract

In order to supply a monolithic memory with minimum of power, the storage is provided with a circuit which drives power to the selected chips of the memory only when its input is high. The driver comprises a Darlington stage connected to the output of a differential amplifier in such a way that the level of the output is specified with five percent accuracy.

This text was extracted from a PDF file.
At least one non-text object (such as an image or picture) has been suppressed.
This is the abbreviated version, containing approximately 88% of the total text.

Page 1 of 2

Voltage Gate Driver

In order to supply a monolithic memory with minimum of power, the storage is provided with a circuit which drives power to the selected chips of the memory only when its input is high. The driver comprises a Darlington stage connected to the output of a differential amplifier in such a way that the level of the output is specified with five percent accuracy.

The input signal drives a differential stage T1, T2, which is fed by current source T3, T4.

Assuming that the input signal is low, T1 is off, while T2 is saturated. As T5, T6, D1, D2 are matched together, the following relationships are obtained: V(A) = V(ref) + 2 V(D)

V(B) = V (A) - 2 V(D) where V(D) is the V(be) of the transistors T5, T6 which is equal to the forward voltage drop of diodes D1, D2.

Thus, the voltage V(B) is defined with an accuracy which is that of the reference voltage. The error in the determination of the output level (V(C)) is restricted to the V(be) variance of the transistor T7. V(C) = V(ref) - V(beT7).

Assume that the input signal is high, so that T1 is on, while T2 is off. The diode D5 prevents T1 from saturating, so as to reduce the switching time at the next period of power on. The output voltage should drop lower than 1 V. Should it be higher than Q.7 V, then a current circulates through R19 and the base of T8, producing the saturation of the transistor. As T8 is on, the voltage V drops at the level of the output, so that T7 is no longer conducting. Thus, the o...