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Low Offset Voltage Comparator

IP.com Disclosure Number: IPCOM000042786D
Original Publication Date: 1984-Jun-01
Included in the Prior Art Database: 2005-Feb-04
Document File: 2 page(s) / 34K

Publishing Venue

IBM

Related People

Frankeny, RF: AUTHOR [+2]

Abstract

This voltage comparator has a high AC input impedance and low offset voltage, and requires no negative supply voltage. Transistors Q5 and Q6 are current sources that are matched by conventional techniques. Q6 feeds diode configured transistor Q4, such that when Vin equals Vr, the currents as shown in the figure exist (this assumes that the VBE values of the transistors are matched as would be the case in an integrated circuit configuration). Thus, when the two signals are equals (Vin equals Vr), 2IB is available to drive transistor Q7. Since the betas of the transistors track, the emitter current of Q7 equals 2(I-IB). Resistor R3 is sized to give gain, and, therefore, transistor Q8 is saturated and Vout equals the collector-to- emitter voltage of Q8 when Q8 is saturated.

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Low Offset Voltage Comparator

This voltage comparator has a high AC input impedance and low offset voltage, and requires no negative supply voltage. Transistors Q5 and Q6 are current sources that are matched by conventional techniques. Q6 feeds diode configured transistor Q4, such that when Vin equals Vr, the currents as shown in the figure exist (this assumes that the VBE values of the transistors are matched as would be the case in an integrated circuit configuration). Thus, when the two signals are equals (Vin equals Vr), 2IB is available to drive transistor Q7. Since the betas of the transistors track, the emitter current of Q7 equals 2(I-IB). Resistor R3 is sized to give gain, and, therefore, transistor Q8 is saturated and Vout equals the collector-to- emitter voltage of Q8 when Q8 is saturated. Looking at the voltages, the operation of the comparator can be further analyzed. In the equations that follow, the suffix numerals denote the applicable transistors. VE1 = Vr + VBE2 + VBE4 - VBE3 Therefore, VBE1 = VE1 - Vin = (Vr - Vin) + VBE2 + VBE4 - VBE3 Assume VBE matching on Q1 - Q4 Therefore, VBE1 = (Vr - Vin) + VBE2 Since IE1 = IE2 if IE3 = IE4 Then, VBE1 = VBE2 when Vr = Vin Therefore,Vr - Vin = VBE1 - VBE2 If Vin < Vr, VBE1 > VBE2 and IE1 > IE2 and the base drive for Q7 is reduced. When IB7 = O, then Q7 is off. This occurs when IC3 = I or IE3 = IE4 + 2IB Likewise, IE1 = IE2 + 2IB Then, where e is the base of the natural system of logarithms, h is approximat...