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Comparator with wide input voltage range and fixed current consumption

IP.com Disclosure Number: IPCOM000130304D
Published in the IP.com Journal: Volume 5 Issue 11A (2005-11-25)
Included in the Prior Art Database: 2005-Nov-25
Document File: 3 page(s) / 113K

Publishing Venue

Siemens

Related People

Juergen Carstens: CONTACT

Abstract

In order to compare two voltage levels moving in a wide range, a circuit with a well defined current assumption is desirable. Present solution is a simple comparator and a resistor divider in order to shift the voltages to levels applicable to the comparator inputs. The disadvantage of this solution is the current consumption through the resistor divider, which depends on the applied voltage. If the voltage range is wide the quiescent current swing proportionally increases. The following idea proposes a circuit shown in fig. 1, where: • M1 and M2 are low voltage PMOS (P-channel metal-oxide semiconductor) • M3 is a DMOS or an high voltage NMOS (N-channel metal-oxide semiconductor) • R1 and R2 are matched resistors. At the switching threshold the differential voltage is V1 - n * VBE - 2 * VGS2 - R2 * (I2-I1) = V2 - VGS1 - VBE Where VGS1 = R1 * I1 and the voltage difference is

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S

Comparator with wide input voltage range and fixed current consumption

Idea: Marco Piselli, IT-Padova

In order to compare two voltage levels moving in a wide range, a circuit with a well defined current assumption is desirable. Present solution is a simple comparator and a resistor divider in order to shift the voltages to levels applicable to the comparator inputs. The disadvantage of this solution is the current consumption through the resistor divider, which depends on the applied voltage. If the voltage range is wide the quiescent current swing proportionally increases.

The following idea proposes a circuit shown in fig. 1, where:

* M1 and M2 are low voltage PMOS (P-channel metal-oxide semiconductor)
* M3 is a DMOS or an high voltage NMOS (N-channel metal-oxide semiconductor)
* R1 and R2 are matched resistors.

At the switching threshold the differential voltage is V1 - n * VBE - 2 * VGS2 - R2 * (I2-I1) = V2 - VGS1 - VBE Where VGS1 = R1 * I1 and the voltage difference is

2() BE 1

  R 1 V - V 2 I R V - V 1

1 V

2

=

  ⋅ + ⋅

 ⎛ + ⋅

      GS GS 2 2

2

⎜ ⎝

 1 - n

R

 + ⎟ ⎠

if

 V - I I - I I =

0 V

=

1

   1 2

=

    GS 2

    V R

1

2

⇒GS GS 1

V 1 +

GS V

=

2

l

  + ⋅

  C µ

   I k'

   V I w

      TH 0

⋅OX n

=

      TH 0

 1 - I I +

0 V

=

2

  I k'

R

  ⋅ ⋅

()

     TH 0

1

If this condition is satisfied, the threshold differential voltage is:

 R - 1 V V 1 - n I R V - V

2

=

1

() ⎟ ⎠

   + ⋅

1

  + ⋅

      GS BE

⎜ ⎝

2

   2 2

R

By imposing R1 = R2, the threshold differential voltage is:

  22() BE 2 1 V

=

By generating an appropriate current I2 it is possible to have a threshold voltage compensated over temperature.

The quiescent current is I2 when V1 - V2 > R2 * I2 or I0 + I2 when V1 - V2 <= R2 * I2.

In t...