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# Integrated Bipolar Current Source

IP.com Disclosure Number: IPCOM000079958D
Original Publication Date: 1973-Oct-01
Included in the Prior Art Database: 2005-Feb-26
Document File: 3 page(s) / 40K

IBM

## Related People

Jacquart, C: AUTHOR

## Abstract

Shown is a simplified circuit of a low-power current source designed for supplying an integrator with a positive or negative current, the value of which can be adjusted with an external resistor. Furthermore, this current can be switched off.

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Integrated Bipolar Current Source

Shown is a simplified circuit of a low-power current source designed for supplying an integrator with a positive or negative current, the value of which can be adjusted with an external resistor. Furthermore, this current can be switched off.

It comprises a positive current source Il and a negative current source I2, which can be switched through a diode bridge D1 to D4 into a diode clamping circuit composed of diodes D5 D6 and a reference voltage source VR. Three modes of operation can be distinguished: - When inputs A1 and A2 are at a low level, current 11 flows through diode D1; current I2 flows through diode D3, diode D6 and resistor R. Diodes D2, D4 and D5 are reversed biased.

The output phi 1 is at a voltage level equal to the reference voltage level minus one diode voltage drop (D6). V phi(1) = V(R) - V(DB). - Diodes D2 and D4 are in the nonconducting state if the voltage level at input A1 and A2 is: Low level : V(A1) = V(A2) >/- V(R) - 2 V(D) where V(D) is a diode voltage drop.

In this state, a current I/-/ flows through resistor R coming from the integrator. This current is equal to: I = V(D6)/R assuming that the offset voltage of the integrator is 0. - When input A1 and A2 are at a high level, current I1 flows through diode D2, diode D5 and resistor R; current I2 flows through diode D1. Diodes D1, D3 and D6 are reversed biased.

The output phi 1 is at a voltage level equal to the reference voltage plus one diode voltage drop (D5). V phi(1) = V(R) + V(D5).

Diodes D1 and D3 are in the nonconducting state if the voltage level at input A1 and A2 is: High level: V(A1) = V(A1) >/- V(R) + 2V(D) where V(D) is a diode voltage drop.

In this state, a current I/+/ flows through the resistor R, going to the integrator. This current is equal to: I/+/ = V/D5//R assuming that the offset voltage of the integrator is 0. - When input A1 is at a low level and input A2 at a high level, current I1 flows through diode D1; current I2 flows through diode D4; diodes D2 and D3 are reversed biased. In this conditi...