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Voltage To Current Converter

IP.com Disclosure Number: IPCOM000089646D
Original Publication Date: 1977-Dec-01
Included in the Prior Art Database: 2005-Mar-05
Document File: 3 page(s) / 46K

Publishing Venue

IBM

Related People

Schulz, RA: AUTHOR

Abstract

A precision voltage-to-current converter drives a grounded load, minimizes transients, has ultra-high output impedance, does not require output device matching, and can accept a differential input.

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Voltage To Current Converter

A precision voltage-to-current converter drives a grounded load, minimizes transients, has ultra-high output impedance, does not require output device matching, and can accept a differential input.

There have been several operational amplifier voltage-to-current converters, or current sources, described which perform satisfactorily in their intended application. This article describes a new voltage-to-current converter which has this combination of capabilities: 1. Can work with grounded load. 2. Voltage transients at the output terminal do not cause disturbances in the operational amplifier. 3. Output device output impedance is multiplied by the open-loop gain.
4. No output device matching is required. 5. Differential inputs can be accepted.

A general configuration for this voltage-to-current converter is shown in Fig.
1. In Fig. 1, supply voltage E can be of either polarity, whichever is more advantageous to driving the specified load. Load current 1 is given by I = (e2 - e1) R4/(R3R). which is independent of E providing that (R2 + R)/R1 = R4/R3.

This configuration has the advantage that it can develop a large voltage across the load when the load current is unipolar. Somewhat similar circuits can deliver bipolar currents to the load, but are capable of applying only half the load voltage if the load current is unipolar.

The circuit of Fig. 1 is not in final form as shown, but leads to the circuits of Figs. 2 and 3. The voltage V(EE) in Fig. 2 represents a non-precision negative power supply.

When the following condition is met,

R2 + R overR1 = R4 over R3. then variations of the supply voltage V(EE) have no effect on the output of current I. Similarly, for Fig. 3, the positive power supply voltage V(CC) has no effect on the output current I under the same conditions.

Analysis of the circuit in Fig. 3, assuming an ideal operational amplifier and that (R2 + R)/R1 = R4/R3, gives the following result: I = R4(e2-e1) over R3R. as long as e2 >/= e1. That is, the output current is un...