JPS61108972A - Current measuring circuit - Google Patents

Abstract

PURPOSE:To adjust an offset voltage, drift, etc. easily by connecting switches between the input of a current to be measured and the input of a current/voltage conversion circuit and between the output of an integrating circuit and the output of a compensation circuit respectively. CONSTITUTION:The current/voltage conversion circuit 12 inputs the current to be measured and outputs a voltage corresponding to the current value. The integrating amplifier circuit 22 amplifies the converted voltage, integrates the amplified voltage and outputs a voltage corresponding to the mean value of the current to be measured. The compensation circuit 20 inputs the output voltage of the circuit 22 and compensates the offset voltages of the circuits 12, 22 and the drift of the output. When the output of the circuit 22 is applied to the circuit 20 through the switch 27, the circuit 20 compensates the offset voltages of the circuits 12, 22 and the drift. When the switch 27 is opened and the current to be measured is applied to the circuit 12 through the switch 16, the current is measured under the state that the circuits 12, 22 are compensated.

Description

[Detailed Description of the Invention] 9) Industrial Application Field The present invention relates to a current measurement circuit, in particular a circuit in which a minute current flows in a steady state, and a relatively large pulsed current flows periodically or irregularly. This invention relates to a circuit for measuring the average current of a current.

(b) Conventional technology For example, in a micro combinator for a watch, a program is run by an interrupt that occurs every second to perform timekeeping operations, and otherwise the system clock is in a halt state where the operation is stopped. . In the halt state,
Only the oscillation circuit and frequency divider circuit operate, the other circuit parts stop operating, and the current is very small (approximately 1 μA).
). On the other hand, when performing timekeeping operation, each circuit is in operation for a short period of time, so the current is approximately 17FL.
It flows about eight times. Therefore, although the current ratio during halt and operation reaches 1000 times, the average current is about 5 μA.

Generally, to measure current and read it as a digital value, it is common to convert it to a voltage value using a current-voltage conversion circuit, and then convert it into a voltage value. The conventional current-voltage conversion circuit was developed in 1974.
As described on page 304 of "Practical Electronic Circuits Handbook (21)" published by CQ Publishing Co., Ltd. on May 20, 2015, a circuit using an operational amplifier is used.

When using this operational amplifier to measure the current flowing through a microcomputer for a watch, a measuring circuit as shown in FIG. 2 is used.

In Fig. 2, the first stage operational amplifier (1) converts the current to be measured flowing through the circuit under test (21, i.e., the watch microcomputer) connected to the - input terminal into a voltage, and is connected to the - input terminal. By using the feedback resistor (3), a current of 1 μA is converted to a voltage of 1 mV and output.The second stage operational amplifier (4) amplifies the output of the operational amplifier (11), and has an input resistance of IKΩ ( 5) and 10Ω
It has a gain of 10 times due to the feedback resistor (6), and outputs a voltage of IOV when the current to be measured is 1 mA. That is, since the linear region of an operational amplifier is normally ±IOV, it is designed so that when the maximum value of the current to be measured is 1 mA, the maximum voltage is 10V. The third stage operational amplifier (7) integrates the output voltage of the operational amplifier (4) using a capacitor (8), and its gain is set to IV when the current to be measured is 1 μ by adjusting the input resistance of IKΩ. (9) and 10
A feedback resistor of 0Ω (100 times larger by cutting 1), and an integrating capacitor (8) with a maximum value of 1mA of the current to be measured.
100 μF is used in order to set the time constant to about 10 seconds considering that the flow period is 1 second. By using such a measuring circuit, it is possible to measure the average current flowing through the watch microcomputer as described above, and with an average current of 5 μA, the output of the Obean Fugo 7) is 5 V.

(／→ The problem that the invention aims to solve However, the second problem is
The measurement circuit shown in the figure accepts input currents ranging from 1μ8 or less to 17FIA, and in particular handles small currents at the level of 1μ8 or less, so the operational amplifier used must have good characteristics with small temperature drift. Therefore, an expensive operational amplifier must be used. In addition, in order to adjust the offset voltage, the first stage operational amplifier (1)
However, a variable resistor 01) is required, and its adjustment takes time and effort.

(B) Means for Solving the Problems The present invention has been made in view of the above-mentioned points, and includes a current-voltage conversion circuit that converts the current to be measured into a voltage, and an integral that integrates the converted voltage. a correction circuit that corrects offset voltage, drift, etc. using the output of the integrating circuit;
By providing the first switch means provided between the current to be measured and the input of the current-voltage conversion circuit, and the second switch provided between the output of the integration circuit and the output of the correction circuit, the offset voltage can be reduced. It also facilitates adjustment of drift, etc.

(E) Effect According to the above means, when the output of the integrating circuit is applied to the correction circuit by the second switch means, the correction circuit corrects the offset voltage, drift, etc. of the current-voltage conversion circuit and the integrating circuit, and then When the second switch means is opened and the current to be measured is further applied to the current-voltage conversion circuit by the first switch means, the current-voltage conversion circuit and the integration circuit perform measurement in a corrected state. .

(f) Example FIG. 1 is a circuit diagram showing an embodiment of the present invention.

The current-voltage conversion circuit (121) is composed of an operational amplifier (131) that performs current-voltage conversion and an operational amplifier a41 that amplifies the output voltage of the operational amplifier α3. A switch (161) is connected to connect and cut off the flowing current to be measured.
The input terminal is used as an input for the current to be measured, and I
KΩ feedback resistor (171 for a measured current of 1μ)
Outputs mV voltage. On the other hand, the operational amplifier ([41 is
- input resistors of IKΩ connected to the input terminals (181 and I
The feedback resistor a9 of OKΩ has a gain of 10 times, and the output of the operational amplifier (131) is amplified ten times and outputted. In other words, this current-voltage conversion circuit (121 is normally connected to the linear fa region of the operational amplifier). is ±IOV, so the design is such that an output voltage of IOV is obtained when the maximum value of the current to be measured is 1 mA.
4) - input terminal has a correction circuit (output of 201 is IOK)
Resistance in Ω c! It is applied via D, and the output voltage of the operation 7710 can be changed by the output of the correction circuit. The current-to-voltage conversion circuit (to which the output of 121 is applied is an integrator circuit with an operational amplifier C3, an input resistor of IKΩ connected to the - input terminal of the operational amplifier (c), and a feedback resistor of 100Ω □□ □ and an integrating capacitor Oe, and its gain is 100 times.

Also, capacitor c! The capacitance of e is 100 μF so that the time constant is about 100 seconds, considering that the maximum value of the current to be measured, 1 mA, flows at intervals of 1 second. Therefore, when the average value of the output of the current-voltage conversion circuit r12 is 10 mV, that is, when the average value of the current to be measured is 1 μ, the output of the operational amplifier (C) is 1 V. The output of the integrating circuit is applied to an A-D converter circuit (not shown), etc., and is displayed as a current value on a digital mounting device, etc. In addition, the output of the integrating circuit (221) is applied to the input of the correction circuit (2) via the switch Vθ.The correction circuit gm is connected to the operational amplifier (c) and the IKΩ It consists of an input resistance - and a capacitor connected between the -input terminal and the output, and its output is applied to one input terminal of the operational amplifier I via the resistor Q11.This correction circuit (■) It acts so that the output of the integrating circuit (2) applied through the switch block becomes the ground potential.

Next, the measurement method and operation will be explained. First, switch (1
61 converts the circuit under test +151 into a current-voltage conversion circuit a
It is disconnected from the input of z and connected to another potential, for example, ground potential. In this state, the input of operational amplifier a3 is zero, so the output of the integrator/route ■ should be at ground potential, but due to the influence of the offset voltage of each operational amplifier (13 (141 (2) , the output of the integrating circuit (■) will be biased towards the medium voltage side or the one voltage side.Therefore, when the switch (5) is closed, the biased output voltage of the integrating circuit (■) will be applied to the correction circuit CIIN.For example, the output If the voltage is biased toward the medium voltage side, ten voltages will be applied to the - input terminal of the operational amplifier (c), so
The output of the operational amplifier (G) becomes one voltage, which is applied to one input terminal of the operational amplifier α4 via the resistor Qυ. The operational amplifier I is connected to the resistor (1711 and feedback resistor ti9).
Since the gain is 1 with respect to the output of the correction circuit (2), the output of the operational amplifier α4) becomes a voltage with an absolute value equal to the output voltage of the correction circuit (2). This ten voltage is amplified 100 times by the integrator circuit (221) and output in the negative voltage direction, so the output of the integrator circuit +221, which was biased toward the medium voltage side, decreases.Then, the output of the integrator circuit Qz becomes When it reaches ground potential, it is balanced, and the difference between the output voltage and the input terminal of the operational amplifier at that time is held in the capacitor (7).

In this equilibrium state, even if the switch (2) is opened, the output of the operational amplifier is held by the voltage charged in the capacitor (7), so the output of the integrator circuit is held at ground potential. Even after the switch fin opens, the correction is continued by the action of the correction circuit 4.In this state, by switching the switch 11bl and connecting the circuit under test α9 to the input of the current-voltage conversion circuit α2,
Measurements can be made that are not affected by offset voltage or temperature drift.

In addition, the correction operation of the correction circuit (■) by closing the switch amount is performed for about 2 to 5 seconds immediately before measurement;
Do it about 0 times.

(g) Effects of the Invention As described above, according to the present invention, when measuring minute currents using an operational amplifier, troublesome adjustments such as offset voltage and temperature drift are eliminated, and adjustments can be made extremely easily in a short time. It is. Furthermore, accurate current measurement can be performed without using an expensive operational amplifier.

[Brief explanation of the drawing]

FIG. 1 is a circuit diagram showing an embodiment of the present invention, and FIG. 2 is a circuit diagram showing a conventional example. Explanation of main drawing numbers (121...Current-voltage conversion circuit, (151...
Circuit under test, ■...Correction circuit, (221...Integrator circuit, (161(27+...Switch).

Claims (1)

[Claims] 1. In a current measurement circuit for measuring a current to be measured consisting of a small current and a relatively large pulsed current, the current to be measured is input and a voltage corresponding to the magnitude of the current value is output. a current-voltage conversion circuit, an integral amplifier circuit that amplifies and integrates the converted voltage, and outputs a voltage corresponding to the average value of the current to be measured; inputting the output voltage of the integral amplifier circuit; a correction circuit for correcting offset voltage and output drift of the current-voltage conversion circuit and the integral amplifier circuit; a first switch means provided between the current to be measured and the input of the current-voltage conversion circuit; a second switch means provided between the output voltage of the integral amplifier circuit and the input of the correction circuit, the output voltage of the integral amplifier circuit is applied to the correction circuit, and after correction, the A current measuring circuit characterized in that a current is applied to the current-voltage conversion circuit.