Current to voltage conversion circuit. Current to voltage converter using one operational amplifier. Measuring current in the positive pole of the load

Ministry of Education of the Russian Federation

Novosibirsk State Technical University

Department of SSOD

Course project in the discipline:

"CIRCUIT ENGINEERING"

Current to voltage converter

Done: Checked:

Goldobina Elena Pasynkov Yu.A.

Group: AO-91

Faculty: AVT

NOVOSIBIRSK-2001

1. Introduction

2. Technical data for design

3. Block diagram converter

4. Transformation equation

5. Error analysis

6. Schematic diagram

7. Calculation of instrumental errors

8. Conclusion

9. List of references used

10. Specification of elements

Introduction

Currently, there are various converters of physical quantities, for example: voltage to current, resistance to constant voltage, frequencies into voltage.

Converters of one quantity to another are widely used in radio electronics, microelectronics and data acquisition and processing systems. When constructing such converters, operational amplifiers are used. This allows you to significantly increase the output impedance of the circuit, thereby reducing the impact on the operation of subsequent links.

2. Technical data for design.

a) Basic data

b) Additional

3. Block diagram of the converter.

The converter circuit can be structurally represented as follows:

2) – amplifier

I BX – input current

U OUT – rated output voltage.

4. Equation for converting current to voltage.

Resistance R3 equal to the parallel connection of R1 and R2 is included in the circuit to eliminate errors from input currents.

Corrective resistance Rcor - included in the circuit to eliminate errors from resistor tolerances (Rcor = 10 Ohm)

The output voltage is directly proportional to the current, shunt resistance, and scale amplifier gain:

Calculation of circuit elements:

Initial data:

.

Selecting an operational amplifier.

Let's choose an operational amplifier with a small temperature drift E cm in order to minimize the error from the influence of drift.

Let's take op-amp 140UD21. (TKE cm =0.5·10 -6 V, Iin =0.5nA, ΔIin =0.5nA, K=1000000 Uout =10.5V M sf =110 dB).

Calculation of resistors.

Let's choose a shunt with a rated voltage U shnom = 30 mV.

Shunt resistance , therefore, the input resistance of the converter is 3 mOhm, which corresponds to the specified parameters.

The voltage at the amplifier input is equal to Unom. At the output it is necessary to obtain a voltage U out = 1V. Therefore, the closed-loop gain

.

I R – current flowing through resistances R1, R2.

where, I input_about is the input current of the operational amplifier, K is the gain without feedback.

Deciding this system, find the resistor values.

R1 = 60 Ohm R2 = 1900 Ohm.

5. Error analysis

In this scheme, there is only an instrumental error, since the methodological error associated with the source resistance is zero (we assume that the source is ideal, i.e. internal resistance equals ∞).

Therefore, we will consider only instrumental errors:

1. Error due to resistor tolerances.

This error is eliminated by introducing a correction resistance of 10 ohms into the system.

2 . Error from TKS resistors

3. Error due to drift E cm.

The impact of this error will be discussed below.

4. Error from E cm amplifier.

This error is eliminated using trimming resistor R4.

5. Error from input currents.

This error is eliminated by including in the converter a resistance R3 equal to the parallel resistance R1 and R2.

6. Drift error ΔIVX.

The impact of this uncertainty is also discussed below.

7. Error from common mode rejection ratio.

The impact of this error will be discussed below.

7. Calculation of errors

Output voltage equation:

Let's calculate the following errors:

a) Error from shunt resistance tolerance

The tolerance error of the shunt resistance is 0.05% or 15 nOhm.

In other words

R shreal is the real resistance of the shunt.

U hreal – voltage at the amplifier output at R w = R hreal

b) Error from TKS resistors:

Let's choose resistors R1, R2 from the C2-29V series.

For this type of resistors

error d 1 from TKS R 2

error d 2 from TKS R 1

c) Error from TKE SM

d) Error from ΔI BX.

e) Error from the common mode rejection ratio.

Total error

This value satisfies the specified error. Therefore, the correct choice of an operational amplifier with a small zero offset drift is confirmed.

8. Conclusion.

This voltage-to-current converter circuit is quite simple, but at the same time it provides the necessary conversion accuracy (conversion error no more than 0.05). These qualities allow this circuit to be widely used in measurement systems and signal processing systems.

9. List of used literature:

1. Lecture notes by Yu.A. Pasynkov on circuit design for 2001.

2. Horowitz P., Hill W. “The Art of Circuit Design”

3. Kunov V.M. Operational amplifiers. Directory. Novosibirsk, 1992.

11. Technical characteristics of elements.

Shunts.

A shunt is the simplest current-to-voltage measuring converter. It is designed to expand current measurement limits. At the same time most of of the measured current is passed through the shunt, and less - through the measuring mechanism of the device. Shunts have low resistance and are mainly used in circuits DC with magnetoelectric measuring mechanisms.

The shunt is a four-terminal resistor. The two input (power) terminals through which the shunt is connected to the circuit being measured are called current, and the other two, from which the voltage U supplied to the measuring mechanism is removed, are called potential - Fig. 3.1.

I u I M

Rice. 3.1. Shunt connection diagram.

The shunt is characterized by its nominal value I nom and rated output voltage U nom. Their relationship is determined by the nominal resistance of the shunt:

R w =U nom/I nom.

Part of the measured current is taken into the measuring mechanism of the device I:

I u = I R w / (R w + R u)

Where R u– resistance of the measuring mechanism. If it is necessary that the current I u was in n times less current I, then the shunt resistance should be:

R w = R u / (n-1)

Where n = I /I u- shunt coefficient.

Shunts are made of manganin, the resistance of which varies slightly with temperature. Shunts can be built into the device (at currents up to 30 A) or external. External shunts are manufactured calibrated, designed for certain currents and having one of the standard output voltage values: 10; 15; 30; 50; 75; 100; 150 and 300 mV. Serial shunts are available for currents up to 5000A. Accuracy classes of serial shunts range from 0.02 to 0.5.

For multi-range magnetoelectric devices

The sensitivity of a measuring transducer is the ratio of the change in the output signal to the change in the input signal that caused it. The ratio S=ΔY/ΔX is the average sensitivity of the converter over the interval ΔХ, and the limit to which this ratio tends at ΔХ→ 0 is the sensitivity of the converter at point X:

S ═ lim S cp ═ -- .

ΔX→0 dX

If the Y and X quantities are homogeneous, then the sensitivity is dimensionless. There are absolute and relative sensitivity of the transducer. Absolute sensitivity is S=dY/dX, and relative sensitivity is S 0 =(dY/Y)/(dX/X). For example, the sensitivity of a strain gauge transducer is defined as the ratio of the relative change in electrical resistance ΔR/R to the relative deformation Δl/l.

If the transformation function is linear, then S is const and does not depend on X. For example, if y = ax + b, then S = a.

If the transformation function is nonlinear, then S≠S cp and depends on X. For example, if y=ax 2 +b, then a=2ax.

Response threshold– this is a minimal change in the input value that causes a confidently discernible increase in the output value of the converter against the background of noise, zero offset, characteristic hysteresis and other interfering factors.

Input and output resistance determine the degree of matching of the converter with the signal source and the load. So, if the signal being converted is voltage, then Zin should be maximum, and if current, then minimum. In general, the input impedance should be such as to minimize the power consumed from the signal source.

Performance characterizes the ability to quickly respond to

change in input signal. In general, the dynamic properties of the converter are characterized by a differential equation connecting the output and input quantities. Solving this equation with x(t) known gives the value y(t). The order of the equation and its coefficients are determined by the structure and parameters of the converter. In practice, this technique is practically not used in direct form due to the complexity of solving high-order differential equations.

More often, to describe the dynamic properties of converters, characteristic functions are used, which can be obtained experimentally by applying a special test signal to the input, for example, a stepwise or harmonic one. The response of the converter to a stepwise input action of unit amplitude is called the transient function of the converter h(t). Very often, when analyzing dynamic processes, a complex converter is broken down into its simplest dynamic units. Transitional functions of the main ones

does not depend on temperature. The temperature coefficient of a device with additional resistance is less than the temperature coefficient of the measuring mechanism in R u / (R u + R d) once.

In multi-range devices, additional resistors are made in sections - Fig. 3.3.

In Fig. Figure 4.8 shows a simple version of a voltage-to-current converter using just one op-amp. Due to the feedback action, the input voltage and the voltage drop across the resistor are equal. The same current flows through the load as through the resistor therefore. The load current does not depend on, provided that the op-amp operates in linear mode (does not saturate).

Conversion factors.

Input impedance.

For an inverting converter:

For a non-inverting converter:

where is the input resistance for the common-mode signal of op-amp A.

Output resistance of inverting and non-inverting converters:

Rice. 4.8. Two variants of PNT schemes.

Output bias current of inverting and non-inverting converters:

where is the input bias voltage of the op-amp, is the input bias current of the op-amp.

The maximum output current is limited by the op-amp supply voltage and load impedance.

For an inverting circuit:

For a non-inverting circuit:

where is the output saturation voltage of the op-amp.

The maximum output current can also be limited by the built-in protection of the op-amp itself. In this case, to increase the current, you can connect a power amplifier to the op-amp output (Fig. 4.9).

Non-inverting circuit in Fig. 4.8 has a high input impedance, since the input signal is supplied directly to the op-amp input. The input resistance of the inverting circuit is equal to the resistance of the resistor, which can be relatively small. In addition, in an inverting circuit, the control voltage source must provide the entire output current. To get a large coefficient

conversion while maintaining an acceptable resistor resistance, a divider can be included in the feedback circuit (Fig. 4.9). This method has a drawback - the transmission coefficient of the feedback circuit decreases, and this reduces the linearity and accuracy of the conversion, and also reduces the output impedance.

The output resistance in this case is equal to:

those. decreases by a factor.

When operating a large inductive load (for example, a relay or motor coil), take care not to exceed the permissible parameters of the op-amp due to the occurrence of large back EMF. To protect the op-amp and other elements, additional diodes are included. In addition, with an inductive load, circuit stability problems arise. Inductance in the feedback circuit adds an extra pole in the frequency response, which can cause instability and lead to self-excitation of the device. To combat this, a correction capacitor and resistor are included, shown in Fig. 4.9.

The inclusion of another op-amp turns the original circuit into a PNT with a differential input (Fig. 4.10).

For floating control voltage sources, the circuits shown in Fig. are used. 4.11, and the advantage of circuits b) and c) is that they supply current to a grounded load. Due to the feedback action, the voltage drop across the resistor is equal to the input voltage. The current flowing through the resistor must also flow through the load, which leads to the desired result.

Output impedance for circuit a):

and for schemes b) and c):

The total bias referred to the input for circuits a), b) and c):

where is the gain of op-amp A,

CMRR is the common mode rejection coefficient of op amp A, is the input bias voltage of op amp A, is the input bias current of op amp A.

Output voltage for circuits a), b) and c):

Rice. 4.9. The use of a power amplifier and divider in the feedback circuit.

If circuit a) has floating power sources, then point P can be connected to the common wire in order to ground the input signal and the load.

Leakage resistance between the floating terminals of the signal source and ground does not affect the operation of circuit c). However, it affects the operation of circuits a) and b), since part of the output current is diverted from the current-setting resistor by leakage resistance

Figure 1.2 shows the main inverting circuit for switching on an op-amp.

Fig.1.2. The main inverting circuit for switching on the op-amp

The output of the op-amp is connected to the inverting input by a feedback resistance R OS. The signal is supplied to the inverting input through a resistor R 1 . Based on the properties of the op-amp (infinite gain), we conclude that at a finite output voltage, the potential difference in the A And IN equal to zero. Because point potential IN is equal to zero (connection to ground), then the potential of the point A is also zero. This fact gives grounds to consider the point A apparent ground, since this point has no direct connection with the ground.

It follows that the current in the input circuit is determined only by the resistance R 1 : i= u VX / R 1 . Due to the infinite input resistance of the op-amp, the current does not branch off to the amplifier input and flows completely through the OS resistance R OS. From here:
. Substituting the current value here, we get:
. Therefore, the gain is:

(1.1)

The input impedance of the cascade is R 1 .

1.1. Summing amplifier

The presence of an apparent ground point allows you to build summing amplifiers using op-amps (Fig. 1.3).

Fig.1.3. Summing amplifier

Due to the fact that the potential at the point A is zero, the input currents do not affect each other and are determined only by the parameters of the input circuits:

These currents are summed up in the feedback circuit:
.

Let's substitute the current values:
, from here:

(1.2)

By changing the resistance values, you can set the weighting coefficients with which the input voltages are summed. In particular, if all resistances are equal, we obtain the net sum of the input voltages.

1.4. Basic non-inverting op-amp switching circuit

In Fig. 1.4. The basic non-inverting circuit for switching on an op-amp is shown.

Fig.1.4. Basic non-inverting op-amp circuit

Based on the same premises as in previous cases, we will analyze the operation of this scheme.

1)
.

3)
.

4) Equating the currents, we get:
.

5) From here we finally get the gain:

. (1.3)

As can be seen from (1.3), the gain of the non-inverting gain cannot be less than one.

1.5. Repeater

A special case of a non-inverting amplifier is a repeater (Fig. 1.5).

Fig.1.5. Repeater on op-amp

The transmission coefficient of such a cascade is equal to unity. It has very high input impedance and low output impedance. Such properties allow it to be used as a buffer cascade to eliminate the influence of one part big scheme to another.

1.6. Current to voltage converter

The simplest current-voltage converter is, as you know, a resistor. However, it has the inherent disadvantage that for the connected current source its input resistance is not zero (recall that short circuit mode is normal for the current source, since the current source has a large output resistance, which should be much greater than the load resistance ). The circuit shown in Fig. 1.6 is free from this drawback and provides accurate conversion of current to voltage:

u 2 = −R i 1 . (1.4)

Dot A has a quasi-zero potential, so the input resistance of the device is zero, and the current i 1 flows through the resistor R, providing output voltage (1.4).

Fig.1.6. Current to voltage converter