# Current Amplification Circuit

• Engineering
• PhysicsTest
In summary, the conversation discusses a circuit used for current amplification, which includes a micro with an internal opamp and a schematic for amplifying current I1. The output, Vout1, is amplified voltage and is given to the micro. The amplification factor is not constant and varies depending on the input current. The circuit also includes a DC offset added to Vout1. It is noted that the circuit is not an amplifier, but rather an attenuator with a DC offset. The possibility of amplifying the signal without an opamp is discussed, with the suggestion to use an instrumentation amplifier. The conversation also touches on the calculation of the offset in the circuit.

#### PhysicsTest

Homework Statement
I wanted to understand the current amplification circuit
Relevant Equations
KCL, KVL equations
I have taken portions of a schematic for amplifying the current I1, this I1 current is the phase current from one of the legs of the inverter. The output vout1 is the amplified voltage, again given to a micro which has internal opamp. Is my circuit understanding correct?

When I give I1 = 10 Amp
Vout1 = 0.1457
Vout2 = 0.03
Amplification factor = 0.145/0.03 = 4.833

--- Operating Point ---

V(n001): 3.3 voltage
V(vout1): 0.145714 voltage
V(vout2): 0.0300002 voltage
I(I1): 10 device_current
I(Rshunt): 10.0001 device_current
I(R3): -7.71428e-005 device_current
I(R2): 6.62338e-005 device_current
I(R1): 0.000143377 device_current
I(V1): -0.000143377 device_current

when I1 = 20 Amp
Vout1 = 0.162857
Vout2 = 0.0600002
Amplification factor = 0.162857/0.0600002= 2.7
--- Operating Point ---

V(n001): 3.3 voltage
V(vout1): 0.162857 voltage
V(vout2): 0.0600002 voltage
I(I1): 20 device_current
I(Rshunt): 20.0001 device_current
I(R3): -6.85714e-005 device_current
I(R2): 7.4026e-005 device_current
I(R1): 0.000142597 device_current
I(V1): -0.000142597 device_current

The amplification factor is not constant, as expected it to be constant. Please advise.

#### Attachments

• Cur1asc.txt
1.1 KB · Views: 81
That is not an amplifier.
That is an attenuator, with a DC offset.
d·Vout1 < d·Vout2

DaveE
I am sorry for naming convention the input side is vout2 and the output is vout1 which is greater than vout2. Vout1 is given to micro input.

PhysicsTest said:
I am sorry for naming convention the input side is vout2 and the output is vout1 which is greater than vout2. Vout1 is given to micro input.
I get that. But Vout1 is an attenuated version of Vout2.
Look at a voltage step on d·Vout2 and you will see the d·Vout1 step has less amplitude.
Vout1 has a small DC offset added. That does not constitute gain.

PhysicsTest and DaveE
Baluncore said:
I get that. But Vout1 is an attenuated version of Vout2.
Look at a voltage step on d·Vout2 and you will see the d·Vout1 step has less amplitude.
Vout1 has a small DC offset added. That does not constitute gain.

Does it mean the above diagram for pulse input where the difference of Vin = 30mv and Vout = 20mV? Now i have to figure out how to amplify the signal without opamp. Is it possible?

PhysicsTest said:
Now i have to figure out how to amplify the signal without opamp. Is it possible?
Electronic power amplification is not possible without an active device, a negative resistance, or a non-linear component.

First specify the problem, then identify available power supplies, and then find an amplifier solution.
You have a very low impedance source, so start looking for a low input impedance voltage amplifiers.

Look for amplifiers with common mode input voltages that include the negative rail.
Take a look at the LM10 which has a voltage reference to provide an offset, and works near the negative rail. https://www.ti.com/lit/ds/symlink/lm10.pdf

DaveE and PhysicsTest
PhysicsTest said:
I wanted to understand the current amplification circuit
You want to measure a 5 amp current. You need economy and low power, so the last thing you need to do, is to duplicate or amplify that current.

You use Rshunt = 3 mΩ, to convert the current to a low voltage. Then, that low voltage must be amplified. That amplifier must amplify the differential voltage across Rshunt, and not be sensitive to common mode variations, or changes in the resistance of the connections to Rshunt. That is a job normally done by an instrumentation amplifier.

PhysicsTest
Yes now i understand it is not amplification circuit, when i was going through the documentation
Userguide on page 29, fig29. I was trying to calculate the offset as mentioned by the document
"The voltage drop on the shunt resistor, due to the motor phase current, can be either positive or negative, an offset is set by R1 and R2"

Is the offset calculation of 0.225V is correct? i simplified the circuit by removing the opamp as no current is flowing into it.

PhysicsTest said:
Yes now i understand it is not amplification circuit, when i was going through the documentation
Userguide on page 29, fig29. I was trying to calculate the offset as mentioned by the document
"The voltage drop on the shunt resistor, due to the motor phase current, can be either positive or negative, an offset is set by R1 and R2"
View attachment 321259

Is the offset calculation of 0.225V is correct? i simplified the circuit by removing the opamp as no current is flowing into it.
No.

1) When you replace the current source with a voltage source equivalent, you can't remove the resistor ##R_{shunt}##, it should be relocated in series with ##R_2##. This is the Thevenin/Norton source transformation, an extremely powerful and simple tool in circuit analysis.

2) However if ##R_{shunt} \ll R_2##, then ##R_{shunt} + R_2 \approx R_2##. Then you would get your equivalent circuit as an approximation.

3) Your offset calculation shouldn't be a simple ratio of resistance, it's a voltage divider.

Try again

Ok i try again

This is the original circuit for the comp. I modify this circuit as
Ckt1

The voltage drop of Ishunt across Rshunt is V = Ishunt * Rshunt, hence i replace the current source with the voltage source, the modified circuit will be
Ckt2

Applying the Kirchoffs laws

$$I = \frac {(VDD - V_{Shunt})} {R_1+R_2}$$ ->eq1
$$V_{out} = V_{shunt} + IR_2$$ ->eq2
simplifying the equation is
$$V_{out} = \frac{(VDDR_2+V_{shunt}R1)} {R_1+R_2}$$ -> eq3

Please correct if i am wrong i will apply Norton theorem.

One doubt i am getting now is can i consider Vshunt as voltage source?

PhysicsTest said:
One doubt i am getting now is can i consider Vshunt as voltage source?
That is what it is, a differential voltage source.
Why do you think that might not be the case?

DaveE
PhysicsTest said:
The voltage drop of Ishunt across Rshunt is V = Ishunt * Rshunt, hence i replace the current source with the voltage source
PhysicsTest said:
One doubt i am getting now is can i consider Vshunt as voltage source?
Nope, that is what Thevenin's theorem is all about. You can't make a resistor just disappear in the circuit transformations. What you can do is convert parallel branches with just a current source and just an impedance into a branch with a voltage source in series with an impedance, they will have different values according to the transformation rules. This isn't difficult to learn, prove, or remember, IMO. Read the link I posted earlier. This is absolutely worth learning ASAP, it will be incredibly useful for you.

However, your answer is the correct approximation when ##R_{shunt} \ll R2## as I stated before. This is almost always a useful simplification for current sensing circuits.

The exact solution is
$$V_{out} = I_sR_{shunt} \frac{R1}{R1+R_{shunt}+R2} + V_{dd} \frac{R2+R_{shunt}}{R1+R_{shunt}+R2}$$

BTW, thanks for posting good drawings explaining the circuit in question, that's really helpful. There's no wasted effort asking what you are referring to!

## 1. What is a current amplification circuit?

A current amplification circuit is an electronic circuit that is used to increase the strength of an electrical current. It is commonly used in electronic devices such as amplifiers, sensors, and transistors to boost the strength of a weak current signal.

## 2. How does a current amplification circuit work?

A current amplification circuit works by using a combination of resistors, capacitors, and transistors to amplify the current signal. The input current is passed through the circuit and is amplified by the transistors before being outputted at a higher strength.

## 3. What are the applications of a current amplification circuit?

A current amplification circuit has a wide range of applications in various electronic devices. It is commonly used in audio amplifiers, power supplies, sensors, and communication systems to boost weak signals and improve overall performance.

## 4. What are the advantages of using a current amplification circuit?

The main advantage of using a current amplification circuit is that it allows for the amplification of weak signals without introducing significant noise or distortion. It also allows for improved signal strength and better performance in electronic devices.

## 5. Are there any limitations to using a current amplification circuit?

One limitation of using a current amplification circuit is that it requires a power source to operate, which can be a limiting factor in some applications. Additionally, improper design or implementation of the circuit can lead to instability or distortion in the amplified signal.