Is My Understanding of the Current Amplification Circuit Correct?

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.
  • #1
PhysicsTest
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26
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?
1674400087706.png


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.
 

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  • #2
That is not an amplifier.
That is an attenuator, with a DC offset.
d·Vout1 < d·Vout2
 
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  • #3
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.
 
  • #4
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.
 
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  • #5
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.
1674408285913.png

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?
 
  • #6
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
 
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  • #7
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.
 
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  • #8
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"
1674838497053.png

Is the offset calculation of 0.225V is correct? i simplified the circuit by removing the opamp as no current is flowing into it.
 
  • #9
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
 
  • #10
Ok i try again
1674873767803.png

This is the original circuit for the comp. I modify this circuit as
Ckt1
1674873837918.png

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
1674874096750.png

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.
 
  • #11
One doubt i am getting now is can i consider Vshunt as voltage source?
 
  • #12
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?
 
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  • #13
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!
 

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