# Feedback Proof in Current Sensor

Discussion in 'Homework Help' started by ac_dc_1, Feb 2, 2013.

1. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
Consider the following current sensor represented in the figure which produces in the output a tension Vout proportional to the current IL which passes through the 0.02 Ohm resistor.
Prove that the montage has negative feedback.

So assuming that the tension in the inverting input is increasing the output tension of the OpAmp decreases and therefore Ib and Ic of the transistor also decrease and the transistor is at cut.
So the tension in the inverting input starts decreasing and the output tension starts increasing and therefor Ib and Ic increase and we have a drop voltage in the 1k resistor which causes the tensio in the inverting input to decrease and so on...

Is this correct?

Thanks

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2. ### antonv Member

Nov 27, 2012
149
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So assuming that the tension in the inverting input is increasing the output tension of the OpAmp decreases and therefore Ib and Ic of the transistor also decrease
Why do you think Ib and Ic will decrease? Consider the type of transistor being used.

Do you need to prove it using equations or prose like you did?

3. ### antonv Member

Nov 27, 2012
149
27
Also, for this problem, the transistor does not go into cut-off.

Consider where current flows in the circuit (2 paths if you assume the transistor's base current is 0) and consider the input voltage polarities as a result of these currents.

4. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
So the polarity of the voltages change as a result of where the current goes?
The currents that pass through the transistor pass always in the same direction?

My idea was to understand first why there is feedack,and explain it on prose.

Thanks

5. ### crutschow Expert

Mar 14, 2008
13,502
3,376
You determine the polarity of a transistor by noting that an increase in base current results in an increase in collector current. The direction of the base and collector currents follow the direction of the emitter arrow on the schematic. From that you can determine whether the signal voltage polarity at the transistor output is in-phase or out-of-phase with the input.

6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Remember what the essence of negative feedback is. An error introduced into the system results in a response that tends to reduce the magnitude of the error. So introduce a hypothetical error into the system -- it can be lots of things: the load changing, the beta of a transistor changing, a noise signal injected into the forward path, whatever -- and then walk through the chain of events that happens and show that, in the end, they tend to cancel out a significant portion of the error and restore the system closer to its desired operating point that would have been the case without any feedback.

7. ### antonv Member

Nov 27, 2012
149
27
The polarities stay the same - they just change in amount.

What is the main thing an op-amp does? It changes its output to try to make the inputs the same. It can only succeed if you connect feedback from the output to the inverting input, in this case through the transistor.

The two current paths in the circuit are:
- from the input at the top through the 0.02 ohm resistor to a load that is not shown
- from the input at the top through the 100 ohm resistor, the transistor and the 1k resistor to ground.

Do you know how the second one responds to the first?

8. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Uh, no. While this is a useful description of many (even most) opamp-based circuits, that's not the main thing an opamp does. The main thing an opamp does is produce an output voltage that is a highly amplified copy of the difference in the voltages at its inputs. At the end of the day, it doesn't care whether its inputs are at the same voltage or not, it is just an absurdly high-gain differential amplifier.

The notion of its output changing in order to make the inputs the same is related not to the opamp, but rather to the feedback topology of the circuit. You can have opamps in circuits that don't have negative feedback and you can have negative feedback in circuits that don't have opamps. While the two concepts, that of an opamp and that of negative feedback, are natural bedmates, the two concepts themselves are quite independent.

Which underscores the fact that it is the circuit, not the opamp, the matters (not to say that all opamps are therefore interchangeable, they aren't).

Also, in many cases the use of a transistor in the feedback path results in the signal going to the non-inverting input because of the inverting effect that most transistor-based circuits have on the feedback signal. In this particular case, because it is the transistor's emitter in the feedback path, the typical inversion doesn't happen (because a change in voltage on the transistor's base results in an almost identiical change in voltage on the transistor's emitter).

In addition, how do these relate to the output voltage of the opamp? Assume the current in the 100Ω resistor were too high. What is the chain of events that would ensure and would this tend to reduce the current in that resistor?

9. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0
So if the current in 100 Ohm resistor were to high, we will have a Ie equal to current through the 100 Ohm resistor and the tension in the inverting input will increase because a higher current in the 100 Ohm resistor implies a bigger drop voltage in 100 Ohm resistor and therefore V- increases.Then Vout decreases until we have Vout=V-.The Vout will decrease because the current through the 1k resistor will also drecrease.
But if the V- decreases the drop votage in the resistor of 0.02 Ohm will increase and therefore V+ and so V- also increases and a new loop begins...
Is that it?

Thanks

10. ### antonv Member

Nov 27, 2012
149
27
Keep clear in your mind which current controls everything else: the current through the 0.02 ohm resistor. Everything else changes as a result of what this current does.

11. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
If the voltage in the 100Ω resistor is too high then why would the excess voltage drop across it result in the voltage on V- going up? The voltage on the other side of the 100Ω resistor is constrained by the 0.02Ω resistor, so consider that it isn't changing (plus, that is out input signal and, for this discussion, we are assuming a steady input and asking what happens if an error creeps in elsewhere).

But take a step back and forget about that for a moment. Let's say that the voltage at the V+ pin is 1V. What is Vout when everything is nice and stable and working as intended?

12. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0

If V+=1V then because this circuit produces an output voltage that is a highly amplified copy of the difference in the voltages at its inputs Vout=Ad(V+-V-)=10(1-3)=-20 V ...

13. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
UNITS!!!!!!

Once again, your extremely weak fundamentals is preventing you from making much progress. You really need to address them or you simply are not going to make it in this field.

How are you getting that V- is 3V?

Where are you getting an Ad value of 10?

When I said that the output is a highly amplified copy of the difference in the voltages at the input, I really did mean "highly", as in not 10 or even 10,000, but more like 200,000 and up (recall I then referred to an opamp as an absurdely high-gain differential amplifier).

14. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0

Its said in the circuit that the gain in tension is equal to 10.
My error is related to the fact that i considered that the diferential gain is equal to the gain in tension which is probably not true.
Once again looking at the data given we notice that the sensor works in a range, as far as i have understood, in each when we have

3V of difference in the input we have 0.416 A in the output

36 V if difference in the input we have 5 A in the ouput

So we can only have 1 V in the V+ terminal if we have 4 V (not 3 V i've mistaped the number).

So i can not determine the output tension because i do not know the differential gain.What i do know is that 3V of difference in the input correspond to 0.416 A in the output so V=RI and V=1*10^3 Ohm *0.416 A=1,248 V......

15. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
The Av=10 that they are talking about is the voltage gain of the opamp circuit and it is due to the 1kΩ resistor divided by the 100Ω resistor. I can see how that would be misleading for you. Basically, the bottom of the 20mΩ current sense resistor sets the reference for the amplifier circuit. You then have the input voltage, Vin, which is the voltage at the top of the 20mΩ resistor relative to the bottom of it (i.e., the voltage drop across the current sense resistor. This Vin results in a current I=Vin/Rin where Rin is the 100Ω resistor. This current then flows down through the transistor to the output output resistor, Rout=1kΩ, where the output voltage is Vout = I*Rout = Vin(Rout/Rin). Thus the gane of the voltage amplifier is Vout/Vin=Rout/Rin=1kΩ/100=10.

The gain of the opamp itself is much, much higher than that. I took a quick look at the data sheet and, not surprisingly, it doesn't give the open loop gain of the internal opamp. But reading between the lines on some of the other things leads me to believe that it is about 1,000,000, which is not an uncommon number for modern opamps that aren't expected to operate very fast.

You're pulling numbers out of thin air where they don't exist.

You have NO idea how the input voltage of 3V to 36V corresponds to load current -- you don't have a load!!

The notation 3V to 36V merely indicates the range of supply voltages you can use and get the IC to work without letting its magic smoke out.

MAYBE, the supply voltage is increased in order to increase current to the load, but more likely the supply voltage is FIXED and changes in the load cause the current in it to change.

Again you are grabbing whatever V and whatever R and throwing them at whatever equation is handy.

Have you tried reading the data sheet for this part? It explains, in gory detail, how it works.

http://cds.linear.com/docs/en/datasheet/6106fa.pdf

16. ### ac_dc_1 Thread Starter Member

Jan 27, 2013
74
0

I was not computing the load current ,i was computing the voltage referenced in the scheme as Vout.The numbers did not come from the thin air.I thought that 3V to 36 V was the range of values in the ouput when we have a certain current in the input...I also thought that the 5A stated in the scheme were the input current when we have 36 V output.Since we are talking about a current sensor i thought that a certain value of current in the matched a certain value of tension in the output(not in the LOAD.)

In the datasheet it is said"

This is what explains the feedback in this circuit the changes of the current going through the 0.02 Ohm resistor,which affect the currents that pass through the internal transistor?

17. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
By "thin air", I mean grabbing whatever voltages are noted on the diagram and using them wherever you need a voltage because you need a voltage and there one is and you don't have the skill (yet) to recognize what the notations are talking about, whether they bear any relation to the voltage you need, or how to figure out the voltage you need if they don't.

i.e., thin air.

It does.

Look at the notation at the output. It states 200mV/A. That is the gain of the overall circuit, which is also known as a transimpedance amplifier because the gain has units of V/A, or resistance (impedance in the general case) and the "trans" part means that the voltage and the current involved in that ratio are not at the same point - so a current at one point in the circuit results in a voltage at another point in the circuit.

For every ampere of input, you will see 200mV at the output.

When you have 5A of current, you have 1V at the output (of the circuit, not of the opamp).

Yes. Remember the analysis assumption for working with opamp circuits. When they are in the linear region (not saturated), the output voltage is effectively equal to the input voltage. This is a direct consequence of the very high gain of the opamp. If you have an opamp that has a gain of 1 million, then it will only take a dozen microvolts of difference between the two inputs to saturate the thing. Hence, if it isn't saturated, the voltages at the two inputs must be within a dozen microvolts or so of each other. Even with all the non-ideal affects seen in real opamps, the differences are usually less than a dozen millivolts or so.

Not that the "V-" they are using in that last equation is NOT the inverting input of the opamp, but rather the negative supply voltage. In many cases, this would just be GND.