A Question about another Differential Amplifier

Hello,

I am new here and I wonder if I can get help in solving the problem which is shown and explained on the attached image below

Thank you.

This is homework so we don't do that for you.
You need to show us your best effort, and then we can help you with that.
Have you looked up differential op amp circuits?

Is this just for simulation or do you need it to be an opamp circuit?
Hint: For simulation you can use a behavioral source.

I_dc 300A ? Confirm that.

Regards, Dana.

danadak said:

I_dc 300A ? Confirm that.

Regards, Dana.

Click to expand...

It is generating 0.1V across the 1mΩ (shunt) resistor so I think it is correct.

First, thank you all for your time in taking care to reply or comment.

And you are totally right. My question looks indeed as if it were homework. Sorry, I didn’t update yet my profile; I am 73
By the way, I used to ask others when I face a new situation during my design as a way to stimulate my brain to find out its optimum practical solution before some others will do.

For this case, I tried using one opamp but I realized that it ends always to closed roads (so I guess no one here can give a solution by using one opamp only ) But, a few hours after I posted the question (while trying to sleep in bed), I found out that it could be done with two opamps (so some of you may likely know how to do it ). But being a rational/practical man, I have to test what I have in mind first (by using the simulator LTspice, for example) before approving it. So, I will be back soon later.

For your curiosity, it is about an automatic battery charger. In brief, V_dc is proportional to the sinusoidal voltage of the mains supply (220V, 50Hz). And I_dc is the maximum peak of the battery ‘pulsing’ current (actually, it is about 60A for R1=5m) and the peaks of the current pulses follow the sinusoidal amplitude of V_dc, that is V1-V2.

Cheers,
Kerim

I am very sorry. It seems I wrongly simplified the schematic of the problem above. So just to complete the topic, I attached the updated one below. Now, it looks like the case of a hi-side differential amplifier (though R2 is not a sensor here). Although special ICs are made for hi-side sensors, they may not be suitable for a relatively high voltage on it (here, it is V1-V2 on R2 where V2 is close to ground). Also, I avoid, as possible, using special ICs while I have a lot of conventional opamps.

Anyway, it is up to you now to try sharing your ideas or not about a practical solution. On my side, I am sure I will find one which could be rather simple and suitable for the project I am working on.

Best wishes,
Kerim

A simple, conventional differential amp with a single opamp and four resistors is all you need.

See the "Operational amplifier as differential amplifier" section in this article:

The two input resistors should be equal.
The two feedback & reference resistors should be equal. The ratio of those to the input resistors defines the gain.

The point the ref resistor (Rg) connects to defines the output voltage when the inputs have zero difference.

Use close tolerance resistors to minimise errors, or a lower value for one input resistor plus a preset in series with it, so the total resistance can be varied above and below the theoretical value, to trim out errors.
eg. Rather than a 10K, you could use 9k2 plus a 2K preset.

(You can actually use different values for the input resistors, and also for the two feedback resistors, as long as the ratio of input to feedback is the same - but it just complicates things and may introduce more errors).

rjenkinsgb said:

A simple, conventional differential amp with a single opamp and four resistors is all you need.

See the "Operational amplifier as differential amplifier" section in this article:

Differential amplifier - Wikipedia

en.wikipedia.org

The two input resistors should be equal.
The two feedback & reference resistors should be equal. The ratio of those to the input resistors defines the gain.

The point the ref resistor (Rg) connects to defines the output voltage when the inputs have zero difference.

Click to expand...

Thank you for replying.
I am afraid that this simple conventional differential amp works only if its two inputs are independent of each other.
Here if V2 decreases, V1 decreases too. That is V1 is a function of V2.
But I also understand that as long you don't have time to simulate what you proposed (as I did already before I started this thread), you have to believe that the conventional differential amp also works here.

KerimF said:

I am afraid that this simple conventional differential amp works only if its two inputs are independent of each other.

Click to expand...

Well, that is confusing.
What do you mean "independent of each other"?

On your schematic, you say you want V_out = K(V1-V2).
That is what a differential amp will do.

KerimF said:

Thank you for replying.
I am afraid that this simple conventional differential amp works only if its two inputs are independent of each other.
Here if V2 decreases, V1 decreases too. That is V1 is a function of V2.
But I also understand that as long you don't have time to simulate what you proposed (as I did already before I started this thread), you have to believe that the conventional differential amp also works here.

Click to expand...

This is nonsense. A differential amplifier imposes no such restriction on the inputs. Any particular amplifier that you can buy may have range restrictions on the inputs, but nothing requires them to be or not to be related. The transfer function just does what it is designed to do.

KerimF said:

I am afraid that this simple conventional differential amp works only if its two inputs are independent of each other.
Here if V2 decreases, V1 decreases too. That is V1 is a function of V2.

Click to expand...

The only restriction on that style of differential amp is that the opamp +input voltage must always be within the common mode range of the device chosen.
That voltage is produced by the +input and reference resistor divider, the ratio of which depends on the gain K.

(And, obviously, the output voltage after gain must be within the opamp o/p voltage range on the supply in use).

Any general opamp which includes V- in its common mode range, and is rated for 30V single supply should work fine.

As long as the resistor values are chosen correctly, the output voltage relative to the ref point will be equal to (V1 - V2) * gain.

rjenkinsgb said:

The only restriction on that style of differential amp is that the opamp +input voltage must always be within the common mode range of the device chosen.
That voltage is produced by the +input and reference resistor divider, the ratio of which depends on the gain K.

(And, obviously, the output voltage after gain must be within the opamp o/p voltage range on the supply in use).

Any general opamp which includes V- in its common mode range, and is rated for 30V single supply should work fine.

As long as the resistor values are chosen correctly, the output voltage relative to the ref point will be equal to (V1 - V2) * gain.

Click to expand...

That's putting a pretty fine point on the matter.

KerimF said:

I am very sorry. It seems I wrongly simplified the schematic of the problem above. So just to complete the topic, I attached the updated one below. Now, it looks like the case of a hi-side differential amplifier (though R2 is not a sensor here). Although special ICs are made for hi-side sensors, they may not be suitable for a relatively high voltage on it (here, it is V1-V2 on R2 where V2 is close to ground). Also, I avoid, as possible, using special ICs while I have a lot of conventional opamps.

Anyway, it is up to you now to try sharing your ideas or not about a practical solution. On my side, I am sure I will find one which could be rather simple and suitable for the project I am working on.

Best wishes,
Kerim

View attachment 140254

Click to expand...

What does "the node AUX could be connected at will" mean?

I just ran the simulation, and it seems your circuit is a hot steaming mess with few if any distinguishing characteristics.


Maybe you'd like to try again for a circuit that actually does something, or tell me what kind of simulation you want to run. I'm at a loss for understanding your intentions.

To crutschow, Papabravo and rjenkinsgb, I apologize and owe you all a great thank for giving me the chance to discover a repeated ‘silly’ mistake in all the tests I did.

In other words, you did give me the help, I needed, by being patient with me.

Thank you again.
Kerim