4-20mA Current loop

WBahn

Joined Mar 31, 2012
26,076
=> Vin+=(V_in+Vsense)*(R2/(R1+R2))
Again, ask if your answer makes sense!

You know that if R1 = 0 Ω that Vin+ = Vin and that if R2 = 0 Ω that Vin+ = Vsense.

Using your new equation, if R1 = 0 Ω then Vin+ = Vin + Vsense and if R2 = 0 Ω then Vin+ = 0.

So you KNOW that this is wrong and there is NO point going any further. Anything else is pure wasted effort!
 

WBahn

Joined Mar 31, 2012
26,076
It looks like you need to take a huge step back and work on your basic circuit analysis skills. You can't hope to solve problems without a command of the fundamentals.

So let's look at, once again, the question of what Vsense (the voltage on the very bottom node in your schematic) is when there is 4 mA of current flowing downward through Rsense and what Vsense is when there is 20 mA of current flowing downward in it. This is very basic stuff and you need to be able to answer questions like these with very little thought. Remember, the sign of a value matters big time.
 

WBahn

Joined Mar 31, 2012
26,076
Youre right.

=> Vin+=Vsense+(I*R2)
Then I dont know what to do...
Okay, so this equation is correct. Now you just need to figure out what I is.

What is the voltage across the series combination of R1 and R2?

HINT: If you have Va on the top of Req and Vb on the bottom of it, then the voltage drop from top to bottom is Va - Vb.
 

WBahn

Joined Mar 31, 2012
26,076
Look at the hint. The voltage across R1 + R2 is the voltage at the top of R1 minus the voltage at the bottom of R2.

What is the voltage at the top of R1?

What is the voltage at the bottom of R2?
 

Thread Starter

johndow

Joined Jan 16, 2016
29
Thank you that you helps me!
I through R1 and R2 is Vin - Vsense/(R1+R2)

And:
Vin+=Vsense+( ( Vin - Vsense/(R1+R2) )*R2)
 

WBahn

Joined Mar 31, 2012
26,076
Thank you that you helps me!
I through R1 and R2 is Vin - Vsense/(R1+R2)

And:
Vin+=Vsense+( ( Vin - Vsense/(R1+R2) )*R2)
Again -- watch out for order of operations!

Vin - Vsense/(R1+R2) is the same as Vin - [Vsense/(R1+R2)].

Write expressions that accurately represent what you mean to say!

If you don't, you are relying on others to guess what you meant -- you are also relying on yourself to guess what you meant at some point later in the problem. Plus you will invariably enter things incorrectly into programs and spreadsheets and other tools that use text-based equation entry.

Vin+=Vsense + ( (Vin - Vsense)/(R1+R2) )*R2

Part of your problem is the degree of sloppiness you are applying to your work. Nothing but demons lie there.

Now ask if this equation makes sense. Does it limit like you know it must for R1=0 and for R2=0?

If so, then there is a very good chance it is correct and you can move on. If not, then you know it is wrong and you need to rework it.
 

Thread Starter

johndow

Joined Jan 16, 2016
29
Write expressions that accurately represent what you mean to say!
I wil take care in the future!
Vin+=Vsense + ( (Vin - Vsense)/(R1+R2) )*R2
Part of your problem is the degree of sloppiness you are applying to your work. Nothing but demons lie there.
Yes I know that Problem :p
Now ask if this equation makes sense. Does it limit like you know it must for R1=0 and for R2=0?
I thin Ith R2=0 it should be (V_in+Vsense)/R1. V+=[(V_in+Vsense)/R1]*R1.
Or what do you mean exactly with R10= and R2=0?
 

WBahn

Joined Mar 31, 2012
26,076
Come on. Take the time to do things methodically and one step at a time. This is clearly an area where you are struggling, so don't go rushing in and missing steps.

You have

Vin+=Vsense + ( (Vin - Vsense)/(R1+R2) )*R2

If you set R1 = 0, what does this equation reduce to?

Look at the circuit. If R1 = 0 (if it is made a short circuit), what will Vin+ be?

Do these agree?

If you set R2 = 0, what does this equation reduce to?

Look at the circuit. If R2 = 0 (if it is made a short circuit), what will Vin+ be?

Do these agree?
 

WBahn

Joined Mar 31, 2012
26,076
Okay, so now we have an equation that gives us Vin+ as a function of Vin and Vsense.

You know that, in normal operation, Vin+ equals Vin- which equals 0.5 V (but let's call this voltage Vthres). So you have two unknowns, R1 and R2.

We need two different operating points for this equation to solve to two unknowns.

When Vin = 1 V what is Vsense (if your circuit is behaving as you want it to)?

When Vin = 4 V what is Vsense?
 

Thread Starter

johndow

Joined Jan 16, 2016
29
(Vin+) - (( Vin - Vsense/(R1+R2))*R2)=Vsense

hm I dont know how t geht vsense, because it is two times in the equation.
 

WBahn

Joined Mar 31, 2012
26,076
(Vin+) - (( Vin - Vsense/(R1+R2))*R2)=Vsense

hm I dont know how t geht vsense, because it is two times in the equation.
What current do you want flowing in Rsense when Vin is 1 V?

What value of Vsense will result if this amount of current is flowing in Rsense?

What current do you want flowing in Rsense when Vin is 5 V?

What value of Vsense will result if this amount of current is flowing in Rsense?
 

Thread Starter

johndow

Joined Jan 16, 2016
29
What current do you want flowing in Rsense when Vin is 1 V?

What value of Vsense will result if this amount of current is flowing in Rsense?
Vin 1V then 4mA. Vsense is 0.2V
What current do you want flowing in Rsense when Vin is 5 V?

What value of Vsense will result if this amount of current is flowing in Rsense?
20mA at Vin 5V. This is 1V Vsense.


This would be: (Vin+ - Vsense)/(Vin - Vsense) = (0.5V - 0.2V) / (1V - 0.2V) = 0.375=(R2/(R1+R2))
 
Last edited:

WBahn

Joined Mar 31, 2012
26,076
Vin 1V then 4mA. Vsense is 0.2V
NOOOOO!!!!!!!!

We have been through this several times.

If Vsense is 0.2 V and the other end of Rsense is at 0V, then that means that the current through Rsense is AWAY from the negative terminal of the only power source in the circuit, meaning that the only power source in the circuit is being charged by a circuit that has no other source of power.

Vsense HAS TO BE NEGATIVE!!!!!!!!!!!!!!!
 
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