Op-Amp: Voltage to Current Converter
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- Опубліковано 8 лип 2024
- In this video, how to design the voltage to current converter circuit using op-amp has been discussed.
Why Voltage to Current Conversion?
Many times in the industrial applications it is required to transport the voltage from one location (from a sensor) to the other location (Data Acquisition System). During the transportation, the voltage used to get dropped across the cable and at the other end output voltage used to be lower than the actual voltage.
This problem can be avoided by using the voltage to current converter. The current will remain constant throughout the cable length and it is less prone to the external noise.
Apart from that in many lab experiments, for testing many circuit components this voltage to current converter circuit is very useful.
These type of current sources are also known as the voltage controlled current source. Because the output current can be controlled by changing the input voltage.
Application of Voltage to Current Converter:
1) Testing the circuit components (Diode, Zener Diode)
2) Driving the LEDs
3) To convert the sensor voltage into the current (4-20 mA) in industrial applications.
The timestamps for the different topics covered in the video is given below:
0:25 Why Voltage to Current Conversion?
2:32 Passive Voltage to Current Converter and it's limitation
3:27 Floating load Voltage to Current Converter using Op-amp
4:46 Voltage to Current Converter Applications
5:50 Grounded Load Voltage to Current Converter using Op-amp
This video will be helpful to all students of science and engineering in understanding how to design the voltage to current converter using the op-amp.
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The timestamps for the different topics covered in the video is given below:
0:25 Why Voltage to Current Conversion?
2:32 Passive Voltage to Current Converter and it's limitation
3:27 Floating load Voltage to Current Converter using Op-amp
4:46 Voltage to Current Converter Applications
5:50 Grounded Load Voltage to Current Converter using Op-amp
can we consider there circuit as VCCS????
your accent is hard, but man you definately know your stuff and are good at prepairing a class and organizing the knowledge. thank you!.
lmao indian accent is the only accent where the words are spoken to the utmost clearance
turn on the captions mate
I have almost lost my hope to get pass in this subject.. but after watching your videos I am now confident to sit on exam hall... Thank you very much..
did u pass, becaus ei failed
I have done good sir.. I will pass.. Thanks a lot.. Honestly.. I have watched only your videos and appered in exam.. and exam was good..
application of the ckt described was great. thanks!
Sir, you are amazing! This is a better circuit then in the book in my studies :D
Great teaching by a genius person.
As always wonderful explanation
Thanks for these Presentation!
THANK U SIR FOR EXCELLENT LECTURE
amazing explaination...👍👍
very nice... I m preparing for IES and dmrc Lmrc and other j.e exams please upload full detailed videos... thnks
Great explanation
Thank you sir For this voltage
thanks for uploading this videos 😍
you are just just just amazing and awsome
very good work
Thanks bro. So with only the Vin and a commercial R I can have the desired IL. Because IL depends on I1 and I2 and these are related with Vin right, but I don't get that how IL independent from RL, I mean IL would be IL= Vin-0/RL?
as i can see the voltage at non inverting terminal is also depending on output voltage , so virtual ground concept is looking doubtful.please explain.
At 4.59, what is the voltage across the LED? And what is the value of Vout?
Sir, can you tell me what are the applications for op amp voltage to current converter?
Best explained
Thank you sir
Tq...bro...it's..usefull..🤗🤗
is this circuit able to measure a bidirectional current?
how to measure 1 A current by minimizing output voltage to zero?
Excellent 👌
Need help....basically I am not aware of electrical and electronics....,can any one tell me that how to store voltage alone? Can it be converted to current? The source from where i am getting voltage doesn't have tetminals....,pls suggest. TIA
Thank you Sir for your prompt reply.
But one question still remain answered that how you come to know that Ckt is operating in Linear region and it is not saturated? Is there any systematic way to investigate the mode of operation (linear or saturated) when both type of feedback is present in Op-amp? Or should we always conclude that when both type of feedback are there we should assume , by default op-amp is operating in linear region? I will be very thankful to you if these question is being answered in crystal clear way as you do in most of your lecture
It actually depends on the amount of positive and negative feedback. But if the amount of feedback is not known then you can analyze the circuit assuming the circuit is operating in the linear region. And then you can check the output. If the output is less than the supply rail for the given values then it is operating in the linear region. But if the output saturates then the op-amp is operating in the saturation region.
Your calculated load current,, where ur converted Vin in to I out current
Excellent
What is the difference between floating load and grounded load V-I converter.
What is the application of both?
In a floating load, the load is not actually grounded. While in case of the grounding load, the second terminal of the load is connected to ground.
The floating load V-I converter can be used when you want to provide the constant current to any two-terminal device like LED. (Where your purpose is to provide a constant current to the LED).
The grounded V-I converter can be used when you want to provide the constant current to another circuit or device.
can't thank you enough
Sir, please make videos on voltage to frequency and F/ V converters.
that's call a VCO, voltage-controlled-oscillator. Basically you just need a way to adjust the resonant tank.
can u give the reason why you have used the the concept of virtual short in the lecture at t= 8:14, while this op-amp have both type feedback negative as well as positive feedback?
Because still, the op-amp is working in the linear region. (Output of the op-amp is not saturated to the supply rail)
Thank you Sir for your prompt reply.
But one question still remain answered that how you come to know that Ckt is operating in Linear region and it is not saturated? Is there any systematic way to investigate the mode of operation (linear or saturated) when both type of feedback is present in Op-amp? Or should we always conclude that when both type of feedback are there we should assume , by default op-amp is operating in linear region? I will be very thankful to you if these question is being answered in crystal clear way as you do in most of your lecture.
In the last example, does that mean no matter what RL you put IL is independent of RL?
Yes.
Omg thanks so much
Sir,what happens if the load resistor value RL is very high...?
In that case IL cannot be maintained right...?
Please reply...
Yes, op-amp needs to be in the linear region.
@@ALLABOUTELECTRONICS Thank you sir...
draw differential voltage to current converter and hence show that the load current IL depends on the difference between input voltage and resistance R not on load resistor RL. Solved it please.
Sir why we need to a resistor to the inverting terminal except feedback resistance?
To set the specific current through the load. The current through the load resistor is Vin / R. By changing the value of R, you can change the current. I hope, it will clear your doubt.
What if RL is open circuit ??
I can’t find any difference between non inverting amplifier and this circuit but output of two circuits shows significant differences . How same circuit get different output? Can you help me
The important thing is where the output is measured. And what is measured? (The current or voltage).
The configuration is the same, but here the load resistor is connected in the feedback. (Floating load voltage to current conversion).
So, through the input voltage, the current through the load is controlled.
I hope it will clear your doubt.
Does the virtual short appears only when there is negative feedback..? Please let me know that..
Yes, it is applicable only when the op-amp is operating in the linear region. (with negative feedback)
@@ALLABOUTELECTRONICS so , why the voltage at that point is equal to the voltage at non inverting input ( Vin ) ....? Why not the output voltage ( V0 ) present at that point...?
@@ALLABOUTELECTRONICS please let me know that.. I was too confused.. 😭😭
please make a full video on opamp and a video of numerical.
There are series of videos on the op-amp. You can check that on the channel page.
Here is the link for the entire playlist on the op-amp:
ua-cam.com/video/kiiA6WTCQn0/v-deo.html
Superr sir
Please suggest a reference book for opamp lessons ?
You can refer op- amp by ramakant gayakwad.
I tried V to I converter with grounded load using R=1K and 1K pot as load. The load current remained constant for the full range of load resistance for up to 8VDC input only. Above that input voltage setting the load current varied with load resistance. Can u explain.
What was the supply voltage ??
@@ALLABOUTELECTRONICS Supply voltage was +/-15V
@@surendransunil7025 Have you measured the RL while varying it and the output voltage? Which op-amp you have used? And what is the maximum output driving current of that op-amp?
Please check the maximum output driving current of the op-amp.
@@ALLABOUTELECTRONICS I used op amp 741. I did not measure RL or the output voltage.
What if the position of the diode id reversed..
i did not understand 5.31. How the output voltage become Vin + Vd ?
Applying the concept of virtual short, the voltage at the inverting terminal will also equal to Vin.
Now, if you apply the KVL in the feedback path, from Vout to Vin, then you will get the expression. I hope it will clear your doubt.
Thank u....
Please make video of current to voltage converter using op amp
Very soon it will be uploaded.
can we consider there circuit as VCCS????
YES.
Why we have taken feedback to both terminals
The reason might be to eliminate the V1 from the final IL equation. If there is no positive feedback path the V1 term persists.
How to calculate the output if the load is floating?
Vo = Vin + (IL * RL)
Sir,
Did current pump and Voltage to Current converter are both same ...?
The current pump circuit is also one type of voltage to current converter.
@@ALLABOUTELECTRONICS Thank u sir
Hello Sir 5:21 is this same as Log Ampr?
In the log amplifier, the input is applied at the inverting input terminal. Here. input is applied at the non-inverting terminal of the op-amp.
Thank you for reply sir ..!! one small doubt Sir then will log amplifier works only when an input is connected to Inverting terminsl?
Yes.
You have taken all the resistor values as R 😌
😍 tq
When a sensor measures temp how does it give o/p in terms of voltage?
Using the bridge circuit, it can be converted into a voltage. If you take RTD, its resistance changes with temperature. If it is connected at the one end of the bridge, then the change in the voltage between the two ends of the bridge can be calibrated in terms of voltage. In some sensors, this circuit is part of the sensor and it directly gives output in terms of the voltage.
In theory it is OK but simulations don't think like that. When RL changes, IL also changes, probably in real life will be same.
sir plsss give a vid about arduino and its parts and raspberry pi
I will make the video on it.
Jodi Paris taholee practical kore dhakha
Still confusing asf
in THE
I'm sorry. But who can re-dub it with international english ?
Sala school a ja
No offense sir but i have been rewatching this video 10 times over and i still cannot understand your accent
Please use the subtitles while watching the video. It will help you.
Thank you sir
thank you sir