Proportional only controller in LED Driver using Op-Amp

  • Engineering
  • Thread starter PhysicsTruth
  • Start date
  • #1
PhysicsTruth
116
18
Homework Statement:
Construct a LED driver with ##K_P=10## for -0.2 - 0.2 V input. Correct it to 0-4V using an offset bias.
Relevant Equations:
Using the LED Driver Circuit equation -
##V_{control} = V_{LED/anode} - V_{Th}##
I have a proportional-only controller LED driver circuit -
life_4.png

I need to fine-tune its ##K_P## to 10 for a significant input of -0.2 to 0.2 V in the steady state. In that case, should I use a potentiometer with 10:1 setting? The LED (actuator) can drive voltages between 0 to 4V, so I need to add a resistor bias offset voltage to the controller as well. Can someone help me in these? I know that I need to utilize the Thevenin circuit, so should I just make a resistor bias from my positive rail supply to ground? Cam someone help me out?
 

Answers and Replies

  • #2
PhysicsTruth
116
18
Homework Statement:: Construct a LED driver with ##K_P=10## for -0.2 - 0.2 V input. Correct it to 0-4V using an offset bias.
Relevant Equations:: Using the LED Driver Circuit equation -
##V_{control} = V_{LED/anode} - V_{Th}##

I have a proportional-only controller LED driver circuit -
View attachment 291664
I need to fine-tune its ##K_P## to 10 for a significant input of -0.2 to 0.2 V in the steady state. In that case, should I use a potentiometer with 10:1 setting? The LED (actuator) can drive voltages between 0 to 4V, so I need to add a resistor bias offset voltage to the controller as well. Can someone help me in these? I know that I need to utilize the Thevenin circuit, so should I just make a resistor bias from my positive rail supply to ground? Cam someone help me out?
I guess I can use a potentiometer with 10:1 setting connected from the LED cathode and forming a voltage divider bias with the negative terminal of Op-Amp, and finally to ground. But, I'm having some problems in setting up the offset bias. How do I actually connect this bias to correct it to 0-4V for -0.2 to +0.2 V input, with ##K_P = 10## ?
 
  • #3
Baluncore
Science Advisor
12,083
6,201
I suspect you have not presented the original question as asked.
I cannot separate your assumptions from the question.

You should regulate the current through a LED, not the voltage across it.
The circuit you give is for a voltage controlled current sink, which is good.
The 4 V appears across Rfb, the current sense resistor, not the LED.

What is the power supply voltage? Is it sufficient for 4V + Vled ?
Do you know what maximum current will be needed by the LED?
Can the op-amp source that current or will you need something like an emitter follower?

Kp = 10; Ten what? mA/volt or volt/volt ?
Why not fix the maximum voltage across Rfb to 0.4 volt ?
Then design the required control voltage level shift.
 
  • #4
PhysicsTruth
116
18
I suspect you have not presented the original question as asked.
I cannot separate your assumptions from the question.

You should regulate the current through a LED, not the voltage across it.
The circuit you give is for a voltage controlled current sink, which is good.
The 4 V appears across Rfb, the current sense resistor, not the LED.

What is the power supply voltage? Is it sufficient for 4V + Vled ?
Do you know what maximum current will be needed by the LED?
Can the op-amp source that current or will you need something like an emitter follower?

Kp = 10; Ten what? mA/volt or volt/volt ?
Why not fix the maximum voltage across Rfb to 0.4 volt ?
Then design the required control voltage level shift.
Kp is in terms of volt/volt, which is thus across Rfb. The Op-Amp is enough to source the current required. In order to set Kp at 10, I've used another resistor from the LED, which is approximately 9 times that of the feedback resistor. This does give me a output voltage of 10 times that of the input voltage through the output terminal of the Op-Amp, but this works for only positive voltage inputs. Negative voltage inputs seem to saturate the output. Can you suggest an alternative in this case?
 
  • #5
Baluncore
Science Advisor
12,083
6,201
Who says you must use Kp = 10 v/v ?
Again, what is the positive power supply voltage ?
Is there a negative supply ?
 
  • #6
PhysicsTruth
116
18
Who says you must use Kp = 10 v/v ?
Again, what is the positive power supply voltage ?
Is theer a negative supply ?
Well, I'm supposed to build the proportional only controller such that for the input voltages of -0.2 to 0.2 V, I get an output around -2 to 2V from the output terminal. Now, the voltage required to drive the LED is around 0-4V, as I'm expected, but -2V wouldn't drive the LED. So, I need to create that offset bias to change it to 0 to 4V, so that the forward voltage for the LED would be 0 to 4V, and not -2 to 2V. Also, the positive rail supply is +8V, and there is a negative one with -8V supply.
 
  • #7
PhysicsTruth
116
18
Who says you must use Kp = 10 v/v ?
Again, what is the positive power supply voltage ?
Is there a negative supply ?
Can you suggest what resistor values should I use for the voltage divider bias from the output of Op-Amp to the anode of LED, in order to shift the voltage level by 2V?
 
  • #8
Baluncore
Science Advisor
12,083
6,201
... Now, the voltage required to drive the LED is around 0-4V, ...
But the LED brightness is proportional to current, not linearly dependent on LED voltage.
LED voltage is dominated by the LED wavelength, plus a small increase with current.

Again, who says you must use Kp = 10 v/v ?
Is it in the original question, or is it just your assumption ?
 
  • #9
Baluncore
Science Advisor
12,083
6,201
Can you suggest what resistor values should I use for the voltage divider bias from the output of Op-Amp to the anode of LED, in order to shift the voltage level by 2V?
There must be an easier way.
 
  • #10
PhysicsTruth
116
18
But the LED brightness is proportional to current, not linearly dependent on LED voltage.
LED voltage is dominated by the LED wavelength, plus a small increase with current.

Again, who says you must use Kp = 10 v/v ?
Is it in the original question, or is it just your assumption ?
I'm just instructed to obtain a Kp of 10, and since it deals with obtaining 0-4V input to the LED from -0.2 to 0.2V input, I assume it's a V/V Kp.
 
  • #11
PhysicsTruth
116
18
There must be an easier way.
Well, I'm absolutely new to feedback control systems, and I've been asked to use a voltage divider for level shifting in this case. It would be great if you could help me out with the Thevenin equivalent.
 
  • #12
Baluncore
Science Advisor
12,083
6,201
Without current limiting, a forward voltage of 4 V will destroy most LEDs.
Likewise the reverse voltage maximum is 5 V, at very low current.
Your ±8 V supply is more than sufficient to destroy the LED.
You will need to know the maximum LED current before you can design the circuit.
 
  • #13
PhysicsTruth
116
18
Without current limiting, a forward voltage of 4 V will destroy most LEDs.
Likewise the reverse voltage maximum is 5 V, at very low current.
Your ±8 V supply is more than sufficient to destroy the LED.
You will need to know the maximum LED current before you can design the circuit.
Well, I've a red LED, and it can support a maximum of 4V forward voltage with saturation current around 6mA. I say this because I've tested it out using a 4V reference.
 
  • #14
Baluncore
Science Advisor
12,083
6,201
Well, I've a red LED, and it can support a maximum of 4V forward voltage with saturation current around 6mA. I say this because I've tested it out using a 4V reference.
Then it is not a simple LED, it has a series resistance, Rs.
Red is a wavelength of about 640 nm.
The photon energy, in eV = 1239.84 / nm
So the voltage needed to make red photons is about v = 1.94 volts.
You measured current at 6 mA with total v = 4 V;
Vr = ( 4 - 1.94 ) = 2.06 V across the resistor Rs.
Ohms law; Rs = 2.06 / 0.006 = 343 ohms.
But there will be almost no light emitted until V ≥ 1.94 V.
Are you still sure you want to regulate LED voltage rather than control current ?
 
  • Like
Likes PhysicsTruth
  • #15
osilmag
Gold Member
195
43
Well, I'm absolutely new to feedback control systems, and I've been asked to use a voltage divider for level shifting in this case. It would be great if you could help me out with the Thevenin equivalent.
A level shifter requires transistors. I believe you would need a pnp junction transistor to shift the negative output.
 
  • #16
PhysicsTruth
116
18
A level shifter requires transistors. I believe you would need a pnp junction transistor to shift the negative output.
It does. But, using three resistors and a divider bias, we can shift the voltage level as well, if I'm not wrong. I'm unable to arrive at the values in that case.
 
  • #17
PhysicsTruth
116
18
I've solved the problem regarding the saturation of the negative input voltages. If only someone could guide me how to design the voltage shift from -2 - 2 to 0 - 4 using three resistors and a divider bias
 
  • #18
PhysicsTruth
116
18
Then it is not a simple LED, it has a series resistance, Rs.
Red is a wavelength of about 640 nm.
The photon energy, in eV = 1239.84 / nm
So the voltage needed to make red photons is about v = 1.94 volts.
You measured current at 6 mA with total v = 4 V;
Vr = ( 4 - 1.94 ) = 2.06 V across the resistor Rs.
Ohms law; Rs = 2.06 / 0.006 = 343 ohms.
But there will be almost no light emitted until V ≥ 1.94 V.
Are you still sure you want to regulate LED voltage rather than control current ?
Yes, I was doing it wrongly. I created a separate controller for the LED driver, and it's working fine now. Only, I need to get an output of 0-4V from the controller part (without the LED, it functions as a non-inverting amplifier), instead of a -2 V to 2 V output from -0.2 to 0.2V input. Can you give me a hint regarding how to setup the voltage divider with 3 resistors? I've been trying out configurations and have done some shifting, but can't obtain the 0-4 level-shifted output.
 
  • #19
Baluncore
Science Advisor
12,083
6,201
It is real bad practice to use a power supply rail as a voltage reference, mainly because it will add noise to the signal. Changing the power supply will require a recalibration.

The offset will require an accurate voltage or a current reference. You should be aware that there are a number of op-amps that have an 0.2 V buffered reference voltage, and can operate without a negative supply. The LM10 was the first example. There are many more now.
https://www.ti.com/lit/ds/symlink/lm10.pdf
See section 8 for applications.
 
  • #20
Baluncore
Science Advisor
12,083
6,201
  • Like
Likes osilmag and DaveE

Suggested for: Proportional only controller in LED Driver using Op-Amp

  • Last Post
Replies
21
Views
474
  • Last Post
Replies
1
Views
378
Replies
8
Views
196
  • Last Post
Replies
12
Views
1K
Replies
12
Views
748
  • Last Post
Replies
5
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
7
Views
861
Replies
44
Views
2K
  • Last Post
Replies
2
Views
390
Top