LED Driving Circuit

Thread Starter

student_897

Joined Nov 2, 2023
12
Good afternoon,
our Professor explained to us the attached circuit to drive an LED, but did not explain it very well.

Could someone, kindly explain to me that circuit? I know the basic principle of constant current driving purposes, but I'd like to ask you further explanations.


Thank you very much in advance.
View attachment 307027
 

Thread Starter

student_897

Joined Nov 2, 2023
12
Yes.

What are your questions? What are the values for the zener diode and R3? What does R2 connect to?
Hi,
The Zener is a 3v9. R3 is approximately 90Ohm.

My question is related to the reason why we employ two transistors, a PNP and an NPN to get a constant current instead of only using an NPN and controlling the emitter.
Is this configuration more stable?
Instead, the concept of using a Zener diode at the input of the opamp is clear.
Finally, why do we employ that diode between the base and the collector of the NPN?

Thank you again!!
 

dl324

Joined Mar 30, 2015
16,654
1699459015962.png
My question is related to the reason why we employ two transistors, a PNP and an NPN to get a constant current instead of only using an NPN and controlling the emitter.
The NPN serves no useful purpose. With R2 not connected, it doesn't do anything useful.
Is this configuration more stable?
More stable than what?
Instead, the concept of using a Zener diode at the input of the opamp is clear.
The opamp plus PNP transistor is configured as a constant current source. Do you understand how it works?

43mA is a little high for a standard type LED.
Finally, why do we employ that diode between the base and the collector of the NPN?
It would protect the base-emitter junction from breaking down if the voltage on R2 can be negative enough.
 

Thread Starter

student_897

Joined Nov 2, 2023
12
View attachment 307034
The NPN serves no useful purpose. With R2 not connected, it doesn't do anything useful.
More stable than what?
The opamp plus PNP transistor is configured as a constant current source. Do you understand how it works?

43mA is a little high for a standard type LED.
It would protect the base-emitter junction from breaking down if the voltage on R2 can be negative enough.
Sorry very much for not inserting a very important mark in the figure!!!!
R2 is connected to a 1kHz square wave generated by an external function generator.

With more stable I didn't mean anything in particular, I tried to guess something useless indeed....

The LED typical current the professor investigated is about 50mA.
I guess that I understood how it works but an additional explanation would be gratefully accepted, clearly :)

Ok for the diode.

Thank you!
 

dl324

Joined Mar 30, 2015
16,654
R2 is connected to a 1kHz square wave generated by an external function generator.
What is the voltage swing on the square wave? Can it go below -5V?
With more stable I didn't mean anything in particular, I tried to guess something useless indeed....
We need something to compare with to determine if the circuit you posted is more or less stable.
The LED typical current the professor investigated is about 50mA.
Without a part number, most would assume that it's a typical LED with a 20mA continuous forward current rating.
I guess that I understood how it works but an additional explanation would be gratefully accepted, clearly
Do you know the function of every component and why its value was chosen? If you do, you've mastered that circuit.

Ask questions for anything you don't understand.

It would be helpful to know the part number for the opamp and the power supply voltages being used so we can determine if the opamp can function as expected.
 

Thread Starter

student_897

Joined Nov 2, 2023
12
What is the voltage swing on the square wave? Can it go below -5V?
We need something to compare with to determine if the circuit you posted is more or less stable.
Without a part number, most would assume that it's a typical LED with a 20mA continuous forward current rating.
Do you know the function of every component and why its value was chosen? If you do, you've mastered that circuit.

Ask questions for anything you don't understand.

It would be helpful to know the part number for the opamp and the power supply voltages being used so we can determine if the opamp can function as expected.
The Vpp = 5V with zero mean value, thus Vmax = +2.5V and Vmin = -2.5V.
The supply voltage is +-12V, for a TL071IP opamp.

I understood basically everything in the circuit but the capacitor C4.
Indeed, when I test it and insert this capacitor (100nF), I get a very noisy signal.
Removing the capacitor instead, I get a not noisy signal but with an attenuated amplitude, that's what I did not understand.
The noisy signal mantains its period of 1kHz but I get a completely unexpected amplitude, of about Vpp = 7.5V

Thank you
 

dl324

Joined Mar 30, 2015
16,654
The Vpp = 5V with zero mean value, thus Vmax = +2.5V and Vmin = -2.5V.
What do you think the purpose of the diode is?
I understood basically everything in the circuit but the capacitor C4.
Indeed, when I test it and insert this capacitor (100nF), I get a very noisy signal.
Removing the capacitor instead, I get a not noisy signal but with an attenuated amplitude, that's what I did not understand.
What do you mean by noisy? Is C4 being used in a way that the other capacitors aren't?

Why are there two caps across R1?
The noisy signal mantains its period of 1kHz but I get a completely unexpected amplitude, of about Vpp = 7.5V
Where are you measuring this?
 

Thread Starter

student_897

Joined Nov 2, 2023
12
What do you think the purpose of the diode is?
What do you mean by noisy? Is C4 being used in a way that the other capacitors aren't?

Why are there two caps across R1?
Where are you measuring this?
If you mean the Zener, we use it for voltage stabilization and limiter.

With noisy I mean that the upper part of the square wave is correct, the lower part instead is completely noisy, as the attached figure.
The two capacitors (1uF and 100nF) I guess that are used always for voltage stabilization purposes.

I measure this signal at the LED anode node --> thus between the two collectors of PNP and NPN, with an oscilloscope.

1699462075873.jpeg
 

dl324

Joined Mar 30, 2015
16,654
1699462576453.png
With noisy I mean that the upper part of the square wave is correct, the lower part instead is completely noisy, as the attached figure.
Can you attach a picture of the waveform? What is the value of C4?
The two capacitors (1uF and 100nF) I guess that are used always for voltage stabilization purposes.
They aren't always used. Why do you think you need two different values?
 

dl324

Joined Mar 30, 2015
16,654
If you mean the Zener, we use it for voltage stabilization and limiter.
Sorry, I forgot that that diode didn't have a component designator. I meant the diode I labeled D1.
Sincerely, I think that 1uF would be enough for disturbances elimination and thus voltage stabilization purposes.
We sometimes use two different values of ceramic capacitors in parallel due to parasitic inductance in the larger capacitor affecting its high frequency response. It's much more common to use 0.1uF for decoupling high frequency noise on the power supply caused by switching logic and to use a 10nF cap in parallel if high frequency noise is anticipated. You don't really have that in this circuit. Power supply noise isn't going to have much impact on the current source.

Is the yellow arrow with 1 the zero voltage level?
 

Thread Starter

student_897

Joined Nov 2, 2023
12
Sorry, I forgot that that diode didn't have a component designator. I meant the diode I labeled D1.
We sometimes use two different values of ceramic capacitors in parallel due to parasitic inductance in the larger capacitor affecting its high frequency response. It's much more common to use 0.1uF for decoupling high frequency noise on the power supply caused by switching logic and to use a 10nF cap in parallel if high frequency noise is anticipated. You don't really have that in this circuit. Power supply noise isn't going to have much impact on the current source.

Is the yellow arrow with 1 the zero voltage level?
D1 serves as a protection.

Sorry, I confused the capacitors values.

Yes, it is!

What do you think? I cannot get over this problem...

Thank you!
 

dl324

Joined Mar 30, 2015
16,654

Thread Starter

student_897

Joined Nov 2, 2023
12
Protection against what?
Which ones?
If the circuit is operating properly, can you get a negative voltage on the anode of the LED? Can you increase horizontal resolution so we can see what's happening when the waveform isn't correct?
Protection to reverse voltages for which base-emitter junction goes into breakdown, thus damaging the BJT.

I confused the capacitor values when answering to your question, I meant that I would use only the 0.1uF, sorry.

Yes, here it is!
1699464793057.jpeg
 
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