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My OpAmp is only outputting 1.6V - what am I doing wrong?

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PowerJunkie

New Member
Hello,

I'm trying to connect an OpAmp to boost the overall voltage signal in my project. I'm using a 741 OpAmp as it's the only one they had at RadioShack.

My OpAmp only outputs a grand total of 1.6V, regardless if I feed 3V, 5V or 8V of AA batteries into it. I tried playing with different ratios with the two different resistors, but it's the same with different ratios and even if I just scrap all of the resistors altogether.

What am I doing wrong?

Here's a schematic of my circuit:



I'm extremely new to all of this so I'm hoping you guys can help me! I'm going NUTS here trying to figure this out for several days now and I can't come up with anything I can understand on Google or on the boards! I've read that the Output can switch between a "High" output and a "Low" output depending on the value of the Non-Invert Input, but I can't seem to get it working. I've tried to put several resistors on the Non-Invert Input but it doesn't change a thing.

Thanks so much for ANY help!
 

crutschow

Well-Known Member
Most Helpful Member
The reason it's always 1.6V output is because of the LED (Light Emitting Diode) load. It's a diode and it's forward drop is 1.6V which changes only slightly with current. Basically the op amp is putting out all the current it can. If you put a 2k ohm resistor in series with the LED you will see the output of the op amp vary with the input.

If you don't understand this, then read a tutorial about LEDs.

What is the power supply to the op amp?
 

PowerJunkie

New Member
The reason it's always 1.6V output is because of the LED (Light Emitting Diode) load. It's a diode and it's forward drop is 1.6V which changes only slightly with current. Basically the op amp is putting out all the current it can. If you put a 2k ohm resistor in series with the LED you will see the output of the op amp vary with the input.

If you don't understand this, then read a tutorial about LEDs.

What is the power supply to the op amp?
Hmm.. but how come when I input 8V's into the OpAmp, the output (with the Multimeter reading right before the LED) is only 1.6V?
If the input is 8V and it's outputting 8V, but the LED voltage drop is 1.6V, wouldn't the Multimeter be reading around 6.4V?

Thanks so much for the help!
 
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crutschow

Well-Known Member
Most Helpful Member
The multimeter is reading the voltage drop across the LED to ground, which is the same as the op amp output voltage to ground. The 6.4V is being dropped internally in the op amp, which you can't read externally. It's rather a phantom voltage.
 

PowerJunkie

New Member
The multimeter is reading the voltage drop across the LED to ground, which is the same as the op amp output voltage to ground. The 6.4V is being dropped internally in the op amp, which you can't read externally. It's rather a phantom voltage.
I removed the LED and I was getting the full 3V, 5V or 8V (whatever Voltage I inputted to the OpAmp) on the output wires with my Multimeter. As soon as I connect the LED, it drops to 1.6V.

So, the way I understand it, does the LED simply take whatever Voltage it needs? If this is the case, what's the point of adding a resistor to the LED? I used an LED Calculator online and just tried to connect a 68ohm resistor to the negative side of the LED (I always thought you connect the resistor to the positive side of the LED, but the calculator showed it on the negative side), but it doesn't change anything.

If a resistor is in place, shouldn't I be getting 6.4V before the resistor since theoretically there is no load before the resistor?

Sorry if I sound ignorant, I'm clueless at all of this but I'm trying my best to learn here!! :D Thanks again!
 

crutschow

Well-Known Member
Most Helpful Member
Yes, the LED is a very low resistance once it's threshold voltage is reached and will draw all the current is can (until it blows from too much current) if you don't limit is with a resistor. The 741 can only output about 25mA maximum, so that's why having no resistor did not blow the LED.

The 68 ohm resistor is too low to see much of a voltage drop since it's only 1.7V at 25mA. That's why I suggested a 2k ohm resistor.

Do you understand ohms law? I suspect not and thus suggest your read about it and understand it. It's the most fundamental of all electrical equations and is necessary to understanding circuits. Otherwise I'm just spinning my wheels trying to explain this to you.
 

Mikebits

Well-Known Member
Otherwise I'm just spinning my wheels trying to explain this to you.
I actually think your doing okay and bringing him closer, don't give up so easy. Your doing good.. :)
 

MikeMl

Well-Known Member
Most Helpful Member
I removed the LED and I was getting the full 3V, 5V or 8V (whatever Voltage I inputted to the OpAmp) on the output wires with my Multimeter. As soon as I connect the LED, it drops to 1.6V.
An LED acts as a forward biased diode with a forward drop of ~2V. Like a diode, once it begins conducting, it has a very low resistance. If connected to a voltage, some else must be there to limit the current that flows through the LED.

Most LEDs have a maximum current rating of ~30mA, and they are damaged if this current is exceeded by much.

I simulated two identical LEDs, one with a resistor in series with it, the other not. The simulation shows the current through the two LEDs as the input voltage is swept from 0V to 10V (the horizontal X-axis).

Look at the upper plot. No current flows through R1 (or D2) until the input exceeds ~1.8V, at which point the current I(R1) increases linearly to 33mA when the input is 10V. Note that the voltage across D2 V(D) saturates at about 1.8V even though the input voltage goes all the way to 10V. This is the correct way to drive an LED.

Now look at the lower plot. Note that the current I(D1) starts increasing as the input V(in) exceeds 1.8V. By the time the input voltage reaches 3V, the current I(D1) is already 200mA, which is SIX TIMES the max. allowed. This would destroy the LED! Note that even if the LED took it, by the time in the input gets to 10V, the LED current is up to 1.3+A!!!

The only reason you didn't vaporize your LED is because on a good day, the 741 can only put out about 25mA, so it is like it has a built-in current limiting resistor. When you measure the voltage across the LED at the output pin of the 741, it shows the 1.8V just like the upper plot.

Clear as Mud?
 

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Mikebits

Well-Known Member
I think this is a tricky concept to explain to someone. I have been wracking my brain to find an analogy.

The problem with simulation files, is they don't smoke :)

The difficulty in explaining an LED Vr drop is that the resistance is not constant with voltage, yet voltage drop remains constant. Arghh
 

Hero999

Banned
(I always thought you connect the resistor to the positive side of the LED, but the calculator showed it on the negative side), but it doesn't change anything.
Why would it change anything?

It doesn't mateer, as long as the resistor is in series with the LED, it will do its job.
 

audioguru

Well-Known Member
Most Helpful Member
I don't think a lousy old 741 opamp works in this circuit if the supply voltage is less than 6.5V. The inputs work only when they are 3v above the negative supply (ground in this circuit) and 3V less than the positive supply.

With a 6.5V supply, a 1.8V LED and a current-limiting resistor selected for 20mA the max output voltage from the opamp is about only 4V.
 

tcmtech

Banned
Most Helpful Member
LED = Zener diode

Nuf Said.
 

smanches

New Member
Research Ohms law and how it applies to voltage dividers. That will give you the basic principle of what's happening.

When you connect a meter to the circuit, you are not measuring the "output" of a device, but how much voltage is being "dropped" between the your positive probe and negative probe. Ohms law teaches that voltage drop is directly proportional to resistance of the circuit when current is constant (as it is in your circuit); the higher the resistance, the more the voltage drop.

When you disconnect the LED, the resistance of the circuit is infinite (open circuit), so therefore the voltage drop is infinite. Since your meter measures voltage drop, you are seeing the entire voltage, 8v. When you connect the LED, the resistance is now at a set value, so when you attach the probes, you see the voltage that is dropped in relation to the resistance value.
 

Hero999

Banned
I don't think a lousy old 741 opamp works in this circuit if the supply voltage is less than 6.5V. The inputs work only when they are 3v above the negative supply (ground in this circuit) and 3V less than the positive supply.
He hasn't said what voltage the power supply is, that was going to be my next question.

What voltage is the op-amp's power supply?

Is it a dual supply or is it single?
 

smanches

New Member
My OpAmp only outputs a grand total of 1.6V, regardless if I feed 3V, 5V or 8V of AA batteries into it.
I removed the LED and I was getting the full 3V, 5V or 8V (whatever Voltage I inputted to the OpAmp) on the output wires with my Multimeter. As soon as I connect the LED, it drops to 1.6V.
Assumed single supply.
 
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