Op Amp Circuit Dilemmas: Diodes, Current, and Resistors Explained

[Faiq]

Q1. Why do we connect the Diode 2 in the Op Amp- Relay Circuits? (Check the image )
Q2. Why shouldn't a high current pass through an Op Amp?
Q3. Why should there be a resistor in between an LED and an Op Amp output?

 

[Rive]

Q1: that 'freewheeling' diode belongs to the relay, not related with the opamp. You do that to every relay, to handle the voltage spike when you switch off the relay.
Q2: the load (current, voltage) on the opamp output should be lower than the maximal allowable. The 'high' here depends on the type of the opamp.
Q3: that resistor belongs to the LED, not to the opamp. The current on the LED should be limited, according to the LED type & opamp output voltage.

 

[Faiq]

Q1: that 'freewheeling' diode belongs to the relay, not related with the opamp. You do that to every relay, to handle the voltage spike when you switch off the relay.
Q2: the load (current, voltage) on the opamp output should be lower than the maximal allowable. The 'high' here depends on the type of the opamp.
Q3: that resistor belongs to the LED, not to the opamp. The current on the LED should be limited, according to the LED type & opamp output voltage.

Q2. Why is there a maximal allowable value?
Q3. My book stated that large currents through LED can damage both Op Amp (Because it is connected in series with LED) and the LED? Is that wrong?

 

[jim hardy]

Q2. Why is there a maximal allowable value?
Q3. My book stated that large currents through LED can damage both Op Amp (Because it is connected in series with LED) and the LED? Is that wrong?

How about if you post an opamp datasheet so we can show you where in that datasheet are the answers to your questions?

eg

 

[Rive]

Q2. Why is there a maximal allowable value?
Q3. My book stated that large currents through LED can damage both Op Amp (Because it is connected in series with LED) and the LED? Is that wrong?

Q2: It's what the actual device can/guaranteed to bear at most. The values are available in the device datasheet, or more likely: you will chose devices, which fits your design requirements.
Q3: It's true. However, typically it's the LED which is more sensitive to current, at least while it's still a 'LED' and not a dysfunctional piece of short circuit - which would bring down the opamp too if the limiting resistor is not present.

 

[Faiq]

How about if you post an opamp datasheet so we can show you where in that datasheet are the answers to your questions?

eg

View attachment 113669

I don't have a datasheet. I am simply making questions regarding the statements made in a A Levels book.

Sentence 1: "...Once the output of an Op Amp is higher than an acceptable value, the current in the LED and the op amp will be very high and this will damage both of them."In your example, does the term short circuit can also be replaced with high current?

 

[jim hardy]

n your example, does the term short circuit can also be replaced with high current?

yes.

I don't have a datasheet. I am simply making questions regarding the statements made in a A Levels book.

http://www.ti.com/lit/an/sloa011/sloa011.pdf
http://www.physics.unlv.edu/~bill/PHYS483/op_amp_datasheet.pdf

http://www.ece.usu.edu/ece_store/spec/LM324.pdf

 

[richard lareva]

Not sure of your basic electronic knowledge, but there is Ohm's Law and the same law for diodes. For diodes, you want a resistor in series with the power source that will drop all the supplied voltage with the exception of the forward voltage drop of the diode. Since you do not have the data sheet for these parts, you can assume that the LED has about a 2 volt forward bias drop. Subtract this voltage from your supply voltage, assume that the maximum safe forward biased current through the LED is about 20 milliamps, then solve for the value of "R" using the formula R=E/I, "E" being voltage and "I" being current. Any more questions feel free to ask.

 

[jim hardy]

I am simply making questions regarding the statements made in a A Levels book.

I shouldn't have been short.

Since you do not have the data sheet for these parts, you can assume that the LED has about a 2 volt forward bias drop. Subtract this voltage from your supply voltage, assume that the maximum safe forward biased current through the LED is about 20 milliamps, then solve for the value of "R" using the formula R=E/I, "E" being voltage and "I" being current. Any more questions feel free to ask.

As rl says, we're unsure of your familiarity with basics. And we don't know how good are explanations in your A level book.

Q3. Why should there be a resistor in between an LED and an Op Amp output?

Have you yet learned "ideal voltage source" ? It can deliver infinite current.
Of course there's no such thing in the real world, but we do use ideal parts in our imaginary thought experiments.

So as rl explained , an ideal voltage source will deliver whatever current is necessary to satisfy its desired terminal voltage. That current might be quite large.
While that large current won't hurt an ideal part it sure could wreck a real one.
The opamp will do its best to mimic an ideal voltage source .
As a circuit designer you don't want your opamp to die trying, so you include in your circuit enough resistance to cause just the right current for just the right voltage. That protects both the opamp and the LED.

It's that simple.

old jim

 

[Faiq]

I shouldn't have been short.
As rl says, we're unsure of your familiarity with basics. And we don't know how good are explanations in your A level book.

Have you yet learned "ideal voltage source" ? It can deliver infinite current.
Of course there's no such thing in the real world, but we do use ideal parts in our imaginary thought experiments.

So as rl explained , an ideal voltage source will deliver whatever current is necessary to satisfy its desired terminal voltage. That current might be quite large.
While that large current won't hurt an ideal part it sure could wreck a real one.
The opamp will do its best to mimic an ideal voltage source .
As a circuit designer you don't want your opamp to die trying, so you include in your circuit enough resistance to cause just the right current for just the right voltage. That protects both the opamp and the LED.

It's that simple.

old jim

Thank you very much for your perfect explanation. Please see if I understood this correctly. An amplifier has a zero output impendence, so the current coming out of an amplifier is very large. The large current can destroy an LED if a limiting resistor is not present.
Apologies for bothering you this much but I just have one last question. A very low input current is fed to an amplifier and a very high current is obtained. Does this not violate Kirchoff's Law of Charge conservation?

 

[phinds]

An amplifier has a zero output impendence

No, it has a very low output impedance if it is any good, but not zero

so the current coming out of an amplifier is very large.

That depends on the load and in any case is limited to the capability of the power source.

The large current can destroy an LED if a limiting resistor is not present.

Yes, BUT only if it is created by a large voltage. Amplifiers normally amplify voltage, not current.

A very low input current is fed to an amplifier and a very high current is obtained. Does this not violate Kirchoff's Law of Charge conservation?

No, because the extra current is not being created out of thin air, it is coming from the power supply.

 

[Faiq]

No, because the extra current is not being created out of thin air, it is coming from the power supply.

If the Power supply is providing a current 1A, does this mean the highest value attainable by the current at output of amplifier is 1A?

 

[phinds]

If the Power supply is providing a current 1A, does this mean the highest value attainable by the current at output of amplifier is 1A?

Yes, but the way you phrased your question still implies some confusion. The power supply doesn't supply 1 amp unless it is needed, so the proper way to think of it is that the power supply is CAPABLE of supplying 1 amp, in which case the output of the amplifier is limited to 1 amp. And, really, the rest of the amplifier circuit will take some current to operate, so the output would be limited to something less than 1 amp.

 

[Faiq]

Yes, but the way you phrased your question still implies some confusion. The power supply doesn't supply 1 amp unless it is needed, so the proper way to think of it is that the power supply is CAPABLE of supplying 1 amp, in which case the output of the amplifier is limited to 1 amp. And, really, the rest of the amplifier circuit will take some current to operate, so the output would be limited to something less than 1 amp.

Ah so technically the power supply can supply any current, it just depend on the resistance of the whole circuit.

 

[phinds]

Ah so technically the power supply can supply any current, it just depend on the resistance of the whole circuit.

NO. The power supply can supply whatever current it is designed for. Within that limit, it can supply any amount from zero up to that amount. If the limit is 1 amp then that's all it can supply.

 

[Faiq]

NO. The power supply can supply whatever current it is designed for. Within that limit, it can supply any amount from zero up to that amount. If the limit is 1 amp then that's all it can supply.

Oh yeah right. Thank you very much for clarifying this.

 

[phinds]

Oh yeah right. Thank you very much for clarifying this.

Oh yeah right. Thank you very much for clarifying this.

You seem to have a stuck "post" button. I just saw that you posted another thread about 5 times. You should check on that.

 

[Faiq]

Yeah I notified the moderator, he fixed the problem.

 

[phinds]

Yeah I notified the moderator, he fixed the problem.

But you did it again in this thread, is why I was pointing it out.

 

[Faiq]

Okay I wll restart the browser. Maybe that will solve it