Why doesn't the led glow?  Bread Board Circuit

I am trying to glow my LED For Making Project Of Detectore but it doesn't glow.

When i connect this circuit with my full project that time LED Wasn't Glowing After I Disconnected From Main Circuit and Make A Small Circuit For Testing Detector Circuit!!

I Do Small Circuit As Is Shown In Image.

I am using these components and my circuit connection as per image:


10k resistor

9V battery.


closed as unclear what you're asking by Chris Stratton, PeterJ, Rev1.0, Finbarr, Elliot Alderson Dec 24 '18 at 14:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ That's the supply. Question was about the inputs. I can see what you did, but I want you to think about it, and think about if what you are doing is reasonable. And even if it were reasonable (which it isn't) why do you think doing that would cause the LED to light? If it does, it's only because you're leveraging ways in which a simple model breaks down in practice. \$\endgroup\$ – Chris Stratton Dec 23 '18 at 6:55
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    \$\begingroup\$ Which says nothing about why you expect this out-of-bounds circuit to light your LED. What, in your own thinking, do you imagine is the theory of operation that would lead the LED to light? \$\endgroup\$ – Chris Stratton Dec 23 '18 at 6:59
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    \$\begingroup\$ @Nemish Actually Chris is helping to improve the quality of questions and answers, and probably didn't downvote you, so please chill a bit. \$\endgroup\$ – Spehro Pefhany Dec 23 '18 at 7:22
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    \$\begingroup\$ @Nemish, please don't be rude to those trying to help you. You have no schematic in your question so it is difficult to see what you are trying to make and how you expect it to work. That makes your question a poor one. I suggest you read each comment carefully and edit your question to clarify each point raised. \$\endgroup\$ – Transistor Dec 23 '18 at 8:53
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    \$\begingroup\$ Please post a schematic next time instead of (only) making a picture of the breadboard. \$\endgroup\$ – Rev1.0 Dec 23 '18 at 10:36

The LED should go on, at least for a while, if this is connected sensibly. I have the following comments:

  1. There is a schematic creator button, please use that to communicate circuits, though the photo is useful in this case too.

  2. You should have series resistor for the LED. The short-circuit output current of the op-amp is enough to possibly damage an LED, but it probably won't happen instantly.

  3. You appear to be trying to stuff stranded wire into a solderless breadboard. Maybe one or more connections are not being made.

  4. The LED may be backwards (usually the long pin would be towards pin 1 of the dual op-amp), or perhaps your battery is dead.

  5. There are some other comments about common mode range of the inputs, however the LM358 will give a proper output if at least one input is within the common-mode range (which is between about 0V and supply voltage minus 2 or 3V). When it is working normally we would expect BOTH inputs to be within that range.

Edit: As per @ChrisStratton, if BOTH inputs are outside the common mode range, then the output will be low and the LED will not light. That is what you have. So you need to bring at least one input within the common mode range.

Edit: Consider the schematic of one of the amplifiers of the LM358:

enter image description here

In your case you have the (+) input tied to V+ through a 10K resistor, so Q3 and Q4 will be "off" and all the 6uA bias current will flow through Q8, meaning the collector of Q9 is almost at ground potential, Q10 drags the base of Q11 down, turning off Q12, the collector is pulled up by the 100uA source and the output rails high.

However if BOTH inputs are close to the V+ supply, the 6uA current source will no longer work and the output will go low. That is your current situation.

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    \$\begingroup\$ LM358 can't source enough current to damage a LED, in the photo that resistor is in the +ve rail, \$\endgroup\$ – Jasen Dec 23 '18 at 6:58
  • \$\begingroup\$ If you expect this to work with the non-inverting input tied to the + supply and the inverting input tied the + supply through a 10k resistor, the explanation of what implementation details in this particular type of op-amp cause that would make interesting reading. Voltage drop in the resistor from input current? \$\endgroup\$ – Chris Stratton Dec 23 '18 at 7:02
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    \$\begingroup\$ @Jasen Actually the datasheet says 40mA can be considered the maximum source current, which will damage many 5mm LEDs if left on continuously. See note 6, page 6 \$\endgroup\$ – Spehro Pefhany Dec 23 '18 at 7:02
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    \$\begingroup\$ short circuit current: yeah i sit corrected. \$\endgroup\$ – Jasen Dec 23 '18 at 7:11
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    \$\begingroup\$ So does your LED light if you duplicate this circuit, with both inputs connected to the 9v supply, inverting one via a 10K resistor? \$\endgroup\$ – Chris Stratton Dec 23 '18 at 7:14

LM358 cannot work correctly such a high voltage on the inputs - input voltage must be at-least 2V lower than supply positive.

http://www.ti.com/lit/ds/symlink/lm358.pdf table 7.3 "VCM"

Also note that the inputs to this amplifier are current sources, so were the voltages in range (eg connect the inputs to ground instead) the one with the 10K resistor would see a higher voltage than the other.

Also as Spehro Pefhany notes, a resistor should be used to limit the current into the LED 1K would be good.


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