How Does Output Voltage Work in ICs

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Choosing the Right Power-Supply IC for your Application ...

Supply quiescent current is the current required to bias the internal operation of the OPAMP.

I've pasted part of the data sheet page, you want the source or sink output current, and they are not the same for this amp. Design for the minimum capable, 20 ma source, and 10 ma sink (versus nominal).

The output current is dependent on the load and the analog output of the amplifier if used as that function, if used as a comparator, it will saturate to the V+ or V- terminal minus some offset (this is a function of the internal transistors, and any series output resistor). This offset saturation (rail voltage) is a function of the load. For 2Kohm, the spec sheet indicates 1.5 volt from saturation.

The spec sheet gives you input voltage range of your supply, as maximum amount, this determines your output voltage swing.

How are you using this OPAMP, in linear mode, or as a comparator?

The fault is not in the operational amplifier (LM386). Depending on which package you were using it can deliver at least a quarter watt of power and up to one watt directly into a load. Very few red LED can handle this much power without damage.

...i.e. the LED is kaput.

There wasn't enough voltage going to the operational amplifier surprisingly.

LEDs are not voltage-operated devices. They have a maximum current throughput, the value of which will be found in the datasheet. Ordinarily, an LED requires a series resistance to limit the forward current.

It could be that simple, plugging it into the same socket.

Mikerho, I agree with you, the pins for supply and in and outputs are totally different. Your data sheets should explain some difference.

It's an amplifier, the output voltage depends on the input voltage and gain, assuming you have it wired correctly. An opamp needs feedback, generally a resistor from output to inverting input. If it is properly biased, you can figure the output voltage from the input if you remember:

No current flows into either inverting or noninverting input

Open loop gain is very high so the voltage at the inverting and noninverting inputs is the same

The output voltage will be whatever satisfies the first two conditions, provided that it is between the two rails (plus and minus supply voltage minus some amount).

I can't comment any more on your circuit without a diagram. Red LEDs need about 1.7 volts to turn on and then start drawing current, the more current, the brighter the light output.

Gain on/for all Operational Amplifiers follow the same basic formula:

Gain = Rfb/Rin

The Op-Amp drives to equalize the output with/to the input.

If you need more output power you must adjust the Rfb (Feedback resistor) size to the needed value where the Op-Amp output circuit provides what is needed for the circuit application.

Some Op-Amp devices allow external feedback resistance biasing adjustment and some do not as it totally depends on the packaging.

Reviewing the cut-sheets for the particular OP-Amp in question will (should) answer the questions.

If you are going to saturate the OP-Amp to it's maximum output value for extended time periods it is best to oversize the device so that the peak output demand stays within the 80% range if accuracy is required.

As to the voltage-current terminology; it is an age-old issue of individual preferred semantics due to the fact that current through any stable impedance can only be manipulated by changing the applied voltage (pressure) level.

For example:

If an amplifier drives a stereo speaker with 8 ohms impedance the impedance does not ever change unless the speaker fails open or shorted.

Therefore since E=IR or E=IZ and I=E/R or I= E/Z and Power (P) in watts is P=EI then the only way I can increase is if E (voltage) is increased first.

Current (I) is a function of circuit impedance and applied voltage (E).

It is not possible to increase current in any circuit without increasing the applied voltage or decreasing the circuit load impedance.

As long as the Op-Amp is biased properly and used within it's applied voltage limitations there is very little restrictions on the application it can be used for.

TRIVIA: Many of the old tube-type guitar amplifiers utilize a "push-pull" design that applies as much as 500 volts to the speaker output terminals which can be lethal to humans and animals.

Hope this helps. Good luck!

While in general your theory stands for linear loads, taking the speaker example is not the best. While music in general is a mix of frequencies, your speaker "could" act partly as "linear", however when the frequency doubles, the impedance Z doubles also.

Z=ωL were Z = impedance, and ω= 2.π (pi) f (frequency)

(This is why speakers use more power, the lower your frequency is, and where the amp starts "clipping" the first. When clipping, the speaker "sees" almost a DC voltage appear over its coil and then the resistance R comes to play - and R is just the resistance of the coil wire)

TRIVIA:

The 500 volts "can eventually" appear when the amp is on, and initiated with the guitar, or oscillates and only to the speaker terminals, when the speaker is not connected or defective.

Push pull tube amps use to have a end transformer that matches the output impedance with the speaker impedance. (= a ohms value, commonly in reference to a 1000 Hz sinusoidal signal)

With old tube amplifiers, sometimes the end transformer was "unified" with the speaker. Perhaps this leads your thoughts?

Yes. Excellent information! Thanks for the clarification of the frequency affect on speaker impedance.

LM386 is NOT an operational amplifier, it is a mono power amplifier with fixed (but modifiable) voltage gain of 20 (~26 db). So the small gain IS a problem even if you compensated for the different pinout and used two LM386's,(since LM358 is a dual Op Amp) Try adding a resistor between pins 1 and 8 to increase gain and see if it gets any better. S.M.

Why are you using an op amp to drive your led? What is the supply voltage and LED VF and IF? It sounds like your application could be solved much easier by creating a simple buck/boost SMPS than using an op amp. The buck boost controller would control the constant current to the LED and you would set the voltage with conditioning resistors.

LM386 outputs Vcc/2 if the input pin is grounded. It can drive the LED but you cannot get rid of the Vcc/2 volts. The IC can drive a 1/4 W into 8 ohms speaker. If you calculate the current, it is 31mA, good enough for an LED which mostly needs 20mA.

Gajanan Phadte