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# Op Amps gain

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Ok, if the data sheet for a differential op amp says 100dB, then that means a power gain of 10 000 000 000 right? Holy SMOKES! That can't be right. The way I figure that huge number is 100dB = 10B right? And the way to get the ratio gain from B is to put 10 to the power of the bels, so 10 to the power of 10B = 10 000 000 000. Ok so here's my first question: How do you convert the power gain to voltage gain? Because that's what I'm really interested in here (I don't know why the data sheet would tell me the power gain and not the voltage gain

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Ok, if the data sheet for a differential op amp says 100dB, then that means a power gain of 10 000 000 000 right?

No. The voltage gain of most opamps is 100dB. 60dB is a voltage gain of 1,000 and every 20dB more multiplies the gain by 10 so 100dB is a voltage gain of 100,000. Some very high gain opamps have a voltage gain of 120dB which is 1,000,000.

let me in on the big secret of how to convert power gain to voltage gain

An opamp has an extremely low input power and a low output power so is used as a voltage amplifier not as a power amplifier.

Lets say I had an op amp whose output goes to the gate of a mosfet. And lets say the voltage on the negative input of the opamp is 5V, and the input on the positive lead is 3V, and the voltage gain is 1, that would mean the output of the opamp is -2v

Opamp differential amplifiers have resistors connected to the inputs and as negative feedback from the output make voltage dividers for the opamp's inputs. The output would be -2V if the opamp had the resistors to make it a differential amplifier circuit with a gain of 1.

Would that turn on the mosfet?

Most Mosfets need 10V on their gate to fully turn on. Some "logic level" Mosfets need only 5V. You might be able to find or select (from a bunch of them) a Mosfet that conducts a little with only 2V on its gate.

Or does a mosfet need a positive voltage at its gate?

An N-channel Mosfet needs a positive gate to turn on. A P-channel Mosfet needs a negative gate to turn on.
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Thanks for the help audio, but aren't bels arranged in a logarithmic scale? So wouldn't 60dB (6Bel) be a gain of 1 000 000 (10 to the power of 6)?

You said that most opamps have a gain of around 100dB, resulting in a voltage gain of 100 000. Does that really mean the output of an opamp with it's inputs at 5V and 4V (1V total) will be 100 000V? Seem's a little bit big to me, but what do I know  ;D

One last question, if I were to apply a huge voltage like that (100 000V) to the gate of a MOSFET, would it destroy it? Or would it just be turned on like usual and run fine? Basically my question is, is there an upper limit to the amount of voltage you can apply to a MOSFET?

Thanks again, a lot of the stuff I read about I don't fully understand until one of you guys give me a down to earth explanation of it  :)

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aren't bels arranged in a logarithmic scale?

Yes. +6dB is twice the voltage, +3dB is twice the power.

So wouldn't 60dB (6Bel) be a gain of 1 000 000 (10 to the power of 6)?

Yes, it is power gain, but opamps aren't rated for power gain.

You said that most opamps have a gain of around 100dB, resulting in a voltage gain of 100 000. Does that really mean the output of an opamp with it's inputs at 5V and 4V (1V total) will be 100 000V?

Opamps use negative feedback to reduce the voltage gain to a usable amount, reduce the distortion and increase the bandwidth.
The output voltage of most opamps cannot quite reach the power supply voltage with is usually +18V and -18V max.

if I were to apply a huge voltage like that (100 000V) to the gate of a MOSFET, would it destroy it? Or would it just be turned on like usual and run fine? Basically my question is, is there an upper limit to the amount of voltage you can apply to a MOSFET?

Of course a high voltage will destroy the gate of a Mosfet, just look at a datasheet.
The absolute max gate voltage of most Mosfets is 20V.
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Ok one last thing: I have a picture of a circuit here, and there is an opamp used, but there is no negative feedback to lower the output voltage

Lets say the voltage across the resistor in the phototransistor branch is 1V and the voltage across the phototransistor is 11V when there is no light. That means that at the positive input on the opamp there is 11V right? The negative input on the opamp will always be 1.5V in this circuit. So then there is a 9.5V differential to be amplified. On the 358N opamp there is a gain of 100dB (which is 100, 000 times, just like audio said earlier) resulting in a 950 000V output. That can't be right, but there is no negative feedback to lower the output voltage down. Am I missing something? Or would that MOSFET just be blown to smithereens hahah :D

That is my last question, I promise. Audioguru, you seem a wee bit mad at me, so I'm sorry if I'm asking stupid questions....but at least I'm being decent about it. It's not like I'm saying " Hey Guy5, wh3re ArE teh L33t 5chematICs U arE goin6 to snD [email protected]?!!?11one!?"  ;D

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I have a picture of a circuit here, and there is an opamp used, but there is no negative feedback to lower the output voltage

Negative feedback is used to reduce the voltage gain, not reduce the output voltage.
This circuit would be extremely sensitive to unmodulated light because the opamp's frequency response without negative feedback is flat from DC to only about 2Hz or 3Hz.

On the 358N opamp there is a gain of 100dB (which is 100, 000 times, just like audio said earlier) resulting in a 950 000V output.

Of course not, "The output voltage of most opamps cannot quite reach the power supply voltage".
Just look at the darlington emitter-follower transistor at the output of the LM358 that pulls its unloaded output up to 1.2V less than the power supply voltage which is shown on a graph on its datasheet. Therefore with the 12V supply, the max output of the LM358 is 10.8V.
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:)gee i envey you audio guru your knowledge and great diagnostics skills oh if only we can all be like that

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