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 Author Topic: Hard amplifier problem - Pleease anwer  (Read 1178 times) 0 Members and 1 Guest are viewing this topic.
the land of sun+snow
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 « Reply #15 on: January 03, 2013, 01:37:15 am » Bigger Smaller Reset

Another hint, where it says ...
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The Op-amp Golden Rules
From Horowitz & Hill:
For an op-amp with external feedback
it should be saying ...
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The Op-amp Golden Rules
From Horowitz & Hill:
For an op-amp with external "negative" feedback
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Murphy's Corollary: the "real" problem is usually what they don't tell you about, which leads to endless second-guessing. m

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 « Reply #16 on: January 03, 2013, 02:16:26 am » Bigger Smaller Reset

Simulate it in Circuit Lab.
https://www.circuitlab.com/

Then you'll know the answer... input and output waveform graphs, and voltages even.
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Manchester (England England)
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 « Reply #17 on: January 03, 2013, 02:32:03 pm » Bigger Smaller Reset

The thing you have to remember that there are two different things a circuit can amplify, voltage or current. So the question is exactly how good is the person setting the question. Do they know that fact?
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 « Reply #18 on: January 05, 2013, 05:15:20 pm » Bigger Smaller Reset

I asked this question in another forum ..
Here is one of the replies ...

"""""
The output voltage is:

Vo = Vi / ( 1- 1/A)
where A is the open loop gain of the op amp. so A>>1
you can disregard 1/A then you have.
Vo=Vi
That is a voltage follower. In theory the same as the standard one with negative feedback.

But in real life the output will smack against one rail or the other and stay there. Any difference between the two inputs is amplified and increases the difference rather than reducing it.

""""""

So what he is saying ..
If the Op-amp is ideal .. ( I+ = I = 0 ) and ( V+ = V- ) .. Then , Vin = Vout in this circuit ..

and if it is not ideal ,, the op-amp will be saturated and Vout will be equal to the value of the external power supply ..

What do you think of this???
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 « Reply #19 on: January 05, 2013, 05:46:38 pm » Bigger Smaller Reset

Did you simulate it?  (my link above, CircuitLab)

Yes, Vout=Vin.  Value of resistor doesn't matter. You can even put 0R.

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But in real life the output will smack against one rail or the other and stay there.

That's true for all opamps. -- not just this circuit.

For the positive cycle, Vout cannot be greater than V+ (where V+ is your supply voltage).  So if your Vin >> V+, Vout = V+
(and the converse is also true for V-, if you're using a split power supply).

In real life, Vout max is usually some V+ minus (some voltage value, say 0.7V or something, depends on the opamp).
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 « Reply #20 on: January 05, 2013, 06:04:00 pm » Bigger Smaller Reset

Did you simulate it?  (my link above, CircuitLab)

Yes, Vout=Vin.  Value of resistor doesn't matter. You can even put 0R.

Yes I did ..
But the graph didn't make sense ..
Maybe I made a mistake .. I dunno ..

But are you sure .. If the op-amp is idea , Vin=Vout .. Even with the positive feedback!
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 « Reply #21 on: January 05, 2013, 06:51:00 pm » Bigger Smaller Reset

See screenshot
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 « Reply #22 on: January 05, 2013, 08:10:58 pm » Bigger Smaller Reset

Thaaaanks
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Manchester (England England)
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 « Reply #23 on: January 06, 2013, 12:14:09 pm » Bigger Smaller Reset

Yes but as I mentioned before the input impedance is high and the output impedance is low therefore there is a current gain even if there is no voltage gain.
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 « Reply #24 on: January 06, 2013, 02:26:47 pm » Bigger Smaller Reset

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there is a current gain even if there is no voltage gain.

An opamp by definition is a voltage amplifying device, not a current amplifying device.

Don't confuse more output current drive capability with current gain.

In fact, current output capability of any opamp can be further increased by strapping on discrete NPN/PNP booster circuit at it's output... but it's still a voltage amplifier.
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Manchester (England England)
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 « Reply #25 on: January 07, 2013, 04:31:33 am » Bigger Smaller Reset

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Don't confuse more output current drive capability with current gain.
No?
Cair to explain the difference then?
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 « Reply #26 on: January 07, 2013, 04:50:53 am » Bigger Smaller Reset

For it to be gain there should be a proportional relationship.  Input current of 0 and output current of infinity (ideal) doesn't count...
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Manchester (England England)
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 « Reply #27 on: January 07, 2013, 05:07:36 am » Bigger Smaller Reset

What about an emitter follower, that has a current gain but no voltage gain. Using an op amp to make a voltage follower is the same thing.
Just because the current gain is saturated does not make it any the less of a gain. In exactly the same way as a common emitter amplifier being used as a switch has a saturated voltage gain, no proportionality there but there is still gain.
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 « Reply #28 on: January 07, 2013, 11:55:42 am » Bigger Smaller Reset

This is an opamp. It's by definition a voltage amplifier.
Voltage Delta Gain = Vout / Vin

You don't compute Iout/In in an opamp.

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http://Cair to explain the difference then?
Most opamps have limited output current drive. You can use their outputs to drive another opamp, or a simple high impedance circuit.

But their outputs aren't beefy/strong enough to drive for example a 600 ohm load output transformer... and then several feet of cable behind that transformer's secondary.  They're going to get hot and just break down.

There are special opamps that have more current output drive that can drive these transformer loads. Or you can tack on a discrete transistor booster circuits at the end of the opamp to increase its drive capability.

So now you have a high current output driver, but still operating at unity (Voltage Gain = 1) (depending on resistor values/ratio).

You don't call this beefier opamp as having "current gain"... That's implying there is a proportional relationship between it's output current and the input current you feed into it.

If you look at the datasheet for opamps having high output current capability, you'll find in the datasheet it's IOut (Output Current), example +/-26mA, and the load at which it's tested RL=600ohm, Vs=+/-17V.

If you look at a simple/typical opamp datasheet, you won't find this spec listed.

edit:
Here's a 250mA high-speed, high output current buffer opamp. BUF634
Unity Gain, voltage gain=1. They don't call these opamps "current amplifiers" (or having current gain) just because it has a higher output current drive capability.

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The BUF634 is a high speed unity-gain open-loop
buffer recommended for a wide range of applications.
It can be used inside the feedback loop of op amps to
increase output current, eliminate thermal feedback
and improve capacitive load drive.
For low power applications, the BUF634 operates
on 1.5mA quiescent current with 250mA output,
2000V/µs slew rate and 30MHz bandwidth. Bandwidth can be adjusted from 30MHz to 180MHz by
connecting a resistor between V– and the BW Pin.
 « Last Edit: January 07, 2013, 12:14:06 pm by vasquo » Logged

Manchester (England England)
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 « Reply #29 on: January 07, 2013, 01:18:10 pm » Bigger Smaller Reset

I suggest you go back to school and learn what you are talking about. Most of that is total rubbish.

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You don't compute Iout/In in an opamp.
Why not? You need to know about impedance matching in some circuits and that is about current capability.
The fact that you do not compute something is not the same as saying you can not compute this, it is just saying that you do not know that it is computable.

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You don't call this beefier opamp as having "current gain"
I suggest you look up the term beefier in any electronics text book for a definition.

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If you look at a simple/typical opamp datasheet, you won't find this spec listed.
True do you know why? It is not because this parameter does not exist.

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They don't call these opamps "current amplifiers" (or having current gain) just because it has a higher output current drive capability.
No they don't use that term but that does not stop it being what is actually happening.

I feel you are taking your rather limited education too literally.
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