Understanding OP-AMP parameters

[Bjtpower]

Hello Friends,

I want to understand the curve shown in the OP-AMP Datasheet..

1) Open Loop frequency responce
2) Large signal frequency responce
3) GBWP VS Temperature
4) Output voltage vs Output current..??

What this graph shows..

Does it is limitation of X cordinates with the Y-Cordinates

I am thinking about the BasiC LM358 Operational amplifiers

Regards
Marx

 

[KlausST]

Hi,

1) open loop means there is no feedback (resistor). The chart at 1kHz says about 60dB. This means you may expect a gain of 60dB = 1000V/V of gain. Input 1mV RMS sine and expect about 1000mV RMS sine at the output.

2) now this is with feedback. Assume your input signal is 100kHz sine.
Now go to the chart and see that the max output swing is limited to about 1Vpp at 100kHz.
With this Opamp you can not have an output of 10Vpp because of it's limited slew rate. Not at gain=1 and not at gain= 100.

3) I don't see this in my TI LMx58-N datasheet. Gain of internal transistors depends on bias current. I assume this bias current drifts with temperature, then open loop gain also drifts with temperature.

4) the output current capability is limited (usually by the base current and gain of the output transistor). So with lowcurrent you may expect a high output voltage range (low voltage drop). Mind that the diagrams show the voltage drop.

Klaus

 

[Bjtpower]

What is this DB..??
10000V/1V.. AND
How it can be expressed in DB.. Why..??

Marx

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Then again What is the Common mode rejection ratio..

It is shown that CMRR is 80db..

I understand the DB Concept..
20log(1000v/1v)=60db

What is CMRR..??

It should be high or low..??

 

[FvM]

If you don't mind to read a bit more about OPs https://www.analog.com/en/education/education-library/op-amp-applications-handbook.html

Expect most terms to be explained in Section 1: Op Amp Basics.

 

[Bjtpower]

I want to understand the Integartor and Differentiators..

How it can be elaborated in the Following examples.
Vout=(-1/(RinxC))XINT(Vinxdt) (over time 0 to T)

What will be the T..??

How equation will look like in sine wave or Square wave having frequency of 100HZ.


Regards
Marx

 

[KlausST]

Hi,

Please post a circuit. Then we can discuss about it.

Klaus

 

[Bjtpower]

Here is my dought,

i want to understand how integration output comes.. Mathematical approch




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Here is my dought,

i want to understand how integration output comes.. Mathematical approch

Vin is step waveform shown

 

[KlausST]

Hi,

Two issue:
Your negative Opamp supply is GND. And your noninverting input is GND, too.
--> you need an Opamp that can handle this.

The integrator is inverting. So with positive input, the output is travelling towards the negative supply (GND).
So the Opamp output will saturate near GND, now way for it to go negative.
Therefore one needs to assume that at t=0 the output of the Opamp is well in the positve area.

*******
Now that +In of the Opamp is GND .... then -In is GND, too. (regulated by the Opamp)
This means the complete integrator_input_voltage will be across the resistor.
Therefore you may easily calculate integrator_input_current.
Now that the current at the Opamp_-In is considered to be zero....al the current must flow through the capacitor.
C = I × t / V ----> with a constant input voltage you get a constant input current and this causes a constant output rise/fall_rate in V/s.
Therefore rearrange the capacitor formula this way:
V / t = I / C ....where I = V_input /R, the result is in V/s

V / t = V / ( C × R )
Let's assume you have 3V integrator_input_voltage. Then..
V / t = 3V / (1uF × 47k) = 63.83V/s ... or 63.83mV/ ms

The output voltage of an integrateor at t = 0 is undefined, but let's assume it is 4.0V
Then :
t = 0 --> Vout = 4.0V
t = 1ms --> Vout = 4.0V - 63.83 mV = 3.936 V
t = 2ms --> Vout = 3.936V - 63.83mV = 3.872V
...and so on.. until it saturates.


With zero input voltage .... the rise rate is zero, means a flat horizontal line. The voltage freezes.

*****
Now you asked about sine wave input.

It is similarely simple, but I use another approach.
For an inverting amplifier:
V_out = V_in × R_fb / R_in
R_fb = feedback resistor. In your case the capacitor with it's Xc.

So V_out = V_in × Xc / R_in.
In your case with 3Vpp, 100Hz sine wave input:
Xc at 100Hz = 1 / ( 2 × Pi × 100Hz × 1uF) = 1592 Ohms.
V_out = 3Vpp x 1592 Ohms / 47k = 0.102Vpp

Klaus

 

[Bjtpower]

Not Satiesfied..

can you solve in integration wise as formula states..

 

[KlausST]

Hi,

it´s a long time ago....
But maybe like this:


Klaus

 

[FvM]

Not Satiesfied..

can you solve in integration wise as formula states..

Not sure what your problem is. Calculating the definite integral is explained in any math handbook.

Circuit design wise you need to provide a means to set the initial state, e.g. a reset switch to set V_init = 0. And a supply voltage according to the expectable output swing, e.g. a negative supply to integrate a positive input voltage with an inverting integrator.