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Help required in driving capacitive load using opamp?

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hemnath

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Hi,
I have the input signal of 1.8V at 600Khz ramp waveform.

Attached is the circuit used.

I want to replicate input waveform using LM6181 opamp.

When the capacitor value changes at the output of LM6181, the output waveform shape changes. How to avoid this. Please help
 

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betwixt

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It's almost impossible to get an exact replica of the input shape while driving a capacitive load. You could try moving the feedback resistor to the other side of R18 so the changed wave shape is included in the feedback voltage.

Brian.
 

hemnath

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moved the resistor as you suggested. The waveform shape is different. I would like to have the output waveform similar to the input waveform.

Any other help.

Thanks.

- - - Updated - - -

Also Please tell that Shall I use LM6181 to drive capacitive load. In the datasheet it says,

" The amplifier can directly drive upto 100 pF capacitive loads without oscillating and a
± 5 V and ±15 V 10-V signal into a 50-Ω or 75-Ω back-terminated coax cable system over the full industrial temperature
range."
Please confirm that this IC is better to use for capacitive loads? or should I look for another IC.
 

betwixt

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No IC, or any amplifier likes to drive a capacitive load. The combination of capacitance and drive impedance cause phase shifts that can make it unstable, that is why the data sheet spacifies maximum 100pF across the output. The LM6181 is really a cable driver, it is designed to work with a gain of 2 so it can drive a terminated 50 or 75 Ohm cable.

What waveform are you feeding it and what are you seeing at the output?
Are you taking into account the possibility of additional capacitve loading by the circuitry following it?
What are the variable components supposed to do?

Your solution, may be to use the LM6181 as a conventional amplifier and follow it with an emitter follower to isolate the capacitance from the amplifier output pin.

Brian.
 

BradtheRad

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When the ramp wave reaches the end, you have a sudden transition. The op amp must charge (or discharge) your output capacitor very quickly. Therefore the op amp needs to provide high current.

This is the same as having a fast time constant. Very low RC value. In other words your op amp output must be low resistance.

If it is high resistance, then your capacitor charges or discharges more slowly. You lose the sharp peak where the ramp transitions.
 

schmitt trigger

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By "ramp waveform" do you mean sawtooth or triangle?

As Brad mentions, if it were a sawtooth, at that frequency it would require a discharge current that the opamp may not be able to provide. At least not in linear mode.

Assuming an ideal opamp, to discharge 1.8 volt with a 100 ohm limiting resistor requires a peak current of 18 mA. What does the LM6181 datasheet indicate?
 

KlausST

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Hi,

Please show us your input and output signal.
And add comments what you don´t like at the output signal.

Your negative supply has no capacitor. Please add one.
In general for a high precision signal processing you need a carefull layout and parts selection.
Don´t use long wires. Use resistors an capacitors for high frequency (no wirewound resistors...)
Mind that some scope probe (capacitance) may increase the distortion.
Proper ground connection is essential.

Klaus

- - - Updated - - -

Hi,

did you read datasheet:

chapter "7.5 Driving Capacitive Loads"?

Klaus
 

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