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Slew rate and bandwidth distortion

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Junus2012

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Hello friends,

I have designed an operational amplifier with slew rate equal to 40 V/uS and GBW equal to 30 MHz,

but when I connect it as buffer to test the linearity with large input sinusoidal signal I am having distortion starting from 5 MHz. then what is the meaning of having high slew rate and GBW in my design.

Thank you in advance
 

but when I connect it as buffer to test the linearity with large input sinusoidal signal I am having distortion starting from 5 MHz. then what is the meaning of having high slew rate and GBW in my design.
Click to expand...

40V/µs isn't actually high slew rate.

Simple calculation: 30 MHz 1 Vpk sine has maximal slew rate of 2*pi*30e6*1 = 189 V/µs
 
Well neither 40V/uS nor 30Mhz is particularly fast.

Understand that as you approach the bandwidth of the amplifier it's open loop gain approaches zero - in other words it becomes a really bad opamp. Slew rate is simpler: you can easily calculate or simulate whether a given value is enough for a given signal.

And note that while these parameters relate they are distinct. For example LT1357 has less bandwidth (25Mhz) but 15X the slew (600V/uS) of your example opamp.

Bottom line I see you needing about 10X more bandwidth and 10X more slew to reproduce a large 5mhz signal with reasonable accuracy. Still not too hard to find unless you have other difficult requirements.
 
FvM said:
40V/µs isn't actually high slew rate.

Simple calculation: 30 MHz 1 Vpk sine has maximal slew rate of 2*pi*30e6*1 = 189 V/µs
Click to expand...

Thank you FvM for your answer, it means if I dont want to distore the signal I should design the amplifier with slew rate equal to 189 V/uS at least,,,

only please I need the source of this calculation so I can discuss it with my professor.

- - - Updated - - -

asdf44 said:
Well neither 40V/uS nor 30Mhz is particularly fast.

Understand that as you approach the bandwidth of the amplifier it's open loop gain approaches zero - in other words it becomes a really bad opamp. Slew rate is simpler: you can easily calculate or simulate whether a given value is enough for a given signal.

And note that while these parameters relate they are distinct. For example LT1357 has less bandwidth (25Mhz) but 15X the slew (600V/uS) of your example opamp.

Bottom line I see you needing about 10X more bandwidth and 10X more slew to reproduce a large 5mhz signal with reasonable accuracy. Still not too hard to find unless you have other difficult requirements.
Click to expand...

it means if my amplifier has 30 MHz bandwidth doesnt mean I can use the full range as the gain will drop,,,, so what is the meaning of having this value in our design if we cant use it .....is there a relationship between GBW and slew rate ??
 

Hi,

only please I need the source of this calculation so I can discuss it with my professor.
Click to expand...
Mathematically:The rise rate is just the first derivation of the signal.

But if you use a shhet of paper and a pencil and draw the signal in a diagram with time in us at the X axis and Voltage in V at the Y axis you could simply explore it on your own.

I should design the amplifier with slew rate equal to 189 V/uS at least,,,
Click to expand...
189V/us is for 1V amplitude pure sinewave. The higher the amplitude, the higher the slew rate.
Other waveforms have different rise rates.

Klaus
 

Junus2012 said:
Thank you FvM for your answer, it means if I dont want to distore the signal I should design the amplifier with slew rate equal to 189 V/uS at least,,,

only please I need the source of this calculation so I can discuss it with my professor.

- - - Updated - - -



it means if my amplifier has 30 MHz bandwidth doesnt mean I can use the full range as the gain will drop,,,, so what is the meaning of having this value in our design if we cant use it .....is there a relationship between GBW and slew rate ??
Click to expand...

This isn't a black and white.

You can use it and you are using it there just isn't enough of it. Think of gain bandwidth as a quantity or a resource. The trouble is when you only have 30mhz of it your 5mhz signal uses most of it up. Refer to the the open loop gain plot in your opamp datasheet and you'll see the open loop gain is quite low at 5mhz. An ideal opamp needs very high gain, thus your opamp is very poor when dealing with a 5mhz signal.
 

Thank you all guys for your active answer,,,,

it made me more understand,, thank you a lot
 

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