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##### Junior Member level 1
Hello,

How to adjsut gain in required range minimum and maximum i.e 0.6v to 3.2v equals to 2.8V to 4.8V?
This range is required for transconductance of the Mosfet.

1) Your gain needs to be (4.8-2.8)/(3.2-.6)=.77
2) Your offset needs to be 2.8-(.6*.77)=2.332

Vo=2.332+Vin*.77

Thank you!
how to adjust gain less than 1?Can i get any reference circuit?

Hi,

how to adjust gain less than 1?Can i get any reference circuit?

Use two resistors as voltage divider.

Klaus

Use two resistors as voltage divider.

But i need to get the required range. Voltage divider can't give the required range with the change in input of op-amp. i mean i suppose to get a gain 0.33 and output with 2.9VDC to 4.8VDC with the change in input 0.6VDC to 3.2VDC.

Thanks.

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Use two resistors as voltage divider.

But i need to get the required range. Voltage divider can't give the required range with the change in input of op-amp. i mean i suppose to get a gain 0.33 and output with 2.9VDC to 4.8VDC with the change in input 0.6VDC to 3.2VDC.

Thanks.
Use the voltage divider to drop the variance between lower and upper voltages then use the op-amp to offset the output by a fixed 2.332V. Basically, an attenuator followed by an analog adder.

Brian.

Hi,

Using an Opamp as buffer or not...

The equation is: Vout = Vin × 1.9 / 2.6 + 2.4615

Two resistors could do the job. Input --> 368.4R --> output --> 1000R --> 9.1437V

Even with an Opamp you need a voltage equal or higher 9.14V. It should be fix and stable. Is this available?
If not, what other fix and stable voltages are available? Any VRef?

Klaus

Hi,

Using an Opamp as buffer or not...

The equation is: Vout = Vin × 1.9 / 2.6 + 2.4615

Two resistors could do the job. Input --> 368.4R --> output --> 1000R --> 9.1437V

Even with an Opamp you need a voltage equal or higher 9.14V. It should be fix and stable. Is this available?
If not, what other fix and stable voltages are available? Any VRef?

Klaus

You most certainly do not need 9.14V. One solution:

The 2.5V reference must be a low-impedance source. There are other solutions, of course.

Points: 2
Hi barry,

i have tried the above circuit with 2.5 reference voltage, required output is adjusted in a range.

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