Offset voltage on voltage follower

[Tuurbo46]

Hi
I need a bit of help on a circuit im trying to build. I will try to explain
my problem below.

Circuit requirements:

*input to op-amp = 0 to 5v
*output from op-amp = 0.5v to 3.5v


Im am currently using a CA3140E opamp with a 0.5v offset in it, which is
quoted in the data sheet.

Pin wirring on chip:

Pin 1: to left leg of 10k pot
pin 2: connected to pin 6 with no series resistor
pin 3: varible input voltage, 0 to 5v
pin 4: to middle leg of above 10k pot and connected to GND
pin 5: to right leg of above 10k pot
pin 6: connected to pin 2 with no series resistor, and then this voltage is
dropped across 2 resistors in series to produce the output voltage ratio


Im at the stage in the circuit, that when you increase the input voltage
from o to 5v, the output is alway 0.5v higher than the input which is
correct. At this point i drop the output voltage over a potential divider
and i get the correct output voltage ratio over this. The problem starts
when a 1k load resistor is taken from the potential divider. At this point
the voltage drops. When the load resistor is removed the circuit produces
the correct output voltage again.

I think the problem is the circuit cant provide enough corrent, but i cant
see how to change this.

Any ideas?

Or does anyone have any other circuit ideas?

 

[David L. Jones]

"Tuurbo46" <alex@beale55.fsnet.co.uk> wrote in message news:<cg4prr$19h$1@newsg3.svr.pol.co.uk>...

Hi
I need a bit of help on a circuit im trying to build. I will try to explain
my problem below.

Circuit requirements:

*input to op-amp = 0 to 5v
*output from op-amp = 0.5v to 3.5v


Im am currently using a CA3140E opamp with a 0.5v offset in it, which is
quoted in the data sheet.

Pin wirring on chip:

Pin 1: to left leg of 10k pot
pin 2: connected to pin 6 with no series resistor
pin 3: varible input voltage, 0 to 5v
pin 4: to middle leg of above 10k pot and connected to GND
pin 5: to right leg of above 10k pot
pin 6: connected to pin 2 with no series resistor, and then this voltage is
dropped across 2 resistors in series to produce the output voltage ratio


Im at the stage in the circuit, that when you increase the input voltage
from o to 5v, the output is alway 0.5v higher than the input which is
correct. At this point i drop the output voltage over a potential divider
and i get the correct output voltage ratio over this. The problem starts
when a 1k load resistor is taken from the potential divider. At this point
the voltage drops. When the load resistor is removed the circuit produces
the correct output voltage again.

I think the problem is the circuit cant provide enough corrent, but i cant
see how to change this.

Any ideas?

Or does anyone have any other circuit ideas?

The 1K load is changing the value of (loading down) your voltage
divider. You need an op-amp buffer (unity gain follower) on the output
of the voltage divider to drive your 1K load.

Dave