Op amp confusion

I'm using a lf442cn

2 tied to ground for now

a 330pf cap between 3 and 1 before resistors

pin 2 tested against ground = 0v pin 1 tested against ground = 3.97v ?? shouldn't this be 0???

basically this is a buffering circuit

also get the same results with lf411 single supply amp, but as i can only supply a 5v/gnd to the op amp what would be best?

Thanks

Is this saying you have a 100 ohm resistor between 3 and 1? If so, this is positive feedback.

What is "before resistors"? Is it in series

What is "after resistors"?

A buffer would have the input connected to pin 3 (+ input) and pin 1 connected to pin 2 (output to - input)

The LF411 and LF412 have an input voltage range from about the positive supply voltage (pin 8) to about 3 volts more positive than the negative supply voltage (on pin 4). With a single 5 volt supply this leaves you an input range from about 3 to 5.

They have an output swing range from about 1.5 volts below the positive supply voltage to about 1.5 volts above the negative supply voltage. With a 5 volt supply, this includes about 1.5 to 3.5 volts.

but if connected as a buffer, both the input range and output range must cover the signal, so this amplifier will only operate for inputs between about 3 and 3.5 volts. Blah!

Get an LM358 (dual) or LM324 (quad). Both input and output ranges cover about 1.5 volts below the positive supply to the negative supply, or 0 to 3.5 volts.

If you need more signal range than this, you need what is called a rail-to-rail type whose input and output can operate from the negative to the positive supply voltage. An example of this would be an LMC6482, but there are lots of others.

The LF442 is a slightly later, slightly different version of the LF412. It has slightly lower bandwidth, and half the power supply current. They're part of the same useful family of parts, and the same rules apply. None of the LF411 JFET-input family is suitable for single supply operation. A hearty seconding on all of Mr. Popelish's suggestions.

By the way, if you want to describe a simple circuit like this (view in fixed font or M$ Notepad):

___ .---|___|---. | 1K | | VCC | | 2|\\|8 | '---|-\\ 1 | Vout Vin ___ | >----o---o o---|___|---|+/ 1K 3|/|4 | === GND (created by AACircuit v1.28.5 beta 02/06/05

you might want to try downloading the freeware/beerware (if you ever see Andy, you're required to buy him a beer) Andy's ASCII Circuits at the above site. It's a lot easier than listing nodes or pins, and allows newsgroup readers to easily visualize your problem.

Once you get your choice of op amp taken care of, you might want to look closely at the above op amp voltage follower (buffer) circuit. I think this might be a better way to get a voltage follower. The diagram uses the standard dual op amp pinouts (like the LF442, LM358, &c.) The other end of the feedback resistor from pin 1 (output) should be tied to pin 2 (inverting input). Pin 2 should not be grounded. That would cause the output to rail positive for any op amp with any Vin above GND.

Also, do yourself a favor, and read the data sheet. It's a free education in itself.

Good luck Chris

Someone will correct me if I am wrong, but I see an integrator with no input conditioning or noise compensation. You said you wanted a buffer? Take a look at this:

...and rethink your design.

I believe that if you want that circuit to be a voltage follower, then the feedback path should be a short. The circuit you have printed has a gain of 2. With Rf shorted, the gain will be 1.

|\\ -----|+\\ | \\-+-/\\/\\-+-- | / | | +-|-/ === | | |/ === | | | | |______+_/\\/\\-+

is basically how it's running, threw in an 7812/7912 all is fine but will probably run a 7660 at +5/-5 v i really would of liked not to need a -v i did not realise about the limitations on running 5v/gnd but now that i think about it i think i get it..

Thanks guys

Chris wrote:

Hi, Al. The OP said "basically this is a buffering circuit". I was going from his description of his circuit to try and determine what he had.

If you'll look carefully at my ASCII diagram above, you'll see it is a voltage follower with a gain of 1. Since op amp bias current is usually much greater than offset current, many will use identical input resistors to cancel out the voltage drops due to the former. I guess I kind of assumed that was what he was doing from his description. Of course, with a 50pA bias current JFET input, it's kind of a waste of resistors.

But, in the diagram above, since there's no feedback to the input terminal, and no gain divider on the negative feedback, the gain is 1.

Thanks for taking the time to respond. It's always good to know someone's checking. If bad information were to ever get out on the internets, we'd all be done for. ;-)

Good luck Chris

sorry voltage follower

Yes, you're right, the non-inverting feedback default gain of one is if the 1K input resistor is tied to return as part of the feedback network. Oops!

is one of these methods better than the other, would one produce a cleaner signal?

.-----------. | 1K | | | | |\\ | '---|-\\ | Vout Vin | >----o---o o------|+/ |/

___ .---|___|---. | 1K | | | | |\\ | '---|-\\ | Vout Vin ___ | >----o---o o---|___|---|+/ 1K |/

Vin |\\ o----|+\\ Vout | \\-o-/\\/\\---o---o | / | 100 | o-|-/ === | | |/ === 330pf | | | | |______o_/\\/\\---o 100

supply +5v/-5v

Kitchen Man wrote:

Simple and clean, but if there is any input bias current, and the input signal comes from a high resistance source, the bias current will produce a voltage going through that resistance that will be added to the input signal.

This version attempts to cancel the above mentioned bias current voltage drop by causing an equal one across the feedback resistor. But you don't actually put another resistor in series with the input. The feedback resistor is supposed to be the same as the resistance that is unavoidably in the signal source. The only reason to add to that would be if you wanted to include a low pass filter into the design, or to add over voltage clamping diodes at the + input.

This variation is used if the output is intended to drive a large amount of capacitance, which tends to make the above versions unstable.