Can anyone suggest an alternative replacement for a LM7705? part or circuit?

Thanks

 

I see almost no one has stock. I believe these people do:
http://www.sierraic.com/LM7705

 

Yeah that's the problem no stock till late 2022.

 

Did you check my link? They claim to have them in stock or at least available.

 

Can anyone suggest an alternative replacement for a LM7705? part or circuit?

Thanks

That would depend on what your circuit does. How about posting a diagram?

 

Did you check my link? They claim to have them in stock or at least available.

Yeah they show stock but i checked with them and they are out of stock on all types.

 

That would depend on what your circuit does. How about posting a diagram?

It's pretty simple, using a ADA4091-4 op amp with a single 5v+ supply the V- terminal is biased to a -0.232 voltage so the negative rail can reach 0 volts. See the attached data sheet for a description.

Thanks for any help!

 

Oh and the neg has to be limited to < 0.3 volts because of my microcontroller analog input pin limitation.

 

Have you thought of any ways to avoid the need for a negative supply altogether? That would be an advantage, because it eliminates the hard-to-get IC and it avoids any potential damage to your analogue input and it avoids any possible problems with noise from the LM7705.
DO you have a spare A/D input?
If so, you could make a reference for the analogue circuitry at, say, 1.2V (because there are plenty of 1.2V references available), or even just a diode. It doesn't even matter if it drifts with temperature. Then connect the reference to one A/D the signal to another and subtract. You might even have differential A/D inputs. Where do the inputs to the op-amp come from?

 

Have you thought of any ways to avoid the need for a negative supply altogether? That would be an advantage, because it eliminates the hard-to-get IC and it avoids any potential damage to your analogue input and it avoids any possible problems with noise from the LM7705.
DO you have a spare A/D input?
If so, you could make a reference for the analogue circuitry at, say, 1.2V (because there are plenty of 1.2V references available), or even just a diode. It doesn't even matter if it drifts with temperature. Then connect the reference to one A/D the signal to another and subtract. You might even have differential A/D inputs. Where do the inputs to the op-amp come from?

Are you suggesting making a differential input using two ADC Inputs? I'm not sure on how this will help. The circuit is designed to take a 4-20ma signal, drop it across a 250Ω resistor giving 1-5 volts. This signal goes to the opamp + input. A 1v ref is input on the - input. The gain on the opamp results in a 0-5 volt input to the ADC. In order to reach "0" rail on the opamp a small negative bias is used for the negative power rail of the opamp.

 

Are you suggesting making a differential input using two ADC Inputs? I'm not sure on how this will help. The circuit is designed to take a 4-20ma signal, drop it across a 250Ω resistor giving 1-5 volts. This signal goes to the opamp + input. A 1v ref is input on the - input. The gain on the opamp results in a 0-5 volt input to the ADC. In order to reach "0" rail on the opamp a small negative bias is used for the negative power rail of the opamp.

So why do you need the op-amp? Connect one end of the 250Ω resistor to ground, and connect the other end to the A/D input.
250Ω is plenty low enough to drive an A/D input.
If it's a 12-bit A/D subtract 819 from the A/D reading. If you need the full 0-4096 output, multiply by 5 and shift two places right. You will lose some resolution, but it's less than 1 bit.
(If your processor doesn't multiply, save the reading, shift two places left, add the saved reading and then shift two places right)

 

So why do you need the op-amp? Connect one end of the 250Ω resistor to ground, and connect the other end to the A/D input.
250Ω is plenty low enough to drive an A/D input.
If it's a 12-bit A/D subtract 819 from the A/D reading. If you need the full 0-4096 output, multiply by 5 and shift two places right. You will lose some resolution, but it's less than 1 bit.
(If your processor doesn't multiply, save the reading, shift two places left, add the saved reading and then shift two places right)

Because i need an op amp and you should always have a buffer before a A/D converter. I just looking for an answer to the neg biasing

 

you should always have a buffer before a A/D converter.

Who told you that?
Most A/D converters have high impedance inputs. SAR types have a sample and hold capacitor that you have to charge up, but if you leave long enough, or your source impedance is low enough there is no need for a buffer.