Making a LM1458 work as a Unity Gain buffer with an input of 0 to 12v output to a digital voltmeter

Question

Making a LM1458 work as a Unity Gain buffer with an input of 0 to 12v output to a digital voltmeter.

When I connect this cct to the injector timer (voltage) & a Voltmeter to the output the voltage stays at about 2.0V until the input exceeds that. I tried to bias the input but it made little difference. I swapped ic's with the same result. The source resistance is about 190k, the Voltmeter input impedance is about 125k Ohms. The reason for the buffer is because the Voltmeter sinks the source voltage too much. How do I get the input Sensitivity to start at least as low as 0.1V up to 12V, output to follow suit.?

Answer 1

The answer to your question is very simple, and it comes in 3 parts:

1) you can't do it with that op amp,

2) you need an op amp with rail-to-rail inputs and rail-to-rail outputs, and

3) don't believe everything you read on the internet.

Take a look at the data sheet http://www.ti.com/lit/ds/symlink/lm1458.pdf At the bottom of page 2 you'll see "Output Voltage Swing". For a power supply of +/- 15 volts, you can't count on getting better than +/- 12 volts. In other words, the output will only drive to within about 3 volts of the negative supply. Does that sound familiar?

Answer 2

Previous answer is correct. However, there may be a solution to this. Your V+ is connected to +15V. Your V- is connected to 0V. Hence your output voltage swing can't go as low as 0V. If you apply a negative voltage to V- (pin 8) of -3V or below, then you should be able to get your output voltage to drop down to 0 volts. You can do this with a couple of resistors acting as a potential divider. Use two reistors in series.

 ---V15V  -> ..pin 8.. V+ of LM1458
 |

 R

 |------- > OV point (GND)

 R
 |

----- V0V -> ..pin 4... V- of LM1458

Make R a fairly large value so you're not draining too much current from the supply. Make the 100K resistor connect to the GND in the diagram above.

The opamp is still being fed with a difference of 15V between V+ and V-, but V0V is now 7.5 voltages below GND.

Something like this ought to work. I did something similar with a 741 op amp configured as a wien bridge oscillator fed from a 9 volt battery. The key is recognising that the battery is 9 volts potential difference between the terminals, but you can share that 9 volts across two equal sized resistors to give 4.5volts difference across each. And then make your centre point of the resistors your 0V (a 0V common). Voltages are relative, they're all differences between two points.

Answer 3

Here is a circuit suggestion to create the necessary input offset to get the Output data I want. The Output has also been referenced to the offset to effectively subtract the input jump.

Answer 4

@WhatRoughBeast, @Dean The blue circuit above works, the only amendment was to replace Zd with a yellow LED & link its 2.1V ref back to the summing input instead of the separate voltage divider. However in the end I decided to use a LM358 which is designed for a single sided power supply. Needs almost no additional components as per the 1st post circuit. For anybody ending up here for reasons as I did it would benefit to read Application Note 116 (AN-116) from TI or National Semiconductor available from http://docslide.us/documents/lm358.htm. The best answer is the LM358. Thanks guys.