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I need a differential amplifier circuit for my design and I am using the LM358 op-amp. The problem is of negative supply. I need to give negative power voltage to my op-amp(-Vcc). and I only have a 9v battery. I found a way to genrerate negative supply voltage using 555.

Negative voltage using 555

However as soon as I cinnect my differential amplifier to the circuit the negative voltage drops from -7.9 to -6 volts. (Some drop was expected but this is too much for my application).

Is there a way I can regulate or stabilize the negative voltage. ?

Here is the comlete schematic.

schematic

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  • \$\begingroup\$ Is there a reason you are unable to consider two 9 V batteries? That would be the simple approach -- especially looking at your low-valued resistors and the heavy load they may present. \$\endgroup\$ – jonk Jun 10 '17 at 4:18
  • \$\begingroup\$ Dont reinvent the wheel.. THey make complimentary switched cap supplies for RS232 with a few 0.1uF caps and 1 IC. try looking for this charge pump regulator before reinventing the wheel. Otherwise there is no regulation in a charge pump except "Load regulation" so ratio of source to load impedance determines load error. Thus avoid low R values and reduce current as much as possible. Use power diodes for lower ESR and Schottky for lower Vf \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 10 '17 at 4:36
  • \$\begingroup\$ @jonk I did consider doing that, but that drains the batteries too fast (they do not last even for 5 hrs). the reason is possibly very high current flow. \$\endgroup\$ – Harsh Chittora Jun 10 '17 at 4:48
  • \$\begingroup\$ @HarshChittora Do you think it would drain the battery slower doing it this way??? There is no free lunch. Two batteries, at least, would have a chance to last about twice as long as one. But deriving a negative rail is going to siphon down your battery way faster. The external diodes, as your current requirements rise up, quickly siphon off voltage. I'd bet your 9 V battery drops quickly to 8.4 V and, likely still further under hard load. Also, your two diodes may pick off 1 V each under your heavy load, too. You'll need another Dickson stage. But that won't stop your batteries from dying. \$\endgroup\$ – jonk Jun 10 '17 at 4:54
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    \$\begingroup\$ Is there any reason your circuit needs 220R values instead of 220k? That's a problem, but a bigger problem is lack of project specs. In,Process , Output. Load Power Load impedance etc \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 10 '17 at 5:03
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The negative output voltage drops as soon as you connect a load of just a few milliamperes. That's because the negative voltage generator built around the 555 can't supply that much current.

It is a nice proof of concept circuit that shows how voltage inversion can be done in principle, but as you've found out it is not very practical.

There are chips out there designed for voltage inversion that do a much better job than the 555 which is build for timing tasks.

I suggest to order a few ICL7660 style chips. They just need two external capacitors for voltage inversion and will give you a much more stable negative voltage. Also they are more efficient than the 555 based inverter so your battery will live longer.

Beside the ICL7660 there are almost pin compatible alternatives chips you can choose from. They improve upon the ICL7660 in various ways. To name a few:

  • MAX1044 / TC1044S: Higher switching frequency (e.g. less noise)
  • LT1054: Higher current, output drops less when loaded. Has internal voltage-regulator feature.
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