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Matt Young
Matt Young
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If you only have battery power, and need to bias a negative rail on some linear parts, a voltage inverter like this one is a quick and easy way to do it.   

block schematic You

You will need the IC solution, plus three capacitors. The IC switches the input capacitor \$C_p\$ between +V and ground at a rate determined by \$C_{osc}\$. This makes the output voltage appear inverted, and charges \$C_r\$ to the inverted voltage.

If you go this route, you need to be aware of the current you're pulling from the newly created negative rail. These parts can only source about 10 mA best case.

If you only have battery power, and need to bias a negative rail on some linear parts, a voltage inverter like this one is a quick and easy way to do it.  block schematic You will need the IC solution, plus three capacitors. The IC switches the input capacitor \$C_p\$ between +V and ground at a rate determined by \$C_{osc}\$. This makes the output voltage appear inverted, and charges \$C_r\$ to the inverted voltage.

If you go this route, you need to be aware of the current you're pulling from the newly created negative rail. These parts can only source about 10 mA best case.

If you only have battery power, and need to bias a negative rail on some linear parts, a voltage inverter like this one is a quick and easy way to do it. 

block schematic

You will need the IC solution, plus three capacitors. The IC switches the input capacitor \$C_p\$ between +V and ground at a rate determined by \$C_{osc}\$. This makes the output voltage appear inverted, and charges \$C_r\$ to the inverted voltage.

If you go this route, you need to be aware of the current you're pulling from the newly created negative rail. These parts can only source about 10 mA best case.

Source Link
Matt Young
Matt Young
  • 13.9k
  • 6
  • 35
  • 61

If you only have battery power, and need to bias a negative rail on some linear parts, a voltage inverter like this one is a quick and easy way to do it. block schematic You will need the IC solution, plus three capacitors. The IC switches the input capacitor \$C_p\$ between +V and ground at a rate determined by \$C_{osc}\$. This makes the output voltage appear inverted, and charges \$C_r\$ to the inverted voltage.

If you go this route, you need to be aware of the current you're pulling from the newly created negative rail. These parts can only source about 10 mA best case.