# How is it possible to supply an LM833D with +12V and -8V supply?

I was studying a Philips CDR-765 circuit when I came across an output driver circuit consisting of an LM833 opamp supplied with -8VDC on the 4th pin and +12VDC on the 8th pin.

This is the first time I came across an opamp powered with different + and - voltages.

Can anyone explain how this is possible?

• unlikely in this case, but you often see a negative rail "a bit below zero" in older equipment where rail-to-rail operation wasn't possible (or cheaply available), and only the positive range was needed, effectively assuring the op amp doesn't saturate before 0v Commented Jun 26 at 16:21

Think carefully: how can the op-amp tell that the supply is anything but 20V?

As long as the inputs are within the allowed input voltage range, and the output is within the allowed output range, the op-amp will work. Even if 0V/GND was connected to one of the inputs - it would not know that the supplies are relative to this voltage. It's just an input voltage, not a supply voltage.

In the circuit you describe, the op-amp operates just as-if it was supplied with +20V, and the common mode was set to +8V, both relative to the negative supply pin.

Why shouldn't it be possible?

The Op'amp only cares about the voltage between it's 2 supply pins (and that the voltages at input and outputs pins are within the allowed limits compared to the supply pins).

Don't forget that a voltage like "-8V" or "+12V" is just a shortcut to say that the voltage difference between this point and ground is "-8V" or "+12V". But you have no obligation to use a ground as reference for your op-amp.

One classical way to use op-amps, is to tie one supply to GND, and the other one to +Vcc (when you don't need negative voltages, and don't want to provide one for the supplies)

Some example op-amp circuits as followers, all perfectly valid. They all do the same thing, ie output voltage = input voltage. The only difference is the allowed range. If we consider a rail-to-rail (both input and output) op-amp, then for each circuit, the input range is :

• +500 to +510V
• 0 to +24V
• -5 to +12V
• -24 to -12V

simulate this circuit – Schematic created using CircuitLab

Basically, just choose whatever supply voltages respecting the following constraints :

• the input range is between those supply voltages (with enough margin if your op-amp is not rail to rail input)
• the output range is between those supply voltages (with enough margin if your op-amp is not rail to rail output)
• the difference between the input supplies don't exceeds the specification of your op-amp (for example, an op-amp specified for -24 +24V can be used for -5 +36V, or 0 +48V or +12 +60V, but not for -24 +36V)
• often, you want the supply voltage range to be "safe" for the downstream circuit (in case your op-amp saturates). It's not mandatory if you can guarantee that your output will never reach those "forbidden" values
• if possible, use voltage rails you already have in your circuit
• if your circuit saturates, make sure you take into account the right supply voltages in your computations (some formulas take shortcuts like assuming that supply voltage is -Vcc and +Vcc)
• Thanks for explaining. Commented Jun 29 at 13:33