op-amp power consumption

That will depend on the specific op-amp as well as the external load (including the feedback circuit). The current draw on the inputs is quite small, but it varies over many orders of magnitude depending on the part number. The power supply inputs(s) will draw at least as much as the output sources to external loads (the thing ain't got a battery in it). How much more will depend on its internal dissipation which, like I said before, depends on the specific part selected.

Should not be a difficult question Op Amp data sheet will specify the quiesent current. Add to that the the load current and any bias network current.

M Walter

Mark Walter wrote in news:m_x4g.26731$NS6.4054 @newssvr30.news.prodigy.com:

Well I'm using a LM324 as a unity gain buffer. There is a 1 volt battery going through a 1K resistor into the non-inverting. The output is fed back to the inverting input. My supply voltages are +/- 15 volts and the op-amp is drawing 1.776uA on both supplies.

The datasheet doesn't talk about quiesent except it's 1/5 normal op-amps.

Anyway to calculate that 1.776uA without having to hook up a circuit and measuring it?

Coming in late, are we sure this is an op-amp issue and not something else in the circuit?

Secondly, +/- 15 supplies? 30 volts across an LM-324?

Richard Lamb wrote in news: snipped-for-privacy@earthlink.net:

well I'm using MultiSim actually. I was thinking I stated that, but re- reading it, I didn't. These are the only components in the circuit and I thought an LM324 can have up to 30 volts vcc?

The data sheet (at least the summary sheet from National Semiconductors) does talk about the quiesent current. They list it as a maximum of 180 ua per channel. You would have to read further to see if this a room temperature maximum; if it holds over the entire operating temperature range, how dependent upon supply voltage etc.

M Walter

Well, at extreme supply voltages, the thing is going to heat up. Maybe to extreme?

Mark Walter wrote:

Opamps are very efficient, so most of the power will be dissipated in the opamp load.