Drive 2 amps diff rails w/ 1 signal

You don't say if there is any DC offset on the 10 Hz source, so I'll just have to assume not.

Something like this might do it. View in a fixed-pitch font:

___ ___ A.sin(wt) ---|___|---o--------o-|___|--- Vo R1 | | R3 | +18V | ___ | |\| | +18V ---|___|---o--|-\ | R2 | >--' +13.5V --------------|+/ |/| +9V

R3/R1 sets gain (note: this is an inverting amplifier) R1/R2 sets the level shift

e.g. R3=R1 and R1/R2 = 13.5 / (18-13.5) for Vo = 13.5 - A.sin(wt)

Oops. That should be:

___ ___ A.sin(wt) ---|___|---o----|___|--o-- Vo R1 | R3 | | | | +18V | ___ | |\| | +18V ---|___|---o--|-\ | R2 | >-----' +13.5V --------------|+/ |/| +9V

Is there a problem with ac-coupling the 10Hz to the two op amp inputs in parallel?

Since your plan doesn't include a ground for either op amp, it's a little tricky. Easiest would be to feed your 10 Hz source to a series resistor, and two transformer primaries in series. Then each transformer secondary would drive one of your amplifiers. Better than an instrument amp, though, would be a pseudo-ground input to an inverting amplifier (just an op amp with feedback resistor), with the (+) input of that op amp held at your intended DC level.

It's also possible to capacitor-couple, of course, BUT the full power-supply noise gets added on if you do that. Then it gets amplified.

See the datasheet and app notes for the INA117. Art

What type of stock transformer would you suggest that will handle

10Hz? I believe these need to be specially wound, and at significant cost.

I plan to use batteries, so this would not be such a problem.

Thank you for your reply.

Kevin Marsh

Isn't 10Hz a pretty low frequency for a cap? What value were you thinking of?

Kevin Marsh

Well, Kevin, since you have not provided input resistance to the circuit, I guess almost any value of capacitance will work. Depending on the input resistance, of course. The lower the input resistance, the higher the capacitance required. Yes?

John

Not at all since the op amps and network resistors are such high impedance- even a low 10K input impedance and 1Hz cut-in gets C=3D1/ (6.28*1Hz*10K)=3D16uF which is not so big.

In the frame of reference for the upper rail OA, the input looks like sin(wt)-9V, and the similarly for the lower rail OA the input looks like sin(wt) + 18V. So the transfer function of the upper rail OA should be OUT=3D IN + 3/2 x V+, and similarly for the lower rail OA the transsfer should be OUT=3D IN -3/2 x V+. The resulting summer amps look like so:

Please view in a fixed-width font such as Courier.

. . . . . . 18V . | . .------+ . | | . [10K] | . | |\| . .----[15K]-+----|+\ . | | >--+-->OUTU . | .----|-/ | . | | |/| | . | | 9V | . | +--[15K]---' . | | . | [10K] . IN>---+ | . | | . | 9V . | . | . | . | -9V . | | . | .----+ . | | | . | [10K] | . | | |\| . '-[15K]-+-------|+\ . | | | >--+-->OUTL . [10K] +--|-/ | . | | |/| | . | | -18V | . -18V | | . '--15K]--' . . .

simply couple your 10Hz to each opamp via a capacitor (1 for each opamp). If you tell us what the input impdeance is, its then possible to calcualte what cap size is requierd.

I wouldnt consider using a transformer, except in truly exceptional situations.

NT

Well- if you work up an example circuit, the direct coupling is more cost-effective and has greater accuracy than the capacitor coupling.

Thanks for taking time to draw the complete circuit. It has helped to clarify the design for me.

Given the application stated in my OP, is there any particular op amp you would recommend? I am looking to sink around 50 ma into a resistive load.

To derive the rails, I intend to use three 9V cells.

Two are connected in series, with their negative connected (through the amp) to the negative of the third. This provides 118V less -9V or a 9V potential.

Similarly, the positive of the two is connected (through the amp) to the positive of the third for the +9 and +18.

Does anyone see any problem with running the two amps off one set of three cells as described above?

Kevin Marsh

Don't see how you get those voltages with just 3x 9-Volt batteries.

With regard to the battery connection to the op amps, is this what you mean?

+---------+ | | | + | [9V] _|\| | - | \ + _| / | + |/| IC1 [9V] | | - | +---------+ | | | + _|\| [9V] | \ | - _| / | |/| IC2 | | +---------+

If not, can you draw a schematic of what you are talking about?

Ed

Yes, and the mid point would be the virtual ground.

Thank you for your inquiry.

Kevin Marsh

Maybe...but this is what you described:

Please view in a fixed-width font such as Courier.

. . . . . . . . (+18V) . +---------+ . | | . | + | . [9V] _|\| . | - | \ . | _| / . | |/| IC1 . | | . | | . +---------+ . | + (+9V) . [9V] . | - . --------+ . | | + . | [9V] . --- | - (-9V) . /// +---------+ . | | . | + _|\| . [9V] | \ . | - _| / . | |/| IC2 . | | . ----------+ . . (-18V) . .

The suggestion was to use a low-impedance load on the secondary of the transformer (the pseudo-ground pin of an inverting amplifier), which makes the L/R time constant long enough for low frequencies. This is a kind of current transformer application, and the low frequency limit isn't problematic there. As for 'specially wound', it won't hurt you to count off fifty turns of the spool as you drive it with your (slow, battery powered) drill. Stock transformers are only available when and as high-volume production is done for someone else's needs; stock is ALWAYS an undependable source.

It all depends on volume. I presumed 'for experimental purposes' meant the OP was only building one, in which case whatever is quickest and easiset is probably best, ie capacitors. If the OP wants to build

100,00 then I'd do it as you say.

NT