Assymetrical Power Supply

Put the positive input of the op amp on the desired voltage (with a small trimmer) and use the same overall negative feedback for the whole frequency range from 0 - whatever the highest frequency of your signal.

Take care of stability.

joe

Why bother? Do you expect clipping, and does it need to be symmeteric? Until you DO clip, it doesn't matter about offset of the input signal.

You've not told us about the nature of the signal, ie what parameters matter. Possibles include: resistor to -15 rail diode string + bias R zener in nfb with bias R transistor V shift stage ac coupling with dc restoration etc

NT

A differential op amp can be configured with an output referenced to any voltage within the supply range.

In the simplest method, refering to figure 5, simply return R4 to the required output reference. R1=R3, R2=R4. Gain equal to R2/R1.

The inputs are connected to the two nodes that represent the signal being amplified. Normally either V1 or V2 would be the input signal ground. Impedances around the input nodes should be

The signal is within the audio range, bipolar with a 2V swing. The non-inverting op amp is intended to bring it up to 10Vpp.

Thank you for your suggested solutuions.

Could you please elaborate on which one would be most effective, and how it can implemented in terms of part connections?

PS: I like cats too. Mine likes to sit on my keyboard.

Claus Jensen

[snip]

I have concluded that cats are clever enough to know how to get attention... sit on your keyboard ;-) ...Jim Thompson

-- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at

| 1962 | I love to cook with wine. Sometimes I even put it in the food.

Do you mean 2Vpp to 10Vpp?

Start with a plain old gain-of-5 op-amp circuit.

There will be a resistor from the negative input to ground. On paper, move the ground connection to +1.25V, note how things work.

Now take that resistor and split it into one going to ground and one going to +5V, with the junction at the op-amp negative input. Jigger the values until the parallel combination of the resistors is 1/4 the feedback resistance, and the unloaded voltage is +1.25V.

The location of a tweak pot should be obvious.

--
www.wescottdesign.com

Can you add a resistive divider to provide a fake ground @+15V and just supply 0 and +20 ?

--
Les Cargill

If you can use an op amp that drives output close to the (+5) rail, you don't need to apply offset (or capacitor-couple on the output) to get 10Vpp. Many op amps tolerate filtered but unregulated power, up to 40V, so if there's an unregulated supply (other than the +5V) you could make use of that.

If you DO have to apply a DC offset, summing amplifier tricks will do it. But, the result is no longer a ground-referenced signal, you have to take care of level shifting at the output as well as at the input.

Depletion mode P channel FET.

Most Op I have worked with can handle the - rail being higher than the other.. It's the magic dust inside that makes it happen.

So using the +5 and -15 for your supplies to the Op-amp should not give you any trouble obtaining a symmetrical output, if that is what you are really shooting for? your numbers are kind of confusing.

You issue is that 5 Volts is the most you'll get on the output with out going into saturation on the + side of output. Also, you need to pick a op-amp with a rail output feature, many will not reach the supply rail. So at this point, you know you need an OP-AMP that can do Rail to Rail, at least on the output.

You specified 2 volts input bipolar ? To me that is a +/- signal. You also specified 10VPP on output? That can be confusing because if you are looking at 10VPP on a scope at center line then you get 5 volts on the + side and 5 volts on the - side of the screen = 10VPP ? If you expect this to be unipolar on the output then a 10VPP yeilds

10 volts and if this is on the + side of the output, you'll only reach 5 volts+, but you'll be able to do this on the - side output!

So, assuming what you meant was, a 2V +/- input with a 5V +/- output then you can use the standard non-inverting Op-amp configuration, that would be the one where you select two R's in the - feed back to set your desired gain. THe + input will be the input to use from your signal.

The end results would be an output with a +/ 5 Volts which equals 10VPP on the scope at center line. Just a gain point of your input..

If you are truely trying to move the output into a unipolar case, on the + side, 5 volts for the rail is not enough.

I think maybe we need a better lay out of the input voltage swing and output voltage swing to know better.

Have a good day.

Jamie

We still don't know enough about your app to know which is best for you. A common option is resistors from the supply rails to get your chosen offset, a cap from there to ground to reject psu noise, and connect the point to the + input.

they do know how to train people

NT

I will try to rephrase my orignal post as concsisely as I can.

I have a non-inverting op amp powered by +5 and -15V rails.

My audio input signal is swinging +1V to -!V. This places it "above" the quiescent point of the amp resulting in no output.

Accordingly, I would like to apply a DC offset to the signal prior to the amp input so the swing is centered on -10V instead of 0V as it is now.

The output is intended to be 10Vpp centered on -10VDC.

I am unsure of exactly how to best achieve this operation, and am looking for suggestions with enough detail for me to try them out..

I would prefer not to have a coupling cap in the signal path.

Claus Jensen

I presume you mean centred on -5v. Otherwise your -ve rail isn't big enough.

Now that we finally know you have an audio signal... An easy option is to put 2 Rs in series from audio in to -ve rail, with a nice big cap from their junction to ground. The R from audio in to cap needs to be picked according to the audio signal's resistance. We can't specify it without a circuit.

NT

4.7V zener diode (simple, but not good)

Op-amp level shifter. (can have adjustable offset)

If the gain which you already have stage has a current summing node (a virtual ground) sourcing a current into it will cause an offset at the output.

--
  \_(?)_

If I read you correctly..

100k ___ +|___|+--------+ | | | | | + | Offset output 25k | |+5 | ___ | |\+ | +2.5V---+|___|----+-+|-\ | | >+-------+ -5 to -15/10VPP +-----+|+/ .-----. | |/+ | G | + |-15 +---+_-_-_|+---++ + | | | | '-----' === 2VPP GND (created by AACircuit v1.28.6 beta 04/19/05

That is what I get out of your break down. the 2.5V reference can be done via a voltage divider from the

5 volt rail source. But you still need a rail to rail op-amp/..

Jamie

for 20v peak to peak yes, for 10v no.

NT

Thank you. I try the DC shifter since I prefer not to use blocking caps.

Claus Jensen

Applying a shift at the op amp inputs doesn't solve the problem. Your 'shift' means the op amp output is no longer ground-referenced, so a load (from output to GND) will be DC biased.

That's usually not desirable.