Adding a dc offset to a signal

I thought I could feed both signals through series resistors into the non- inverting input of a buffer amplifier, but when I simulate it on LTspice, m y offset is low (1.4V rather than 2V) and my input signal is attenuated.

wave as the signal source

ound to the buffer op-amp, and it behaved correctly

the simulation, my waveforms got all distorted

Your offset voltage V9 should be 1V, not 2V.

1V, divided by the 2:1 divider, multiplied by the 2x gain, = 1V offset.

Cheers, James Arthur

Thanks James, I overlooked the divider, I am trying to get a 2V offset on m y input signal, which the 2nd circuit does because of the gain... and now I 'm thinking that when I swap out the 2V source with a reference, the output resistance of that reference is screwing up my gain... is this just the wr ong approach?

my input signal, which the 2nd circuit does because of the gain... and now I'm thinking that when I swap out the 2V source with a reference, the outp ut resistance of that reference is screwing up my gain... is this just the wrong approach?

I've replaced the series resistor on the the ac signal with a 10u cap, and its isolated the DC offset voltage from that divider, so it added 2V to the output. Thanks! now I tried swapping the 2V source with a reference and it 's acting goofy again.

Then Bob's yer uncle. (If would help if you'd just say "I want a gain of X with a 2V offset.")

Your approach assumes equal source impedances. That's fine as long as it's true. If not true, using 100k input resistors would make the source impedances 100x less important.

Cheers, James Arthur

I thought I could feed both signals through series resistors into the non- inverting input of a buffer amplifier, but when I simulate it on LTspice, m y offset is low (1.4V rather than 2V) and my input signal is attenuated.

wave as the signal source

ound to the buffer op-amp, and it behaved correctly

the simulation, my waveforms got all distorted

Do you care if the output is inverted? If not then the usual summing amp will work fine.

George H. (Oh, well you'd have to put in -2V DC.)

I swapped out the series resistor in-line with the input ac signal with a 10uF cap. This prevented the DC offset voltage from getting divided down.

Now I swapped out the 2V voltage source with a voltage reference part, and after adjusting the DC in-line series resistor I was able to get what I was looking for

although I'm not sure if it was just coincidence that the resistor that worked is the same value as the resistor I'm using to feed the voltage reference (45k)

What I ended up doing:

**I can't have the output inverted**

(View in a fixed font, like Courier)

+5V -+- |R1 [45k] | R3 |\ +----[45k]----+----|+\ .-' U2 | | >---+--> ^ 2.5V | .-|-/ | | | | |/ | === | '--------' 10uF | Vin >---||---------'

That works fine. R3's value shouldn't matter--you could make it a lot larger to improve the low frequency response. You could also add a resistor to ground at the non-inverting input, to create exactly 2 volts offset.

You can also bootstrap R3 to increase a.c. input impedance...

Electronics is fun.

Cheers, James Arthur

Does the downstream circuit really depend on that accurate an offset that you need to use a reference rather than just deriving 2 volts from your power supply? I wonder because opamps have their own offsets, so the result won't be exact.

You could buffer your reference with another opamp to avoid issues with your reference having a significant impedance.

Sylvia.

Could you be more precise in your problem statement? What signal conditions going in? What desired signal conditions at output? ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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             I'm looking for work... see my website.

I can't believe all the hulabaloo about this and am not even really an engi neer.

You put the input to the plus side of an OP AMP, you DIRECTLY couple the ou tput to the minus side through a capacitor high enough to give you flat res ponse at the working frequency, which actually means lower by a couple octa ves. Then you manipulate the DC voltage at the minus side with resistors or Zeners or whatever the hell you want. You can even have a pot to adjust it at will.

Not sure I follow that. Seems the output needs to be directly coupled to the minus input so the output will be 2 volts DC. The input needs to go through the capacitor, not the output. Sorry I'm so stupid.

You're not stupid... jurb6006 is a village idiot. To do controlled offsetting, the non-inverting input of the OpAmp should be a virtual ground, then signals applied via resistors to the inverting input are summed, amplitude-weighted by the resistor values. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 

             I'm looking for work... see my website.

See...

As I opined in...

Message-ID:

"To do controlled offsetting, the non-inverting input of the OpAmp should be a virtual ground, then signals applied via resistors to the inverting input are summed, amplitude-weighted by the resistor values." ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 

 Wonder who was better, AG Loretta Lynch or Monica Lewinsky ?>:-}

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

But doesn't that cause a phase inversion? I think the OP wants the output in phase with the input.

.

So add another OpAmp... OpAmp's are cheap >:-} ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 

 Wonder who was better, AG Loretta Lynch or Monica Lewinsky ?>:-}

the

gh

Assuming you want a solid and stable DC offset it is better to go DC on the input and then use the capacitor in the feedback. That makes the OP AMP pr ovide the DC offset rather than just a resistor.

I assume you know enough to just put a capacitor in the signal path and use a resistor to set the DC after it. Since you posted, I assume such a simpl e solution is not acceptable. It has been done for a century, it is called a coupling capacitor. Like betweent the plate of one tube to the grid of th e next in a multistage amplifier. It was also done extensively in multistag e audio amplifiers, and still is somewhat thought they have gone to as uch DC coupling as possible for better low end response.

Anyway, like this :

The capacitor locks the AC gain of the OP AMP at unity and provides for a l ow impedance output suitable for almost anything. The resistor sets the DC gain based on whatever the source resistance of your DC is. The potentiomet er represents whatever DC source you want to use. You can have it stable or you can have it vary under whatever circumstances you choose.

I do believe he said that somewhere along the line but am too lazy to look back.

I do pay attention sometimes.

I cannot believe a real engineer said that. Do you faintly remember a saying "Don't waste silicon" ?

You must work for the government, they are the only ones who would support doubling the cost of something for no good reason.

Interesting. I haven't seen it done that way. Not sure how the OP wants the circuit to behave. In your solution, the output will track the input offset with the additional 2 volt DC offset. So if the input moves up a volt of offset, the output will move up a volt to 3 volts DC offset. Maybe that's what he wants to do. I don't know.