how to raise the offset

Hello everyone

Its not really homework, but needed for a project in school
Our input signal is AC -2.5V to 2.5V from another circuit, and we need to "raise" this signal to 0-5V
(ex. so -2.5V equals 0 and 0.5V equals 3V, and so on)

Is there an easy way to do this?

We tried something with "summing amplifier", and it works in simulation, but in realisation its toally messed up, we might have done several things wrong..

Here is a screenshot of circuit in multisim
**broken link removed**

U4 is a voltage follower
U5 is voltage follower supplying the 2.5V (we only have 15V supply)
U6 is the summing
U7 is an inverter

Can anyone tell either what we are doing wrong, or even better tell us an easier way to deal with this problem?

I've run the same sim in LTSpice and it's fine. So any error must be in the actual circuit construction or faulty components. Can't think of any easier way to do this (other than omit a buffer amp or two).

I got this response in another topic:
"Well, you've made this waaaaay more complicated than it has to be. First, U4 is supplying 0 volts to R15; just connect it to ground! But here's a simpler solution, assuming your input comes from a relatively low impedance source.

Connect your 50K/10K voltage divider to the input of a voltage follower. Connect a capacitor to that same input. Connect the other side of that cap to your input. VOILA! your output is now offset by the value of your voltage divider."

And it works perfect, just sharing if anyone should find this..

First, U4 is supplying 0 volts to R15

Click to expand...

Indeed it is, as shown. But I see the word 'Input' next to the non-inverting input of U4. So presumably that's where your -2.5V to +2.5V signal is intended to go?

Connect your 50K/10K voltage divider to the input of a voltage follower. Connect a capacitor to that same input. Connect the other side of that cap to your input. VOILA! your output is now offset by the value of your voltage divider.

Click to expand...

That will be accurate if your signal is rapidly varying (fast enough that the capacitor does not have time to charge/discharge significantly during the signal waveform period), but be aware that it will introduce error for a slowly varying signal.

alec_t said:

Indeed it is, as shown. But I see the word 'Input' next to the non-inverting input of U4. So presumably that's where your -2.5V to +2.5V signal is intended to go?

Click to expand...

That is corerct!

alec_t said:

That will be accurate if your signal is rapidly varying (fast enough that the capacitor does not have time to charge/discharge significantly during the signal waveform period), but be aware that it will introduce error for a slowly varying signal.

Click to expand...

How fast are we talking?

First we have a lowpass filter with amplifyer, and then this to raise the offset, and its planned to receive signals from a dynamic microhpone.. At first only clear tones like 200Hz to 1000Hz or something like that
But later we hope to use "human voices" aswell..

How fast are we talking?

Click to expand...

Let's do some rule-of-thumb maths. Consider the example of a 200Hz signal. The period is 5mS. If the time constant associated with the cap is, say, 50mS then in half a signal cycle (2.5mS) the capacitor charge could change by very roughly 2.5/50 ~ 5%. So at the voltage follower the signal could have an error of ~ 5%. In the proposed circuit the potential divider has 10k to ground, so ignoring the signal source impedance the time constant is ~ 10k*C. If you can tolerate a 1% error then the time constant needs to be > ~ 100*2.5mS = 250mS. So C = > 250mS/10k, = > 25uF.