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let me start off by saying I am a computer science student so I'm not familiar with electronics whatsoever. Anyway, I am faced with this problem:

I am given some input (say, a sine wave) centered on 0V (Peak to Peak is, say, -5V to 5V). I want to make my output signal with a Peak to Peak of 5V to 15V, how would I approach this? One attempt at the solution was the following circuit i got from googling - enter image description here

The circuit managed to change the DC offset but the peak to peak is not where I want it to be. Are there any other circuits I could try? I read somewhere that a non-inverting amplifier (with an op-amp) may change the offset so I will attempt that next.

Any thoughts?

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  • \$\begingroup\$ What frequency is that sine wave? 40 Hz? \$\endgroup\$ – dpdt Apr 25 '18 at 0:50
  • \$\begingroup\$ What's the signal feeding into? That 1k pullup is what's sucking the life out of the signal. It's too big. \$\endgroup\$ – Paul Uszak Apr 25 '18 at 0:55
  • \$\begingroup\$ Sorry, the frequency should be 600 Hz (It's 600 Hz in the example but I don't think the frequency matters? I could be wrong though). The signal isn't feeding into anything, im just looking at the output measured at the far right of the circuit \$\endgroup\$ – Jack Bauer Apr 25 '18 at 1:16
  • \$\begingroup\$ Perform an Internet search for an application report titled "Designing Gain and Offset in Thirty Seconds", written by Bruce Carter and published by Texas Instruments (TI document #SLOA097). \$\endgroup\$ – Jim Fischer Apr 25 '18 at 2:30
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Gonna break out the ol' falstad because you used it.

The non inverting op amp circuit you mentioned looks like this: enter image description here

It adds an offset equal to the voltage on the non inverting input (+), and adds some gain equal to -Rf/Rin, where Rf is the top resistor.

enter image description here

This is also an alternative way, but you're gonna lose some amplitude because the arrangement creates a low pass filter.

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  • \$\begingroup\$ That first circuit inverts the signal. \$\endgroup\$ – JRE Apr 25 '18 at 11:25
  • \$\begingroup\$ I just built this circuit and it applies the offset but the signal is inverted, why is it inverted? I thought this was a non-inverting op amp \$\endgroup\$ – Jack Bauer Apr 25 '18 at 18:20
  • \$\begingroup\$ It is inverted because it should be inverted--it is an inverting amplifier. There is a version where it doesn't invert the signal, but you can't add a DC offset easily. \$\endgroup\$ – hatsunearu Apr 28 '18 at 0:38
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Consider a clamping circuit. Such a circuit can relocate the upper or lower extreme of a waveform to a predetermined voltage level.

enter image description here

It sounds as if the following circuit with a 5 volt battery (or floating power supply) is what you are looking for:

enter image description here

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    \$\begingroup\$ Err, what's a 5 volt battery? \$\endgroup\$ – Paul Uszak Apr 25 '18 at 0:52
  • \$\begingroup\$ I'll make that 5.6 volts, to compensate for the diode drop. 4x 1.5V alkaline batteries plus a 5.6V zener diode and limiting resistor? \$\endgroup\$ – Sredni Vashtar Apr 25 '18 at 2:25
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    \$\begingroup\$ @PaulUszak Just use "alibaba 5V battery" and you will find one. Yes, I did that and immediately found two Li-Ion . \$\endgroup\$ – Oldfart Apr 25 '18 at 7:41
  • \$\begingroup\$ @PaulUszak Some NiCd cells are specified as 1.25 V, 4 of those would be a 5 V battery. \$\endgroup\$ – Colin Apr 25 '18 at 9:01
  • \$\begingroup\$ Has anyone ever measured the unloaded voltage of a battery with 1.5V printed on it? \$\endgroup\$ – Paul Uszak Apr 25 '18 at 11:43
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I am given some input (say, a sine wave) centered on 0V (Peak to Peak is, say, -5V to 5V). I want to make my output signal with a Peak to Peak of 5V to 15V, how would I approach this?

Instead of using the resistor in your circuit connected to +5 volts, connect it to +10 volts. This makes the average level at the output +10 volts with the 10 Vp-p sinewave superimposed on top. Thus it attains a peak of +15 volt and a trough of +5 volts. Ensure that R is a fairly high value so that the AC signal is hardly attenuated.

Something like 10 kohms should be OK for your circuit but, beware of loading the output too much - use a buffer if connecting a low impedance load.

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If you don’t want the AC signal going directly into your DC power source, you should replace the resistor with an inductor.

Then you would have a basic Bias tee.

At such a low frequency (600 Hz) your inductor would have to be quite large, however. Something like 1 H. Also, your DC supply should be centered at your mean value: 10 V.

Something like this:

schematic

*Note: sorry for the crude schematic. It’s all I can do from my phone.

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