# Designing a Voltage level shift circuit

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#### ans94

##### Newbie level 5
I have a 3.3V AC voltage source. I want to level shift it by 3.3V (D.C. Level Shift). The circuit should work like a clamper circuit. How can i possibly design a circuit for the same? Is there any IC which does the same?

Hi,

OPAMP as adder: Vout = Vin + 3.3VDC.

Klaus

The op-amp method suggested by Klaus will always work but if the frequency is fairly constant, the load is small and you have a higher DC voltage available you can do it with a capacitor and two resistors making a potential divider with 3.3V at it's center point.

Brian.

[Moved]Designing a clamper circuit

My input voltage is AC and is constantly changing in magnitude. This change in magnitude has to be tracked by the DSP. Since DSP works on DC, I have decided to clamp the voltage by equal amount of DC voltage. eg. if my input is 1.65V AC, then the circuit must clamp it by 1.65V DC in order to get a peak voltage of 3.3V (pulsating dc), or if my input changes to 3.3V AC, the circuit must clamp it by 3.3V DC in order to get a peak voltage of 6.6V.
I tried to implement the same using a capacitor and a diode (positive unbiased clamper circuit), but however the capacitor is unable to clamp it completely since it is not charging to the input voltage value (generally this is seen at low voltages, 1V and below).
Is there any method of implementing it or modifying the circuit i have implemented?
I have to implement this circuit on a pcb as a part of current measurement circuit.

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u can use this circuit...

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Can you simply connect two series resistors, say 10K each across the 3.3V supply and ground so you get 1.65V at their junction. Connect that to the DSP and capacitively couple the signal to the same point.

Brian.

Hi,

yes, like Brian said.

For ESD and startup voltage spike safety I´d add a double series diode (like BAW56 or BAV199) to VCC and GND.

To calculate the cutoff frequency you have to calculate as both resistors as if they are paralleled. (10k each gives 5k)

Hope this helps

Klaus

Thanks a lot. But i did not understand how to capacitively couple the signal to the same point.

The motivation for using a "clamper" hasn't been well explained. I think the design should start with a specification of the intended measurement function.

E.g. you want to watch AC voltage Vpp, averaged rectified value, RMS voltage, whatsoever. There's an optimal circuit for each purpose. Or sample the instantaneous voltage by DSP and need a level shift to match the ADC input range. Different purposes, different circuits. A clamper doesn't seem as the obvious solution for any of it.

I'm also worried about using a voltage clamp as it normally offset the signal by the peak voltage within it. In this instance the voltage should be offset by a fixed 1.65V regardless of the signal content.

But i did not understand how to capacitively couple the signal to the same point
Just connect the signal through a capacitor to the point where the two resistors and DSP input meet. The capacitor will block any DC content in the signal so only the variation passes through and the two resistors will hold the signal center voltage at 1.65V.

The drawback to this method is the capacitor value may be difficult to pick. It depends on what the signal is, a capacitor is better at coupling high frequencies than low ones so if you are measuring signals with wide frequency content, the lower frequencies may measure lower than the equivalent size high ones.

The diodes mentioned by Klaus are not essential but they would be a wise addition. In normal use they never conduct and have negligible effect on the measurement but if the signal level goes much higher than VDD or lower than ground, they become forward conducting and help to protect the DSP from being overloaded.

Brian.

• ans94

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