circuit desgn need help. volt bias to sine wave

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Mrusten

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hi

I have a little project for test equipment but need help to circuit design

I got a sine wave roughly 0-1v varying in the HZ region

this signal need to be the same at the output but added or removed 500mv according to adjustment

and some other features.

I will try to make function blocks here:


Input: sine wave

Adjustment input: pwm 1,4v 2,4khz need to be adjustable sensitivity

Adjustmen on the circuit by pot: end adjustment of the pwm duty cycle 0% (example 0% adjusted to be added +10mv to the sine wave)
between this points its linear so 50% duty cycle is in the middle between those 2 points at all times
Adjustmen on the circuit by pot: end adjustment of the pwm duty cycle 100% (example 100% adjusted to be removed -200mv from the sine wave)

Output: sine wave 0-1v + - 500mv (-500mv and 1,5v


hope it make sense thanks for reading
 
Welcome,

Need a sketch of the waveforms at each point in the circuit.

Your verbiage is really hard to follow...
 
MikeMl said:
Welcome,

Need a sketch of the waveforms at each point in the circuit.

Your verbiage is really hard to follow...
Click to expand...


hei im totally amatuer to designing circuit so i exspect its wrong, i got basic understanding of electronics so about calculations and so on i dont have a clue.

here is my thinking about how it could be done.

about #3 stage input# im not sure, it may be inpossible but the goal is to be able to adjust where the max point is (max output voltage) and where the minimum is (minimum output voltage)

finally at the #output# the orginale signal can be seen at the - - - - - line and the new output signal is the _______ line,
if the #DC volt out# is 0v offcorse the #output# will be the same as #signal input#

the signal should be only added + or - according to #DC volt out#
the examples is example of max biased signal adjusted in #3 stage input#

(basicly a voltage follower with a volt bias?)

there is 500 mv + or - in the examples but bechause my bad drowing skills it dosent look like so, anyway look at the volt scale at the left.



 
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It seems you expect to have a sinewave which could be shifted in level +/- 500 mV.

Could you answer the following?

Peak to peak value to stay the same?

Frequency to stay the same?

How do you expect to control the level shifting? Manually with a pot or with a signal taken from an existing circuit?
 
atferrari said:
It seems you expect to have a sinewave which could be shifted in level +/- 500 mV.
Click to expand...

Yes the sinewave have to be the same as input just shifted (phase shift dosent matter)

atferrari said:
Could you answer the following?
Click to expand...

Yes

atferrari said:
Peak to peak value to stay the same?
Click to expand...

Yes

atferrari said:
Frequency to stay the same?
Click to expand...

Yes sinewave frequency has to stay the same as the input sinewave (varying 1 hz to 200 hz)

atferrari said:
How do you expect to control the level shifting? Manually with a pot or with a signal taken from an existing circuit?
Click to expand...

Signal taken from an existing circuit PWM as in the drawing

with manual pot to adjust sinewave shift at 0% PWM
and another pot to adjustsinewave shift at 100% PWM
 
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Mrusten said:
Level of shifting by PWM as in the drawing but with manual pot adjusted end values (one pot for max dc volt and one pot for min dc volt)
Click to expand...
What is going to vary the PWM duty cycle? If it's another pot then the PWM isn't necessary. If it's a microcontroller then the max and min positions could be set in software so you wouldn't need pots for those.
 
alec_t said:
What is going to vary the PWM duty cycle? If it's another pot then the PWM isn't necessary. If it's a microcontroller then the max and min positions could be set in software so you wouldn't need pots for those.
Click to expand...

The pwm comes from another circuit where its impossible to change anything
 
PWM > filter > DC.
DC endpoints adjustable > pot#1 sets DC value for 0% PWM; pot#2 sets DC value for 100% PWM; 50% PWM = ((pot#2DC - pot#1DC)/2)

Input sine + DC value = output.

Correct?

ak
 
Hola AK,

If the PWM signal is already given, I think that after converting it in a DC signal, a simple pot will ensure the excursion along the whole range.
 
AnalogKid said:
PWM > filter > DC.
DC endpoints adjustable > pot#1 sets DC value for 0% PWM; pot#2 sets DC value for 100% PWM; 50% PWM = ((pot#2DC - pot#1DC)/2)

Input sine + DC value = output.

Correct?

ak
Click to expand...

correct exept... Input sine + - DC value (+ - 500mv) = output.

about 500mv im happy to get more so that part is not critical
 
Mrusten said:
correct exept... Input sine + - DC value (+ - 500mv) = output.

about 500mv im happy to get more so that part is not critical
Click to expand...
What is the amplitude range of the input Sine?
What is the frequency of the input Sine?
What is the amplitude range of the DC value?
 
atferrari said:
If the PWM signal is already given, I think that after converting it in a DC signal, a simple pot will ensure the excursion along the whole range.
Click to expand...

Only if the incoming PWM signal is bipolar about ground, and most aren't. As noted again in post #10, the DC offset can be +/- about GND. First thought (probably a bit weird) is to use the incoming (unipolar above GND) PWM to switch between two voltage sources, both adjustable per post #1. This variable amplitude low voltage PWM waveform is analog filtered, preserving DC accuracy, to give the bipolar DC voltage that is then summed with the input sine. One quad opamp, one voltage reference, one NPN and PNP, a bunch of Rs and Cs. Output sine with be 180 degrees shifted from input sine.

ak
 
MikeMl said:
What is the amplitude range of the input Sine?
Click to expand...
in the hand drawing you can see its 1 volt, but variable so the circuit need to be a voltage follower

MikeMl said:
What is the frequency of the input Sine?
Click to expand...
as i said before, in the hz region more spesific varying between 1hz and 200hz

MikeMl said:
What is the amplitude range of the DC value?
Click to expand...
its up to the circuit designer.
the pwm signal is 1.4v reference to ground
 
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sounds perfect, the output 180 degree shift dont matter


if you deside to make a circuit diagram please use european standard parts as i got some at home, as lm324

thank you so much
 
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I would be so happy if someone could help to design a circuit like this so i can breadboard it for test and finally make a pcb
 
Note that for this to work you will need two power supplies, one positive and one negative, like +/-5 V, +/-12 V, etc.

ak
 
Last edited:
AnalogKid said:
Not that for this to work you will need two power supplies, one positive and one negative, like +/-5 V, +/-12 V, etc.

ak
Click to expand...

I understood that from your prevously post #12. Its no problem.

i would prefer a split powersupply
 
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Here is the Bode plot of the EDN circuit compared to a conventional opamp Low Pass Filter. Note that it is almost a 2nd-order Butterworth response with a cutoff frequency of 2388Hz.



In order to properly filter PWM, the frequency of the PWM would have to be at least 10X the filter cutoff freq. Here is the time-domain response of the two filters to 20Khz PWM that suddenly switches from 10% duty-cycle to 90% duty-cycle at 1ms. Note the settling time, the residual 20KHz ripple, and the -2.5V DC offset intrinsic in the EDN circuit.

 
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