how to isolate two blocks from each other

Discussion in 'The Projects Forum' started by kahafeez, Feb 1, 2009.

1. kahafeez Thread Starter Active Member

Dec 2, 2008
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my problem is that i'm using two voltage dividers in a circuit ..... both r getting effected by each other.... is there any way that i could isolate them from each other..... i've explained the matter in a better way in the image.....

the frequency of the square wave is 300bps.....

plz help....and keep it as simple as possible and briefly explain the solution too..... thanks

problem 2:
at some other place i've a 300bps square wave which has 2.5V as lo and 3V as hi..... i want to change it to a 0-5 square wave..... how would i achieve this???? i thought of achieving it in two steps... first bring the lo voltage back to 0V, like remove the DC component and then amplify it to 5V using a single supply op amp...... plz comment if there is a better solution.....

• sqr.PNG
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2. Wendy Moderator

Mar 24, 2008
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OK, on the first part of the problem you have to remember that only the AC component of the square wave comes through. This would be a 5VP-P, or 2.5P waveform. On the other side of the cap you would have a square wave that goes from 1.5V to 6.5. Your voltage divider would drop that to 20.4% (1KΩ / (1KΩ + 3.9KΩ)), so that square wave would become .5VP. 4VDC + .5 is 4.5V, 4VDC - .5V is 3.5V. This is where your wave form is coming from.

Frankly, I can't make heads or tails from your second drawing. Are you connecting the 250Ω resistor to the chips +5V line? If so you need to state this clearer, don't assume people will know what you are trying to show. Being able to draw a clear schematic is crucial to electronics communication.

3. kahafeez Thread Starter Active Member

Dec 2, 2008
150
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ok thanks Bill, i'm working on a better drawing now.... just downloaded a few softwares used for drawing circuits.... havent worked on any before.....

yes u r right that i'm taking it from the 7414's power supply.....

so wat u r saying is that the capacitor is lifting the wave to 1.5V????
did i get u right ???

pardon me for my ignorance and i sent u a PM cz i posted the thread yesterday same time and no one replied..... and there is another problem too in my 1st post.... called problem 2

4. Wendy Moderator

Mar 24, 2008
20,772
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Check out my blog, I have a set of templates collectively called PaintCAD, it's how I generate my schematics. Cheap and easy.

5. kahafeez Thread Starter Active Member

Dec 2, 2008
150
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finally i made a schematic..... couldn't really figure out how to use the templates of ur Paint CAD Bill...... anyways will try to figure out how to use them.....

right now i need to know the following

1. how to fix the problem that both the voltage dividers work independently and don't effect each other.....

2. please suggest a solution to my problem 2....

thanks...

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6. SgtWookie Expert

Jul 17, 2007
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As far as your 2nd problem, a comparator will make for an easy solution.
R1 is a potentiometer that sets the "trip" level.
R2 is a pull-up resistor for the open-collector comparator output.
R3 provides hysteresis which helps prevent false triggering.

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7. kahafeez Thread Starter Active Member

Dec 2, 2008
150
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Thanks Sargent, appreciate it, will try it in the lab and post the results......

8. Wendy Moderator

Mar 24, 2008
20,772
2,540
Breaking it down, you would have the following out your second schematic...

I figure you would have a 3.9V to 4.1 V square wave out. Are you wanting 1V P-P out (3.5 to 4.5)? The 4.9KΩ would need to be 800Ω instead. The 200Ω is 250Ω and 1KΩ in parallel.

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9. kahafeez Thread Starter Active Member

Dec 2, 2008
150
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regarding problem 2:

the schematic Sargent suggested worked perfect...... the signal was a a pure 0-5V output.... the simulation shows some noise in the upper level bt that wasnt observed in the real hardware..... thanks Sargent.... now could u plz answer my following questions, how did u find the values of the Resistors and how did u set the triggering point ? like is there any equation with which u can set the triggering point. if yes plz tell me for future use...... thanks

@ Bill
thanks Bill, for the solution...... really really appreciate it..... will try in the lab and pot the results.....

10. Wendy Moderator

Mar 24, 2008
20,772
2,540
You're welcome. You can see how drawing a schematic clearly can help solve the problem too. I went through notepads galore when I was learning, but paper is cheap.

11. kahafeez Thread Starter Active Member

Dec 2, 2008
150
0
regarding problem 2:

the schematic Sargent suggested worked perfect...... the signal was a a pure 0-5V output.... the simulation shows some noise in the upper level bt that wasnt observed in the real hardware..... thanks Sargent.... now could u plz answer my following questions, how did u find the values of the Resistors and how did u set the triggering point ? like is there any equation with which u can set the triggering point. if yes plz tell me for future use...... thanks

@ Bill
thanks Bill, for the solution...... really really appreciate it..... will try in the lab and pot the results.....

12. SgtWookie Expert

Jul 17, 2007
22,183
1,728
Good to know you got it working. However, if you look at the output signal again, you should note that it does not go all the way to ground; the lowest the voltage should be going is around 90mV. This is due to the limitations of how much current the output transistor of the comparator can sink, and the resulting voltage drop.
That was expected. The simulation has no capacitance or inductance added. A "real" circuit will have at least a small amount of parasitic capacitance. Even a few pF would be enough to squelch the noise.
To minimize current draw, I simply selected a value for R1 that would result in less than 1mA current from 5v to ground. 5v across 10k Ohms = 0.5mA current.
As far as R1's wiper setting, your stated signal voltage was 2.5v to 3v. Since the supply voltage is 5v, 2.5v/5v=50%, 3v/5v=60%. Since in the simulation the pot was installed upside-down, the wiper was set at 45% instead of 55%. This set the threshold level at 5v x 55% = 2.75v.
Different values of resistance could certainly be used, but current levels greater than 1mA would be excessive IMHO, and current levels less than 0.1mA would make the threshold much more susceptible to noise.

R2 was selected to provide roughly 3.33mA current for the comparators' output to sink; the target was between 3mA and 3.5mA. The comparators' output can sink somewhat more current than that, but as it sinks more current, the minimum output voltage increases.

R3 was chosen to arbitrarily provide roughly 1/10 the current flow of R1, roughly 125mV of hysteresis, or 1/4 of your input signal level's peak to peak voltage. 3V-2.5V=0.5V; 0.5V/4=125mV. Actual hysteresis is somewhat less due to R1's offset from center, but it should still work OK.

13. kahafeez Thread Starter Active Member

Dec 2, 2008
150
0
dude, thanks so much for the explanation...... thanks everybody..... this forum is so great.....