Positive Biased Clamper Help

Thread Starter

kahafeez

Joined Dec 2, 2008
150
Hello Folks, this is supposed to be very easy..... bt i'm still having problems with this.... plz help.... i also want to retain the shape of the signal.....
 

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Thread Starter

kahafeez

Joined Dec 2, 2008
150
its nt really a homework..... its just that one block is producing 0-5 square and the next needs a different level..... the next block needs a 200mV signal that has a 1.1V as low and 1.3V as high..... bt i could do that attenuation through a simple voltage divider..... the problem is abt clamping that 200mv signal to 1.1v.....
 

Thread Starter

kahafeez

Joined Dec 2, 2008
150
its a VCO.....

i just used this schematic.... it gave me better results in the simulation.... will try it in the lab tomorrow.... can anyone plz tell me whats the difference... like how do they work.....
 

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Ron H

Joined Apr 14, 2005
7,014
You don't need a diode. You can do it with 3 resistors and a power supply.
I have calculated the resistor values. Simple algebra will allow you to do the same. See attachment.
 

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Thread Starter

kahafeez

Joined Dec 2, 2008
150
i tried the new schematic too but that brings noise in the ciruit..... i dont know why its coming.... plz tell me a way to clamp a signal to any voltage level bcz my new application requires clamping the voltage to 1.6V.... thanks in advance
 

Wendy

Joined Mar 24, 2008
22,170
How accurate does the 1.6 V have to be? Would 2 diodes, 1.2V, work?

Restating the specs, a sine wave, 0 to 5V. You want 1.3 to 6.3V out?
 

Thread Starter

kahafeez

Joined Dec 2, 2008
150
thanks for replying.... no a square wave 0 to 5V and i need to convert it to a 200mV signal(which i've achieved using a simple voltage divider) bow the new square wave should have its low at 1.6V and hi at 1.8V..... i need it to be accurate.... nt like 100% even 80% would suffice...
 

Ron H

Joined Apr 14, 2005
7,014
i couldnt see any resistor values :-/

and i need to clamp a 200mV signal to 1.1V.....
As I said, you can calculate them. Below is the solution.

EDIT: This circuit will give you an output of 1.1V-1.3V. Your new range can use the same circuit, but with different values.
You don't need a clamp, as long as your input is always 0 to 5V, DC coupled. All you need is an attenuator and level shifter, which is what this circuit does. It will be more stable than a clamp.
 

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Last edited:

Thread Starter

kahafeez

Joined Dec 2, 2008
150
thanks Ron.... bt can u plz tell me which one is the input.... V3 or V5???? and which resistor is deciding the DC level shifting... plz explain ..... i need to know....
 

Ron H

Joined Apr 14, 2005
7,014
dude ur schematic didnt had any equations and no values..... plz tell me equations...
Dude... I edited my last post, Read it again.
Concerning equations, can you do the algebra to calculate voltage dividers and attenuators? If not, I can post them.
 

Wendy

Joined Mar 24, 2008
22,170
OK, as a starting point, would this do what you want? You could also substitute a 1.5V battery for the 1.6V shown.



**********************

Ron, you must be an early bird. When he PMed me I was the only one up and about, so I thought.
 

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Thread Starter

kahafeez

Joined Dec 2, 2008
150
i simulated both the methods.....

both work perfect.....

bt i'd like to ask a few questions:
@ Ron H:

sir plz give me the algebra etc..... cz i need to knw the reason so that i can do it the next time myself.....

@ Bill:

Sir i knw the two resistors are for attenuation.... i knw how to calculate their value bt i need to know that i can get any level by only changing the DC battery level??? and do i need to change capacitor's value for different levels????


thanks
 

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Wendy

Joined Mar 24, 2008
22,170
On mine, think simple voltage divider. 5 VDC to .2 VDC, it is a pure ratio problem. The power supply sets the low voltage, no calculation needed.

Ron's is somewhat similar. You solve it in two steps. Figure out the resistors needed to create your wanted voltage when the input is 0V (the low on the square wave).

Then figure out the resistors needed to create your desired output with the input is +5VDC (the high on the square wave).

You'll find it is another voltage divider problem.
 
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