# Capacitor Multipliers in Pulse Generator circuit

Discussion in 'Electronics Homework Help' started by Aisha.Raj, Feb 16, 2013.

1. ### Aisha.Raj

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Feb 16, 2013

PROBLEM STATEMENT

Without using a microcontroller or a dedicated timer IC, implement a circuit which, when a trigger is applied , will provide a +ve pulse of amplitude 3V, whose width can be varied from 0.1ms to 10s. You may not use any capacitor greater than 1uF or resistor larger than 470kohm. You are provided with op-amps which operate off + or – 10V. Provide a provision to continuously vary the duty cycle from 25% to 75%.

MY SOLUTION

Monostable multivibrator to provide positive one-shot pulses ( stable state is –Vcc) of amplitude +Vcc.
Pulse Width varied by varying R1C1 time constant . So I fixed R1 = 470kohm and varied C1 . But C1 cant take values greater than 1uF . So I want to use a capacitance multiplier to increase the value of capacitance and also this capacitance has to be varied so I will vary the gain(R7/R6) of the capacitance multiplier.
Duty cycle can be varied by varying the R2C2 time constant of the trigger circuit (which is basically a passive RC differentiator-high pass filter to which square waves of time period T>> R2C2/10 is fed by a function generator) . Here too I fixed R2 = 470kohm and varied C2 . Since very large and variable C2 values were required for this , I again want to use a variable gain capacitance multiplier as above.
So essentially I am using 2 different – variable gain capacitance multiplier whose capacitance value b/w i/p and o/p node is 1uF , and the amplified version of which is seen from the input side ( due to Miller’s Effect) given by the equation, Cin =C(1+R7/R6) where either R7 or R6 is a variable resistance.
Then at the o/p stage of the multivibrator , I put back to back zeners to clip the voltages to (Vz + Vf) and –(Vz to Vf) which roughly estimates to be +(3.3+1.1) and –(3.3+1.1) (ie) 4.4V to -4.4V. Zeners used have a forward drop of 1.1 V max and Vz = 3.3 V (1N5226) . This is further clipped by using a +ve clipper circuit of 2 pn- diodes (1n4001 of Vd =0.7V ) and a load resistor Rl = 10kohm .
The o/p of this entire design is taken from Rl (10kohm) .

MY PROBLEMS

1. I do not know how to connect the capacitance multipliers to my circuit such that , for one capacitance multiplier Cin appears as C1 and the other capacitance multiplier Cin appears as C2.
2. Will there be any effects due to frequency which’ll impede the working of my circuit as it should behave?? Because I remember that Miller effect has something to do with frequency. I ask this as to provide the required pulse width and duty cycle I may have to give the input to the trigger from few Hz to few MHz . So will the circuit function good for this ??
3. Also, at the clipper end I used a random value of Rl as 10kohm. Is there a specific way to choose this Rl value???
4. Any other problems I will face while simulating or implementing this circuit that you can see. Any errors in logic or design that you can detect???

I’d be really grateful if you could advice me suitably on the issues I raised . I have been given a deadline – 20th Feb 2013, gotta present the design to my panel on 21st . So I’d be extremely happy if we can figure out something by then ……Sorry for the extreme short notice and tks for your help in advance.

I am attaching the circuit with this post.

2. ### Harald KappModeratorModerator

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Nov 17, 2011
The schematic of your circuit doesn't show. It's probably on a server that requires login. Please use the "attachment" function and post the schematic right here.

3. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
So you need to create a monostable that generates a pulse with an adjustable duration when it's triggered.

Then the assignment says that the "duty cycle must be continuously adjustable from 25% to 75%"... Huh?

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
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Jan 21, 2010
This has been moved to the Homework help section.

If anyone is unaware of the requirements of this section (especially the policy on asking and answering questions) please read the notes...

The missing schematic would be helpful...

5. ### Aisha.Raj

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Feb 16, 2013
sry about the attachement and the wrong section posting!!!

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6. ### Aisha.Raj

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Feb 16, 2013
WE neednt vary duty cycle continuously....a provision to adjust duty cycle along and separate provision to addjust time period is required....i js got that clarified from my prof!!

7. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I don't think the monostable will work. At least, I think D1 is backwards.

Edit: The monostable doesn't reset the capacitor charge when it expires. If it's retriggered soon after the pulse ends, the next pulse will be much too short. I think you need to tie the monostable in with the capacitance multiplier so that the actual capacitor is discharged very quickly at the end of the pulse.

The voltages on the op-amp inputs will go too close to the supply rails for a 741 to behave correctly. You need to change your 10K/90K divider to more like 30K/70K.

You can't connect your voltage limiting zeners straight across the op-amp output. At least put a series resistor in between.

I assume you're going to replace C1 with the capacitance multiplier... Can you show how they will be connected?

I can't see how that capacitance multiplier works. Where is its "output capacitance" connection point?

There's a simpler single-op-amp capacitance multiplier that should do what you want; see the Wikipedia page and various other places.

The concept of duty cycle relates to an oscillator, but does not relate to a monostable. If the monostable is being triggered at a fixed frequency, the duty cycle of the output signal is equal to the monostable's pulse width divided by the trigger interval, but from the monostable's point of view this is not defined as a duty cycle.

Last edited: Feb 17, 2013
8. ### woodchips

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Feb 8, 2013
First comment is to get out of the habit, right now, of joining signal wires in a circuit at a cross. Signal wires join at a tee, only, period.

Secondly, aren't you allowed to put caps in parallel and resistors in series? Makes it much easier!

Bob

9. ### KrisBlueNZSadly passed away in 2015

8,393
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Nov 28, 2011
More thoughts.

I don't know what you're talking about regarding duty cycle. A monostable simply produces a pulse of settable duration in response to a trigger pulse. Duty cycle applies to an oscillator. You seem to think it relates somehow to the trigger circuit. Please explain what you mean.

Regarding the questions in your original post.

1.
You don't need to use a capacitance multiplier in the C-R differentiator on the trigger input. A rising edge (of sufficient amplitude) from an external source will couple through C2 and D6 and trigger the monostable.

The monostable's timing capacitor, C1, is returned to 0V so you can replace it with a capacitance multiplier. I don't know how that capacitance multiplier circuit is supposed to work. There's a simpler single-op-amp version you can use to replace C1.

2.
The Miller effect is caused by collector-base capacitance in a transistor. There are no transistors in your circuit! But you will have to ensure that the capacitor in the capacitance multiplier is properly discharged when the monostable expires, so it will give the right delay the next time the monostable is triggered. If you don't do that, the capacitance multiplier capacitor will discharge slowly and if the monostable is retriggered before it has fully discharged, the output pulse will be too short.
The circuit won't work very well at high frequencies, no. Especially if you use 741 op-amps.

3.
10k is a good value for RI. The exact value won't make much difference because the forward voltage of a diode doesn't vary hugely with current. For example a 1N4148 might drop 0.55V at 1 uA and 0.75V at 100 mA. Not a huge difference in voltage for a current change of five orders of magnitude.

4.
Yes there are several problems; see my earlier post.

10. ### Aisha.Raj

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Feb 16, 2013
My apologies for creating such a big confusion....... my design question has been modified

I now have to design a circuit which provides a train of pulses of width ranging from 0.1ms to 1s of duty cycle that can be varied from 25% to 75%

As suggested i have decided to go with the astable multivibrator configuration. My trigger is actually giving supply to the op amps , ie when i give a signal to an electronic switch , it has to close a circuit which provides the Vcc to my astable multi-vibrator. I was given an idea to use components like CD4051,CD4052 , CD4053 . Can someone explain what they are???? I tried googling it but i couldnt understand much from the datasheets!!!!

But yet again i gotta use a capacitance multiplier to amplify and vary my capacitance value ........
How do i vary dutycycle now ???

I finally have got hold of the capacitance multiplier circuit.......Could someone please find the transfer function of it ( Relation connecting input and output ) ??? I would be grateful if someone can give me the derivation of the transfer function, as i am not able to zero in on it even after several attempts!!!!!

I am attaching the circuit along

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11. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
So the "trigger" is really an enable input?

The 4051/4052/4053 are analogue switches. There are simpler ways to switch the power supply rails to an op-amp - transistors, for example. But it's easy to stop an astable multivibrator by just clamping the capacitor voltage, using a transistor or MOSFET. I wouldn't bother interrupting the supply rails to the op-amp.

Varying the duty cycle is easily done with a comparator (or an op-amp used as a comparator) which compares the ramp at the capacitor with an adjustable threshold voltage. The output will be a square wave with adjustable duty cycle.

I don't see how that capacitance multiplier circuit works. It may do, but you need to document it, or at least cite where it comes from. It has an input and an output; I don't know how that is supposed to fit wtih an astable multivibrator, unless the input controls the multiplication ratio? That would be pretty cool, but I would like to see a description of how it works.

In any case the circuit is incomplete because the input is shown directly on an op-amp input that also receives feedback current through R1, but there is no resistor in series with the input. That's only appropriate if the input is a current and not a voltage.

As I said earlier, there's a simpler capacitance multiplier in the Wikipedia article on Capacitance Multiplier. Why don't you use that?

What are the restrictions on what components you can use? Can you use any ICs apart from "a microcontroller or dedicated timer IC"?

12. ### poor mystic

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Apr 8, 2011

A better solution than a capacitance multiplier might be to charge your capacitor from a constant current source. You can easily cobble one up from an op-amp.
An advantage of doing things this way is that capacitor voltage increases linearly over time.

Last edited: Feb 19, 2013