# urgent homework

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#### cherry

Joined Dec 17, 2006
13
Hi sexy guys and girls out there..

I am to design a square wave generator.

I am given the following components:

1. 230V/6-0-6V transformer with 3-pin plug
2. ac main supply
3. 1 diode 1N4001
5. IC 741 8-pin DIP OpAmp
6. dc dual power supply
8. capacitor 100uF
9. resistors: 100 ohm, 1k, 2.2k, 4.7k, 10k, 47k and 100k.

Requirements:
a. Transformer: Step down the ac main supply using the transformer above by connecting it to the mains.
b. Rectifier: Using the stepped down voltage, produce a negative half wave rectified output using a single diode, resistor 1k ohm and capacitor 100uF.
c. OpAmp Comparator Circuit: Using the rectified (ripple) voltage in (2), produce a square wave (50% duty cycle) using OpAmp comparator circuit. The Opamp IC is to be powered from +/- 15V supplies from the dual dc power supply. The rectified voltage in point (b) is to be fed into V+ and the reference voltage (Vref) is to be fed into V-. The reference voltage is to be tapped from the -15V supply by the Voltage Divider Rule; using the available resistors in point (9) above. The number of resistors used has to be as few as possible.

In addition, I need to draw the circuit in circuit maker software. Anybody knows how to use it?

And also, to answer the following questions:
What is the turn ratio for 230 V/6-0-6 V transformer?
Is the 230 V / 6-0-6 V RMS or peak value?
Show clearly the calculation/selection of resistor values for the comparator circuit in part (c).

#### Dave

Joined Nov 17, 2003
6,969
Have you done anything on this assignment? If so, please tell.

If you are expecting someone to throw you the answers on a plate you will be sadly disappointed. You are fully expected to dictate the course of your own learning.

Dave

#### cherry

Joined Dec 17, 2006
13
look. of course i have.

but i am not sure how does a 230V/6-0-6V work.

is 230V is RMS?

after connecting the transformer to the mains, i took 6V (i.e. red probe on 0V and black probe on 6V) from the transformer.
yet i obtained a 7.246 V reading from my multimeter. aren't i supposed to get 6V from the transformer? why am i getting 7.246 V?

the rectified output voltage i obtained is -8.731 V DC. i got this output waveform (ripple) with min at -9.2 V and max at -7.8 V. Up to this point i am not sure i've got the correct readings.

and as for circuitmaker i cannot find this transformer 230V/6-0-6V... the most i can get is 10TO1 Centre Tapped, which i can obtain 12V from it. How am I supposed to get 6V?

#### JoeJester

Joined Apr 26, 2005
4,390
Cherry,

... the 230 / 6-0-6 tells me you have a step down center tapped secondary with 6 volts between the 6 and 0 points. Most transformers are stated in rms values. When you quote a voltage, it would be nice if you indicated you used an oscilloscope or a dmm. Using the oscilloscope we can infer your stated the values as peak's while using the dmm, we can infer your stated values as rms.

... did you draw your schematic? This is one of the early steps for your project after getting the parts list and the project description.

You've read 7.246V on the secondary, what voltage did you measure on the primary? If you want to figure out a ratio, you'll need two measurements.

#### cherry

Joined Dec 17, 2006
13
hmm... thanks for the help Joe

Actually, i have advanced through the stage where i have to use an opamp as an comparator. the ripple is between -7.8V and -9.3V(peak). >> refer to the attached pix.

Next I have to input the ripple into an opamp and produce a 50% duty cycle square wave. >>see the attachment.

I need a reference voltage, don't I? I set it at -8.6V(DC). It worked fine. But I dunno how to substantiate this result. Any mathematical formula that can be used?

Any help is appreciated

#### cherry

Joined Dec 17, 2006
13
and here's the schematic

#### dragan733

Joined Dec 12, 2004
152
look. of course i have.

but i am not sure how does a 230V/6-0-6V work.

is 230V is RMS?

after connecting the transformer to the mains, i took 6V (i.e. red probe on 0V and black probe on 6V) from the transformer.
yet i obtained a 7.246 V reading from my multimeter. aren't i supposed to get 6V from the transformer? why am i getting 7.246 V?

the rectified output voltage i obtained is -8.731 V DC. i got this output waveform (ripple) with min at -9.2 V and max at -7.8 V. Up to this point i am not sure i've got the correct readings.

and as for circuitmaker i cannot find this transformer 230V/6-0-6V... the most i can get is 10TO1 Centre Tapped, which i can obtain 12V from it. How am I supposed to get 6V?
6-0-6V is RMS value for Ismax. Without charge, always the voltage on the secondary of the transformer is > 6V

Rectified output voltage is always 1,41xVrms, without charge

To get 6V from 12V, on rectified output voltage place place two capacitors 100uF in series, and get 6V from the capacitor that goes to GND.

#### JoeJester

Joined Apr 26, 2005
4,390
Cherry,

Open the properties of your "transformer" and change the turns ratio to get whatever you wish for a turns ratio.

I'm curious on why you used the Capacitor, C1, in your circuit? If you wanted the line to pulse your comparator, all you would need is a diode and a load.

#### cherry

Joined Dec 17, 2006
13
6-0-6V is RMS value for Ismax. Without charge, always the voltage on the secondary of the transformer is > 6V

Rectified output voltage is always 1,41xVrms, without charge

To get 6V from 12V, on rectified output voltage place place two capacitors 100uF in series, and get 6V from the capacitor that goes to GND.
Hi dragan733. Thanks for your reply. You sure know a lot. But I'm just a beginner. I don't really understand what you were saying. Care to explain in detail? Especially about those in bold.

#### cherry

Joined Dec 17, 2006
13
Cherry,

Open the properties of your "transformer" and change the turns ratio to get whatever you wish for a turns ratio.

I'm curious on why you used the Capacitor, C1, in your circuit? If you wanted the line to pulse your comparator, all you would need is a diode and a load.
Hi Joe, I tried that (change the turns ratio to 20TO1 or whatsoever), but to no avail. The software only recognises all those ratio available (i.e. I can't create a ratio that I want). Unless you're telling me that my circuitmaker is rather outdated.

I agree with you, but the capacitor is a requirement. It's a must. We would then have to make use of the ripple voltage to produce square wave.

#### cherry

Joined Dec 17, 2006
13

#### cherry

Joined Dec 17, 2006
13
It's not. Thank you so much. I'll check it out tonight.

Btw Joe, any idea what dragan was saying?

#### cherry

Joined Dec 17, 2006
13
And do you know how much is the current flowing in the mains?
<-- 230V/50Hz?

#### cherry

Joined Dec 17, 2006
13
Are you familiar with the software? Coz I have this problem. I supplied a -8.57 V DC as a reference voltage, and it 'divided' the input waveform by two. But I didn't get a rectangular wave. I obtained a straight line at negative saturation... why? i have attached my drawing.

#### JoeJester

Joined Apr 26, 2005
4,390
cherry,

Yes I am familiar with the software.

Why did you choose -8.57 as the negative input to the op amp?

The peak output from the diode is only (6 * 1.414) - 0.7

Look at the attached circuit. I included the transient response for the various test points.

#### cherry

Joined Dec 17, 2006
13
Hi Joe,

I am unable to open the TSC file you attached.

For some unknown reason I didn't get 6Vrms in the lab. I obtained 10Vpeak, maybe due to the transformer's efficiency? And anyway I have successfully done the simulation just now.

But wonder which voltage level should I consider as a mark? (to calculate duty cycle).

Thanks for the help.

#### JoeJester

Joined Apr 26, 2005
4,390
Cherry,

Anyways, the duty cycle is the time on to the time off. Other then the initial turn-on, the duty cycle from say 50 to 100 mS is pretty consistent.

Graph one is the outputs I viewed on your schematic.

Graph two is the approximate levels the comparator changes state. You will note the time is almost equal ... aproximately 9.6 mS off, and 10.4 mS on [assuming the positive time is on].

The pdf are the contents of the file you couldn't open.

#### cherry

Joined Dec 17, 2006
13
Ya, I measured the input under the rated RC load. Does it affect the input?

Anyways, the duty cycle is the time on to the time off. Other then the initial turn-on, the duty cycle from say 50 to 100 mS is pretty consistent.
But then when do we consider it is 'on'??

Graph one is the outputs I viewed on your schematic.

Graph two is the approximate levels the comparator changes state. You will note the time is almost equal ... aproximately 9.6 mS off, and 10.4 mS on [assuming the positive time is on].

Thanks. I got the same waveform now. You seemed to be using 0V (onwards) as 'on'?

Merry Xmas!

#### JoeJester

Joined Apr 26, 2005
4,390
Cherry,

Merry Christmas to you too.

Typically you measure risetime from 10% to 90% of the leading edge of the pulse, fall time from 90% to 10% of the trailing edge and pulse width from 50% of the leading edge to 50% of the trailing edge.

You would have to expand a graph out in the attempt to get the pulse width at the 50% points. With that fast of a risetime, the duty cycle won't be much different from any point along the leading and trailing edges no matter which point you sample, although 50% is the point where they are suppose to be measured.

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