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Author Topic: Please help me understand this circuit schematics (Door Alarm)  (Read 19509 times)
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promach
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« on: September 12, 2013, 04:51:35 16:51 »

I have a working assembled door alarm kits.
But I do not understand the working concept and principle behind.

Anyone ?

Please find attached the schematics and my door alarm.
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titi
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« Reply #1 on: September 12, 2013, 06:50:14 18:50 »

Hi promach,

The first stage (Q1) is a High Frequency oscillator, they used for R3 a quite large value (10k) so the oscillator is quite instable.
You can tune the stability with VR1, the wire connected to the door knob acts as an antenna, if some body is very close or touch the door knob, like R3 is large the oscillator stop oscillate.
When the oscillator is working (nobody touch) L1 produce negative voltage from +9v of the battery from D1 Side, so Q2 is satured, so top of R4 is at 9v and Q3 is bloqued.
If somebody touch, oscillator HF dont generate negative voltage for Q2 and Q3 is satured via R4, so LED is ON.
This validate IC1b and IC1a that is a SLOW oscillator (OSC1), OSC1 activate OSC2, that is BF Oscillator that produce a BEEP.
So when you touch the door Knob you produce a BEEP...BEEP...BEEP...

This sytem (Q1 Stage and Q2 + Q3) can be used to activate a ligth or something else without contact,
by example if the wire is connected to a nude PCB cooper side and on the other side a piece of glass or plexiglass, when your hand is close of the glass the light becomes ON.

I hope this helps to understand the working concept.

Here is an other concept.

http://www.pablin.com.ar/electron/circuito/varios/proximi2/index.htm

Best regard.
« Last Edit: September 13, 2013, 05:24:21 17:24 by titi » Logged
Parmin
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« Reply #2 on: September 13, 2013, 12:38:33 00:38 »

Ah yes, the old touch detector alarm.

You can also use capacitive touch system for similar job, it will be much easier for the hardware is replace almost totally with software.
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« Reply #3 on: September 13, 2013, 01:34:46 01:34 »

Ah yes, the old touch detector alarm.

You can also use capacitive touch system for similar job, it will be much easier for the hardware is replace almost totally with software.
LOL
I see you are a well travelled sonsivri... the old hotel door alarm...  I had a open door policy
don't stay in the room.. much better in the bar.
we was issued with one when we was working away from the uk for the MOD. later model was based on a pic chip... a door clamp was the better option...
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promach
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« Reply #4 on: September 13, 2013, 05:19:19 17:19 »

Hi promach,

The first stage (Q1) is a High Frequency oscillator, they used for R3 a quite large value (10k) so the oscillator is quite instable.
You can tune the stability with VR1, the wire connected to the door knob acts as an antenna, if some body is very close or touch the door knob, like R3 is large the oscillator stop oscillate.
When the oscillator is working (nobody touch) L1 produce negative voltage from +9v of the battery from D1 Side, so Q2 is satured, so top of R4 is at 9v and Q3 is bloqued.
If somebody touch, oscillator HF dont generate negative voltage for Q2 and Q3 is satured via R4, so LED is ON.
This validate IC1b and IC1a that is a SLOW oscillator (OSC1), OSC1 activate OSC2, that is BF Oscillator that produce a BEEP.
So when you touch the door Knob you produre a BEEP...BEEP...BEEP...

This sytem (Q1 Stage and Q2 + Q3) can be used to activate a ligth or something else without contact,
by example if the wire is connected to a nude PCB cooper side and on the other side a piece of glass or plexiglass, when your hand is close of the glass the light becomes ON.

I hope this helps to understand the working concept.

Here is an other concept.

http://www.pablin.com.ar/electron/circuito/varios/proximi2/index.htm

Best regard.

Why is first stage (Q1) a High Frequency oscillator ?  Is it a derivative of Colpitts Oscillator without the feedback loop?
What is the purpose of R3 ? Why will large R3 value lead to unstable oscillation and how VR1 can help stabilising the oscillation ? Besides, why will L1 produce negative voltage at D1 side when the oscillator is working (nobody touch) ?
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titi
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« Reply #5 on: September 13, 2013, 07:44:42 19:44 »

Hi promach,

Why is first stage (Q1) a High Frequency oscillator ?
This kind of circuit is called Miller oscillator, it is a very simplified oscillator generally used with crystal,
usually L1 has a capacitor to make a LC circuit, in this circuit C is stray capacitance.
How this circuit can oscillate without feedback loop ?
The feedback is done by Miller capacitance.
The Miller capacitance is a parasitic capacitance between Collector and Emitter
See this for more explanations : http://jaunty-electronics.com/blog/2012/08/low-noise-oscillator-design/

Is it a derivative of Colpitts Oscillator without the feedback loop?
Yes, as you can see here (except C1 on first drawing of the site) http://wikipedia.qwika.com/en2fr/Colpitts_oscillator

What is the purpose of R3 ?
R3 strict minimizes the current for the operation of the oscillator.
The oscillator generate electro-magnetic waves via L1, when you put your hand close the door knob,
you absorb energy that tend to increase current in oscillator.
But R3 is so high, that if current increase, voltage between 9v and the emitter Q1 decrease and oscillator turn off.
When I use the term "instable" it is not for the frequency but for the operation of the oscillator (that stopped or run)

Why will large R3 value lead to unstable oscillation and how VR1 can help stabilising the oscillation ?
The oscillator has just enough power to oscillate, VR1 is used to find the good biasing that permit the oscillator to oscillate, this permit to tune how far it can detect you hand from the door knob (sensibility of detection)

why will L1 produce negative voltage at D1 side when the oscillator is working (nobody touch) ?
When the oscillator is working Q1 and L1 act as a Buck–boost converter that produce enough base current in Q2 to put it in conduction.
see : http://en.wikipedia.org/wiki/Buck%E2%80%93boost_converter
When I tell negative voltage, this is seen from 9v rail.

Hope this help a bit more.

Best regards.

 
« Last Edit: September 14, 2013, 12:12:27 00:12 by titi » Logged
promach
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« Reply #6 on: September 18, 2013, 03:49:36 15:49 »

I use some simulation software and replace the bare wire loop with a capacitor.

I have problem understanding how L1 and Q1 act as buck-boost converter.
Besides, how do I make current flows through diode D1 in the simulation ?

Please find attached the simulation capture.
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titi
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« Reply #7 on: September 18, 2013, 08:10:28 20:10 »

Hi promach,

don't use 1 Farad to simulate the Wire to the the door knob, it's some picofarads and simulator don't simulate the antenna working.
Only special High frequency simulator can simultate correctly this circuit.
To simplify the explaination, it is an oscillator, so L1 and its stray capacitance form a LC resonant circuit.
Across L1 you have a sinus wave like on this page : http://www.sonelec-musique.com/electronique_realisations_indic_niv_hf_001.html
If you put a diode from on terminal of L1 you will produce positive voltage like the previous web page, or negative voltage from the 9v rail, if you reverse the diode like this circuit.
When the oscillation exceed near 1.2v volt peak to peak (0.6v VBE Q2 + 0.6v Vdiode) the transistor Q2 become satured and the R4 voltage = 9v.
If you touch the door knob, no more oscillation, so 0v across L1, Q2 is open, Q3 has a base current through R4 so become satured,
commun of S1 is near 9v, the LED lights,  and Beep...Beep...Beep...

Best regard.
« Last Edit: September 19, 2013, 06:50:23 18:50 by titi » Logged
promach
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« Reply #8 on: September 23, 2013, 06:43:00 18:43 »

As far as I know from http://www.markallen.com/teaching/ucsd/147a/lectures/lecture3/10.php
I am a bit confused with the 1.2V peak-to-peak and saturation of Q2. Can u tell me a little bit more about this ?

Also, when bare wire loop is touched, voltage across L1 is 0V, than Q2 is open. Why and How?

Besides, which high frequency simulator can do the simulation correctly ?

How do we guarantee Q3 will not saturate when I do not touch the bare wire loop and vice-versa/opposite?

Thanks for your help so far.
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titi
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« Reply #9 on: September 24, 2013, 08:36:25 20:36 »

Hi promach,

I am a bit confused with the 1.2V peak-to-peak and saturation of Q2. Can u tell me a little bit more about this ?
If you want that Q2 become satured, you need have a iBase current of Q2 that goes from Base to ground (like Q3 with R4).
The minimum voltage to have a current that goes from Base Q2 to ground is VBE=0.6v and VDiode=0.6v (that give voltage across L1 equal 1.2v)
Generaly we use 0.6v for explaination (for Vdiode or VBE) but its more from 0.65v to 0.7v

Also, when bare wire loop is touched, voltage across L1 is 0V, than Q2 is open. Why and How?
With a big value of 10k for R3, Q1 has just enough power to works, when you touch the bare, the oscillator stopped oscillate (see explainations from previous messages).
If there is no oscillation accross L1, you have no voltage because L1 become a short-circuit for DC voltage, so less than 1.2v across L2, Q2 could not satured.

Besides, which high frequency simulator can do the simulation correctly ?
This kind of simulation is very complexe because when you approaching hand from barre wire you have a hand effect, you change a little the capacitance from LC oscillator, and you absorbs a little power from electromagnetic wave produce by the circuit.
So it is very special to simulate, and I don't know a name for these simulators.

How do we guarantee Q3 will not saturate when I do not touch the bare wire loop and vice-versa/opposite?
If Q2 is saturated, so VBE Q3=Vsat Q2, near 0.2v, so Q3 cannot be satured.
To have the guarantee that Q2 is satured you need have a good inductor that has a good Quality Factor, see http://en.wikipedia.org/wiki/Inductor#Q_factor

This kind of circuit is quite simple (small number of components) but not so easy to understand !

Best regards.
« Last Edit: October 02, 2013, 06:49:09 18:49 by titi » Logged
promach
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« Reply #10 on: October 02, 2013, 04:34:39 16:34 »

I will build a hartley oscillator according to http://www.learnabout-electronics.org/Oscillators/osc21.php and test it out
and then understand its variant which is implemented in this door alarm circuit. What do u think ? It will take me some time but it is fun

Can you give me some explanation regarding what happen after IC1 receives input from pin 6 ?

What are OSC1 and OSC2 ?

What are the purpose of R7, R8 and R9 as well as C3?

What are Q4 and Q5 used for ?
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titi
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« Reply #11 on: October 02, 2013, 06:44:49 18:44 »

Hi promach,

Can you give me some explanation regarding what happen after IC1 receives input from pin 6 ?
Pin 6 is used to enable or disable OSC1, if pin 6 is logic level 1, OSC1 oscillate, level 0 not oscillate.

see: http://www.talkingelectronics.com/projects/DataBook1/DataBook1-42-62.html Audio Alarm, page 59


 
What are OSC1 and OSC2 ?
OSC2 is a square oscillator that make an audio frequency to do beeeeeeeeeeeeeeeeeeeeeeeeeeeeeeep.
OSC1 is a square oscillator that oscillate at very low frequency when is output pin 3 of IC1a is high, OSC2 do Beeeeeeeeeeeeeeeeeeeeeeeeeeeep
So OSC1 enable OSC2 at few herz and produce beep..., beep..., beep..., beep...., beep .......
See: http://www.eleccircuit.com/gate-tone-generator-by-ic-4011/



What are the purpose of R7, R8 and R9 as well as C3?
R7 and C3 define the frequency of OSC1 (very low frequency)
R8, R9, C4 define the frequency of OSC2 (audio frequency)

Remark: f = 1/0,6 R x C (in ohms and farads)

What are Q4 and Q5 used for ?
The CD4011 is a CMOS circuit so it only able to give few milli-amps and can't drive a speaker, so you need amplifier the weak signal.
Q4 and Q5 are use as in Darlington to multiply the Beta of Q4 and Q5.
So few milli-amps give hundreds milli-amps in Speaker, R11 limit current to protect the speaker and Q5.
See: http://en.wikipedia.org/wiki/Darlington_transistor


Best regards.
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promach
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« Reply #12 on: October 22, 2013, 06:55:23 18:55 »

Hi, could you tell me more how to derive the frequency formula of the NAND gate multistable oscillator ?

Also, why is there a resistor R8 (between the preceding NAND gate input and the capacitor) in the second oscillator while the first oscillator does not have ?
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titi
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« Reply #13 on: October 22, 2013, 08:11:41 20:11 »

Hi promach,

could you tell me more how to derive the frequency formula of the NAND gate multistable oscillator ?
The formula to calculate the frequency is : [freq = -2 * ln(1/3) * R * C] to simplify [freq = 2.2 * R * C]
To have more explication see the end of the document in attachment (sorry it is in french)

why is there a resistor R8 (between the preceding NAND gate input and the capacitor) in the second oscillator while the first oscillator does not have ?
This animation show you how CMOS OSCILLATOR works:
http://ressources.univ-lemans.fr/AccesLibre/UM/Pedago/physique/02/electro/multicmos.html
If you look at V1, you can see that the input of first gate has his voltage that reach -VCC/2, some people add a quite large resistor to protect gate from reverse voltage.
This give a better duty ratio because CMOS gate are protected against reverse voltage with 2 diodes, one from input to VCC and one from input to GND.

Generaly this kind of oscillateur works without this resistor. So it not mandatory to add this resistor.

Best regards.

« Last Edit: October 22, 2013, 08:21:55 20:21 by titi » Logged
promach
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« Reply #14 on: November 14, 2013, 03:05:18 15:05 »

Could you tell me about the small and large signal model of colpitts oscillator ?

Also, could you tell me how Q1 and L1 resemble a buck-boost as mentioned in the previous post ?
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titi
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« Reply #15 on: November 15, 2013, 10:34:07 10:34 »

Hi promach,

Could you tell me about the small and large signal model of colpitts oscillator ?
The calculus you get is too complex for somebody that start electronic.
Start with simple things like this:
http://www.sentex.ca/~mec1995/tutorial/xtor/xtor7/xtor7.html See LC Oscillators.
http://www.circuitstoday.com/colpitts-oscillator
http://www.electronics-tutorials.com/oscillators/colpitts-oscillators.htm
http://en.wikipedia.org/wiki/Colpitts_oscillator
Some Videos:
http://www.youtube.com/watch?v=R-bMTlVF0Uk
http://www.youtube.com/watch?v=5QSoFQs78vA

Also, could you tell me how Q1 and L1 resemble a buck-boost as mentioned in the previous post ?
A small drawing is better than long explanations. See drawing in attachment.
The curent is limited by R3 and C2, but don't forget that XC2=1/(C2*Omega), so XC2=1/(C2*2*PI*Frequency), so in high frequency the reactance of C2 is small (XC2= resistance equiv).

Best regards.
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« Reply #16 on: November 15, 2013, 11:48:22 11:48 »

Could you tell me about the small and large signal model of colpitts oscillator ?

Also, could you tell me how Q1 and L1 resemble a buck-boost as mentioned in the previous post ?

If the circuit is working then why do you to understand how this works, Sound a school/collage project to me and you want others do it for you  Grin Cheesy Wink
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« Reply #17 on: November 16, 2013, 12:31:02 00:31 »

Titi,
although I have learn all the above during my training years ago, it is very refreshing to catch up on all this again.
You have a very good way and great patience to describe things friend, and I wish to thank you beyond simply clicking the thank you button for your patience and clear explanation.
I do hope promach and others who gained from this recognize your contributions.

My best regards
Parmin
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