Telephone Ring Detector circuit

Thread Starter

yong

Joined Feb 23, 2008
27
Hi,

Can anybody help to depict the circuit in below link. Refer to page 9.
I would like to understand the purpose of each component, the technical aspect from the AC to DC signal and how the output relay is actually trigger, the purpose of R5 and C3 connected in parallel with Pin 5 and 4 of the phototransisitor of 4n27.Can someone please enlighten. Thanks in advance.

http://www.vellemanusa.com/downloads/0/illustrated/illustrated_assembly_manual_k8086.pdf
 

bertus

Joined Apr 5, 2008
22,277
Hello,

The ring-signal are pulses of 90-120 Volts.
The signal is put back to proprotion for the 4n27.
On pins 4-5 of the 4n27 the pulses will show up.
This will let the relays attract and fall off again.
R5 and C3 will hold the signal a little bit longer so the relays will not flap.

Greetings,
Bertus
 

Thread Starter

yong

Joined Feb 23, 2008
27
Thanks Bertus

For R5 and C3, you mention that the presence of theses components is to allow the relay to stay in the continuous position after the first toggle. Are you able to explain in a technical way?Current will flow from base of PNP T1 right down and split between 2 paths using KCL concept.I believe current flowing thru R5 will charge up C3 but I am not sure how this will actually prevent the relay from flapping.
 

bertus

Joined Apr 5, 2008
22,277
Hello,

This is what happens at a ringpulse :
The ringpulse will let the LED of the 4n27 blink.
The transistor if the 4n27 will conduct.
When R5 and C3 are connected, C3 will be discharged over R3 and the transistor of the 4n27.
When there is no ringpulse the C3 will be charged by R3,R4 and R5 (the timeconstant is much higher).
At the next ringpulse C3 is discharged again and so on.
When no ringpulses are coming anymore, C3 will be fully charged and the relays will fall off afther some time.

Greetings,
Bertus
 

Thread Starter

yong

Joined Feb 23, 2008
27
Any idea what is the 1 Meg ohm resistor connected in parallel with C1 for?

I need help on how to derive the calculation for the values.....anybody can help?
 

bertus

Joined Apr 5, 2008
22,277
Hello,

The 1M ohm resistor is to discharge the capacitor, so you will not get shocked when pulling the circuits from the line.

Greetings,
Bertus
 

bertus

Joined Apr 5, 2008
22,277
Hello,

The capacitor is a resistor for AC without the dissipation.
The resistor will dissipate heat if the value is much lower.
The resistor used across the capacitor is called "bleeder resistor" and is placed there for the safety.

Here is the complete schematic :



Greetings,
Bertus
 

Attachments

Thread Starter

yong

Joined Feb 23, 2008
27
How do we calculate the heat dissipation of the resistor. I would like to prove this theory using mathematical calculation.
 

Thread Starter

yong

Joined Feb 23, 2008
27
Can i also assume the cap will discharge slower if a large resistance is used here?Becos limited current is flowing thru resistor in this case.Therefore disspate less heat.
 

bertus

Joined Apr 5, 2008
22,277
Hello,

The capacitor will not disipate becouse the current and voltage are 90 ° out of phase.

To calculate the power used by the resistor you can use the folloeing formulas:

P = Vrms * Irms.
P = Vrms * Vrms / R
P = Irms * Irms * R.

Here is the page for DC equations.
http://www.allaboutcircuits.com/vol_5/chpt_1/1.html

For AC you must use the effective values (rms) to calculate the power.

If you want to know more theory take a look at the cource notes from this page:
http://technology.niagarac.on.ca/courses/etec1120/notes.htm

Greetings,
Bertus
 
Last edited:

Thread Starter

yong

Joined Feb 23, 2008
27
Hello,

The capacitor is a resistor for AC without the dissipation.
The resistor will dissipate heat if the value is much lower.
The resistor used across the capacitor is called "bleeder resistor" and is placed there for the safety.

Here is the complete schematic :



Greetings,
Bertus
I was jus wondering if since we are already using a 12VDC adapter, do we still need the peak rectifier circuit of D2 C2 and R4? I am not sure if that serves any purpose, correct me if I am wrong.
 

bertus

Joined Apr 5, 2008
22,277
Hello,

The diode D2 is placed there to protect the circuit against the wrong polarity of the power supply.

Greetings,
Bertus
 

bertus

Joined Apr 5, 2008
22,277
Hello,

C2 is a decoupling capacitor (normaly done afther a "polarity safety" diode).
R4 pulls the base of the transistor T1 high when no signal is present.

Greetings,
Bertus
 

Thread Starter

yong

Joined Feb 23, 2008
27
Hello,

This is what happens at a ringpulse :
The ringpulse will let the LED of the 4n27 blink.
The transistor if the 4n27 will conduct.
When R5 and C3 are connected, C3 will be discharged over R3 and the transistor of the 4n27.
When there is no ringpulse the C3 will be charged by R3,R4 and R5 (the timeconstant is much higher).
At the next ringpulse C3 is discharged again and so on.
When no ringpulses are coming anymore, C3 will be fully charged and the relays will fall off afther some time.

Greetings,
Bertus
Just trying to understand, when there is no ring pulse and when c3 is fully charge i assume no current flows from the base of T1. i dun understand how did the relay fallout in this case. can explain the rationale behind?
 
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