Couple of questions regarding this photocell circuit.

Thread Starter

Nealieboyee

Joined May 24, 2017
44
Hi All,
I've got a couple of no name branded photocell or day/night switches. They run from mains and give you a switched live output. I've drawn this schematic from the PCB itself.

My questions:
1: What purpose do R7, and the RC circuit comprising R8 and C1 serve on the AC side of the rectifier?
2: What are R2, R4, R5 and C3 for? I think R2 is for keeping the base of Q1 low and R4 maybe limits the base current, but I'm not certain.
3: Zener diode ZD1 is a 4V diode that (I think) gives you 4V once the Phototransistor gets enough darkness to bring the voltage from divider Q3,R6 high enough to reverse bias the diode above 4V, allowing it to conduct. Is that correct?

As you can see, ZD2 is a 24V diode, so the circuit essentially runs from 24VDC. The relay is also 24V.

Thank you for your help.

Schematic_Photocell_2021-07-27.png
 

ericgibbs

Joined Jan 29, 2010
13,617
1: What purpose do R7, and the RC circuit comprising R8 and C1 serve on the AC side of the rectifier?
hi Neal,
R7 is a current limit
At 50Hz the impedance of the C1 is approx 6k8 ohms. so for 230Vrms mains that will allow a current of approx 33mA, that will be used to power the circuit, the R8 is C1 discharge resistor.

E
 

ericgibbs

Joined Jan 29, 2010
13,617
3: Zener diode ZD1 is a 4V diode that (I think) gives you 4V once the Phototransistor gets enough darkness to bring the voltage from divider Q3,R6 high enough to reverse bias the diode above 4V, allowing it to conduct. Is that correct?
hi N,
The resistance of the PTransistor increases as the the ambient light gets lower, so the voltage drop across R6 will fall and the ZD1 will be biassed On, switching on the relay.

E
 

ericgibbs

Joined Jan 29, 2010
13,617
Hi,
With that type of uninsulated supply from the mains, it can be electrically noisy.
You don't want the relay/light to flashing On/Off with electrical noise,
Also if the ambient light level is close to switching the light On/Off, the circuit could 'dither' On/Off, so it is a simple method of preventing that problem.
E

NOTE: This type of non insulated supply can give a lethal electric shock!
 

LesJones

Joined Jan 8, 2017
3,490
C1 is to limit the current to the circuit. A 470 nF capacitor will have a reactance of about 6.8 K ohms of 50 hz. This will limit the current to about 30 mA.
R7 is to limit the current pulse is the supply is switched on near the crest of the waveform. (Without it there would be a large current pulse trough the capacitor as the rate of change of voltage across it would be high.)
R8 discharges the capacitor when power is removed from the circuit.
C3 and R5 slow the response of the circuit to rapid changes in light level. R2 and R4 form a potential divider to control the light level that that causes the output to switch on and off. R3 gives the circuit hysteresis so it switches the relay on at a lower level (On Q1 base.) than the level at which it switches the relay off.
Your assumptions in question 3 are correct.

Les.
 

Thread Starter

Nealieboyee

Joined May 24, 2017
44
Thank you Eric and Les.

Les, that makes sense, except for R3. How would it give the circuit hysteresis? To me it looked like a simple current limiting resistor for Q2 and Q3. Could you elaborate, please?
Many thanks
Neal
 

Thread Starter

Nealieboyee

Joined May 24, 2017
44
hi Neal,
Can you confirm that the relay is energised when its Daylight and Off when dark,??

E
View attachment 244540
[/QUOTE
hi Neal,
Can you confirm that the relay is energised when its Daylight and Off when dark,??

E
View attachment 244540
It's actually energised during daytime, which opens the NC contact, disconnecting the load. Weird. So at night the relay is deenergised, and the NC contact closes again, connecting the load.
 

LesJones

Joined Jan 8, 2017
3,490
Re post #7
First consider when the the relay is not energised. The voltage across R3 is (220 x 24) /(220 + 22000) = 0.24 volts.
So for Q1 to remain conducting (Holding Q2 off) the voltage on the base of needs to be above 0.6 +0.24 volts = 0.84 volts.
Now consider when the relay is energised. The voltage across R3 is (220 x 24)/(220 + 1600) = 2.9 volts. So now the voltage on the base of Q1 needs to be above 0.6 + 2.9 = 3.5 volts for it to start to conduct (Stopping Q2 from conducting). (I am ignoring the saturation voltage of the transistors and their base emitter current as their minimum hfe value is 110.) You can see this in the graph on Eric's simulation.

Les.
 

Thread Starter

Nealieboyee

Joined May 24, 2017
44
Oh I see. So if I understand what you're saying, the voltage at the base of the transistor needs to be 0.6-0.7v with respect to the emitter. I always figured it was just 0.7 volts regardless of anything on the emitter.
 

LesJones

Joined Jan 8, 2017
3,490
An NPN transistor starts conducting between collector and emitter when the base is more positive than 0.6 to .7 volts WITH RESPECT TO THE EMITTER. (The base emitter voltage also varies with temperature.) The voltages of 0.84 and 3.5 volts are with respect to ground.

Les.
 

Thread Starter

Nealieboyee

Joined May 24, 2017
44
An NPN transistor starts conducting between collector and emitter when the base is more positive than 0.6 to .7 volts WITH RESPECT TO THE EMITTER. (The base emitter voltage also varies with temperature.) The voltages of 0.84 and 3.5 volts are with respect to ground.

Les.
Thats perfect, thank you Les!
 
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