The only possible way for it to pull the relay at 1.3 V on the base (I assume to circuit common) of Q1 is that the zener is conducting in reverse with nearly zero bias, which means it is defective, possibly shorted.

 

The only possible way for it to pull the relay at 1.3 V on the base (I assume to circuit common) of Q1 is that the zener is conducting in reverse with nearly zero bias, which means it is defective, possibly shorted.

But the Zener is connected to the base of the Q2. Q1 turns on....I pulled the base of Q1 down to common using some alligator clips and a resistor the relay energizes but there is a slight delay.

 

hi Yami,
Do you have a link to where you got that awful circuit.?
E
Do the maths for LDR Base drive., considering two transistor Vbe's in series.

Vb= 22* (1.2/(1.2+22))

 

hi Yami,
Do you have a link to where you got that awful circuit.?
E
Do the maths for LDR Base drive., considering two transistor Vbe's in series.

Vb= 22* (1.2/(1.2+22))

I don't know the make this is how I got it. Just a follow up question please refer to the schematic doesn't R4 (connected to the base of Q1) have an impact on that equation?
Thanks

 

Voltage across the voltage divider R3 is 6.4V when its light, when its dark it only goes to 5.6V
When the relay turns on regardless of the amount of light
Voltage across the voltage divider R3 is 5.9V when its light, when its dark it only goes to 4.8V

Whichever....

Even if the LDR (R2) drops to zero ohms, the maximum voltage you should ever see across R3 is
22 * 1.2/(1.2 + 22) = 1.14 volts. That's probably not enough to turn on the Q1/Q2 pair. Are you sure of the component values listed? Lift one leg of R4 and measure the voltage across R3 with and without light.

 

hi Yami,
Typically electrolytic caps have leakage currents in the order 10's of microamps, so the voltage on Q1 Base will be a little lower than the junction of LDR and R3.
I suspect the thinking of the designer, is that R4, with that strange value of 1.36K was an attempt at time delay.???

Even if D1 is removed, ie: shorted the circuit will not work as expected.

E
Re-Check the component values and the copper tracking.

 

Neither transistor can turn on until the voltage across the capacitor is (2 x Vbe + Vz). If the zener conducts even a small amount of current below its nominal voltage, turn-on can be far lower than expected, but for the relay to pull at 1.3 V on the base of Q implies the zener is defective.

 

Hi guys, I made some mistake with some resistor values. I have now checked them with a meter(took one leg out and checked). Also the zener was shorted took it out. What would be a good replacement for that it is TZX 5V. only zeners I have is IN4748A which the breakover voltage is 22V! .
I was able to check the circuit with a 24V powersupply before the zener got shorted - it did work! but the time it takes to turn off the relay is a bit long I'm guessing due to the capacitor parallel to the base.
Here is the revise version of the circuit.

 

hi Yami,
I make the time delay approx 60 seconds.
E

 

hi Yami,
I make the time delay approx 60 seconds.
E

Hi E, how did you get that?
Thanks

 

hi Y,
Look up R/C Time Constant, 100uF, 510k.
E
https://www.electronics-tutorials.ws/rc/rc_1.html

 

T = 1.1 *RC so that's 56 seconds Approx .

Further note, i would replace the transistors with one MPSA14G Darlington, and put the zener in series with the Base feed.

 

https://www.digikey.com/product-detail/en/on-semiconductor/BZX79C5V1/BZX79C5V1-ND/977929 looks like a reasonable Zener.

 

Thanks everyone for your help and time. I would really like to know one more thing, please correct me if I'm wrong. Capacitor (C1) charges through R1,R2 and R4 right? and it discharges through R3 and R4 right? The reason I ask this because I notice that the transistor turns on and off at different rates. So when calculating the time constant should I include the other resistors?
Thanks so much

 

C1 also discharges through the transistors B/E junctions ,,

 

Effectively, yes. Technically you would need to create a Thevenin equivalent of the combination of R1, R2, R3, and R4. The cap charges and discharges through that equivalent resistance to the Thevenin equivalent voltage. And that is neglecting the small base current into Q1.

 

Effectively, yes. Technically you would need to create a Thevenin equivalent of the combination of R1, R2, R3, and R4. The cap charges and discharges through that equivalent resistance to the Thevenin equivalent voltage. And that is neglecting the small base current into Q1.

So Rth should be use when calculating the time constant? T = R*C (R = Rth?)

 

So Rth should be use when calculating the time constant? T = R*C (R = Rth?)

Yes, but the voltage across the cap only needs to vary by a few tenths of a volt to go from off to on (and vis-versa). So the time delay may only be a few tenths of the RC depending on how much the Vth has changed with the change in lighting on the LDR.

So many variables here, even ambient temperature will have some effect..... Best way to determine delay time is experimentation.