Relay driver with Zener cutoff

Question

As seen in the schematic, I need to control a relay using two voltage sources.

The relay goes off when source Nr.1 (4.2V) is less than 3.6V AND when source Nr.2 (7.4V) is less than 6.8V, by the effect of the Zener diodes D1 and D2. The value of the desired cutoff voltage of the Zener diodes has been reduced in 0.7V, in consideration of the transistors Vbe drop.

The Zener diodes resistors (R3 and R4) were calculated on the basis of their maximum power (500mW), using the formula (Vs-Vz)/Iz. Then a series resistor of 22k biases the transistors, whose hFE is 140 (BC549A.) The formula used was (Vs-0.7)hFE divided by the relay current.

Is this well calculated? It is the first time I design a circuit and I'm not sure if the application of the formulas for resistor values is correct and if the transistor base resistors (R1 and R2) will interact badly with the Zener resistors (R3 and R4.)

Any suggestions or different approaches to it will be very much welcomed.

(This question follows indirectly another recent post of mine.)

Answer 1

First of all you dont expect you make a precise cut with simple zener circuit. The second you must know if you want AND or OR logic.

If you do an OR logic you can reduce design to one transistor (both logic to one base). For AND it is a good approach (two transistors in series).

A third relay should have a stable supply voltage (not floated).

A 4-th 22k cant open 40mA for supplying relay with considering Beta=140, you would need about 14V at top of R4.

A 5-th using two resistors in series makes no sense.