Is R2 needed on this circuit?

Question

I am working on a circuit to drive some water pumps. I intend to achieve a double galvanic isolation between the pumps and the CPU to ensure nothing bad will happen to the CPU.

Here is my schematic:

I am using a 4N35 optocoupler to achieve the first stage galvanic isolation between the 5V control signal and the relay driver circuit.

The relay driver circuit consists of a BC337 NPN BJT(Q1) a flyback diode (D1) and a resistor to the base of Q1 (R2).

As Q2 is intended to be used in saturation mode, I learned that Ic = VCC/RC, in this case VCC is the relay supply which is 5V and the coil resistance is 70Ω with 10% error, hence the worst case resistance should be 63Ω, knowing this Ic = 5/63 = 79.3mA.

Knowing Ic is now posible to solve for Ib, Ib = Ic/Beta, I am using the lowest HFE found on the BC337 datasheet, this value is 60.

Hence Ib >= 1.32 mA.

So now I know that I need at least 1.32 mA feeding through the base of Q1 for it to be properly saturated. The thing is that I don't know if R2 is needed on this circuit, because the optocoupler is already sourcing current.

If R2 is needed, how can I solve for its proper value?

Answer 1

Your beta is actually closer to 100 for this use case (Ic = 100mA) The optocouplers CE junction voltage drop should be small, but lets say it drops 0.2V,

So 5V - 0.2V - 0.7V (transistor base) = 4.1V

79.3mA / Beta of 100 = 0.793mA

4.1V / 0.793mA = 1265 Ohm

So R2 should be about 1.2K

You do want to include R2, otherwise the transistor base looks like a diode to the 5V rail, and may damage both the transistor, the optocoupler and possibly your power supply by drawing too much current.

Answer 2

Yes, you need R2 to limit the base current, and the current through the output transistor of the optocoupler. If it weren't there you would risk destroying one or the other of those components.

To calculate the needed value take the saturation voltage of your optocoupler output at your LED current, or look at the curves to see where Vce should be, then add the Vbe voltage of the BJT at your desired base current.

Subtract that value from the relay supply voltage and that's the voltage across R2. Size R2 to give your desired base current plus whatever safety margin you feel comfortable with, as long as it's below the max ratings of your parts.

Answer 3

R2 stops the base emitter of Q1 from simply shorting via diode to ground. It need only prevent the current from being so large something fries while stil being large enough to saturate Q1. You can assume the opto output NPN is also saturated in your calculation.

You know your minimum Ib required. You can look up the max current through opto or Q2 and pick a reasonable Ib in between then calculate how much excess voltage appears across R2 after \$V_{be.Q1}\$ is removed from the supply voltage.