Pull up on NPN transistor (during microcontroller power up)

Question

I am using an ESP8266-01 to control a 3.3v relay via a 2N2222A NPN transistor. The ESP8266 GPIOs use a 3.3v logic level.

The idea is that by setting GPIO2 to hi/lo I can saturate/cut off the transistor and control the relay. This works fine - so long as I connect GPIO2 to the transistor after startup, because GPIO2 needs to be held hi during power up in order to boot normally from flash (http://robertoostenveld.nl/esp-12-bootloader-modes/). I have some concept that I probably need to have a pull-up on the pin, but I am confused with how GPIO2 can be held hi without always having the transistor in the saturated state (ie how can I then make GPIO2 low...). What is the best approach here?

Also - on an unrelated note - I've noticed that in many diagrams of relay driver circuits using transistors, there is a resistor between GPIO and transistor. Any good reason for this? Everything is working as is but maybe there's something I'm missing.

EDIT: To clarify, the problem is that the circuit does work as intended if I let the ESP boot and then connect GPIO2 to Q1. The problem occurs if I try to boot while GPIO2 is connected (ie, as shown in the diagram I provided) in which case the GPIO2 is presumably being pulled low, which results in the wrong boot mode.

Answer 1

You must add a resistor between the GPIO output and the transistor’s base. If the transistor’s base was driven at 3.3V, it would drain a few amperes and either the transistor or the MCU would blow. If it did not blow, it probably means that something is limiting the current on the GPIO output, but I would not recommend depending on this.

If your MCU needs it, you certainly can add a pull-up resistor to GPIO2. As long as its value is high enough (10k is standard for me), the MCU can easily drive the pin low when requested so.

The schematic would be something like this:

^{simulate this circuit – Schematic created using CircuitLab}

Edit: R2 should be much lower than R1, but still high, so I fixed the values and replaces the single transistor with a darlington-like structure to have a high-enough gain to properly drive the relay.

^{simulate this circuit}

Answer 2

What you want to do is to connect your GPIO2 to the relay with this circuit. This will allow GPIO2 to pull up to near 3.3V at the power-on time but not click the relay. Then when the software in the MCU is ready to activate the relay have it output a '0' to the GPIO2 pin. The R1/R2 resistor pair have to be sized in ratio as shown so that the GPIO2 can actually be pulled high. (The circuit in the other answer has these backwards and does not allow GPIO2 to get to anywhere near to a valid high).

Answer 3

Clearly the Base-Emitter {read diode} junction of the NPN is pulling GPOI2 to ground. AND if the GPIO output is high there is no proper current limit for that output.

Even with 10k pullup the 1k series resistor would only have approx 0.3v across it {assuming it is grounded} however there will be around 0.5v Base-Emitter voltage as well so the end result will still be around 0.8v on the IO pin. Will that still create the boot up problem? It certainly is not what I'd call pulled up.

The suggestion of using a PNP transistor might be a good one if you have more voltage to play with, but you still want to include a series resistor for current limiting which will mean less voltage to drive the relay though.

In the end you would be better off using a FET as there will be no gate current and no need of a series resistor. You just need to decide whether to use high-side or low-side switching and therefore P or N channel type FET.

Since you want the GPIO pin High at first but the relay not activated then you should use a P-channel FET in High side switching configuration and pull the gate low to drive the relay. Don't forget the pullup resistor