pull down resistor for npn relay driver

10K would typically be fine. It is not critical.

Bob

 

The correct way around? (Just checking)

 

If the PIC output pin is active (either high or low) at all times, you do not need an additional pull-down resistor from the base to GND. If the PIC output pin might go to an input mode or tri-state (high output impedance), then a pull-down resistor will decrease the circuit's sensitivity to large radiated noise bursts, such as from arc welders on a production floor. The smaller the resistor, the less the sensitivity. BUT, the smaller the pulldown resistor, the more energy the PIC has to produce to turn on the transistor.

Starting with the PIC, its datasheet has its max rating for continuous output current and the associated high output voltage (it will not be equal to Vcc). Minus 0.7 V for Vbe, that gives you enough to calculate the smallest safe value for the series base resistor. Once you have that, calculate the base current and multiply it by 20 (I think 10 is too conservative for contemporary parts) to get the max collector current in hard saturation. If your relay needs less current, you can work backwards starting with the load current to determine a larger base resistor.

ak

 

Yes, it is all about loading effects. A perfect voltage source (power supply, opamp output, PIC output pin, etc.) has a zero ohm output impedance. The output voltage does not change even if the load resistance varies by 1000, or 1,000,000 to 1. It is a perfect voltage source in series with a zero ohm resistor. Why think of it that way? Because in the real world there is no such thing, and the most common difference is that that series resistor has a non-zero value. So it is in series with the load, or the equivalent resistance of whatever the load is, and the two resistances form a voltage divider. Now the voltage across the load varies depending on what the load resistance is. If (working from memory) a PIC output cah source up to 25 mA, but the output voltage sags down from Vcc as the output current increases, that is a sign of a voltage source output with a measurable ESR (equivalent series resistance).

If something else is trying to impress a voltage across the load while the PIC is connected, that low output resistance sorta kinda appears in parallel with the load resistance. Under some conditions, an external voltage source can drive excessive current back into the PIC causing damage.

Separate from that is a current source output, which has a theoretical infinite output impedance. That is, the current through the load does not vary when the load resistance changes. Again it is two resistances in series, but now one of them is (almost) infinity. So no matter what the load resistance is, it is so small relative to the total resistance that it does not change the load current. If this one seems harder to understand intuitively, don't worry. It is, and many people struggle with it.

Tri-state is an (normally voltage source) output stage condition where neither the pull up nor the pull down transistor are on. So there is no very low output resistance in series with whatever is connected to the pin. If that thing is a transistor base (or even worse, a MOSFET gate), then the base is floating, basically disconnected from the PIC, and can be influenced by strong electromagnetic fields (depending on what other components are connected, trace length and shape, etc.). If there is a chance that this condition might occur, then a resistor from the base or gate to GND will prevent a near-infinite impedance condition.

ak