Switching a Relay Circuit with npn Transistor

[spitta]

Hello,
I want to redesign a relay circuit (thru hole to SMT) and I found this existing schematic (attached). The relay coil is energized by switching on a 3904 NPN transistor. I've used similar circuits before, but I've only used a single resistor at the base of the transistor. I'm curious why there is this extra circuitry (D1, R1, & C1), and why those particular component values were used. I'm assuming the reason for this extra circuitry is for protection, but I'm wondering if this would be absolutely necessary when I hook up my own circuit. If this protection is deemed valuable, I would like to know if there is any insight as to why the particular component values (R1 & C1) were used.

Also, what's the worst that could happen if I remove D1, R1, & C1?

[fourtytwo42]

The one thing you have not said is what you intend to drive the circuit with, e.g. an mpu, a switch, how far away etc etc

[spitta]

Yes, the circuit will be driven with a 12VDC switch. However I want compatible with anywhere from 5V to 24VDC. The switch may be up to the length of a house (100ft / 30m or so). I'm assuming the original designer was not worried about the voltage drop across the diode or power loss thru the resistor.

One goal is to save space, so eliminating D1, C1, and R1 would help. Of course I wouldn't want to do this if the circuitry seems necessary.

I have searched online for similar schematics driving a relay circuit to help explain the fundamental operation of the circuit, but can't seem to find one.

[langwadt]

it just slows down the turn off for less flyback voltage to be clamped by D9, It'll work just fine with out

[rfengg]

Also, what's the worst that could happen if I remove D1, R1, & C1?

If you remove D1 you run the risk of punching thru the collector emitter region of the BJT cos of the back emf voltage developed across the transistor Collector - Emitter due to the relay's solenoid.
C1 reduces the back emf as it holds the charge for a longer time across the transistor base and as V=Ldi/dt , the voltage is lower.
If you remove C1 and R1 , your switching times should actually get faster , but keep D1 to reduce the risk of collector - emitter breakdown.

[phil from seattle]

Also, what's the worst that could happen if I remove D1, R1, & C1?

If you remove D1 you run the risk of punching thru the collector emitter region of the BJT cos of the back emf voltage developed across the transistor Collector - Emitter due to the relay's solenoid.
C1 reduces the back emf as it holds the charge for a longer time across the transistor base and as V=Ldi/dt , the voltage is lower.
If you remove C1 and R1 , your switching times should actually get faster , but keep D1 to reduce the risk of collector - emitter breakdown.

D9 is the flyback diode, not D1. I suspect D1 is related to the source signal.  C1 will cause Q1 to remain in saturation longer. R1 bleeds C1 to allow Q1 to turn off sooner.  I'd bet that the original input signal was AC so D1 blocks the negative half of the signal and C1 prevents chatter on the relay during the negative half of the input signal.  Also, R1 will prevent spurious relay closure on power up.

[rfengg]

D9 is the flyback diode, not D1.


My bad......getting to the age where a 9 looks like a 1 and vice-versa