Question on these Resistors

[nickey 0]

In examining my fish tank aerator circuit, I am uncertain about the use and value of a couple of resistors. (I acknowledge I am a tinkerer and not an EE)..

 

1. What is the purpose of R2? is it needed? (I copied its use from a different relay driver circuit)

 

2. Though they both seem to do the job, how would I calculate the correct values of them? (I just started with them first as is)

 

thanks

 

water pump relay.doc

[RobG 1,892]

It looks like R2 is there to make a voltage divider, to make sure VBE is low enough to keep the transistor in off state.

I bet you got that from some TTL design, because in TTL, low level can be as high as 0.8V, and that would be above transistor's cut-in voltage (typically 0.6V for bipolar transistors) making it conduct.

That said, you do not need R2.

 

You do not need R1 either, it is used to limit the current, but in this case it is not required.

MSP430's outputs are current limiting outputs. If you look at the spec sheet, in the "Typical Characteristics

[zeke 693]

Just to back up what Rob is saying...

 

Find out how much current the relay requires to stay active. I think they call it Holding Current. That's how much current the NPN has to sink. Calculate the Collector resistor required. If Vcc = 12V and Ic = 10mA (this is the relay's holding current) then Rc = 12/0.01 or 1200 ohms.

 

I also agree with Rob, the biasing resistors shown are not required because you want to saturate the NPN into the conduction state. The only resistor that you should have is a Collector resistor to limit the current flowing through the NPN's collector.

 

I would also make sure to use the MSP430's internal pullup resistors on the I/O pin driving the NPN.

 

I hope this info is helpful.

[RobG 1,892]

Zeke, you should be looking for actuating current, not holding current.

Also, I would say no to internal resistor, outputs are push-pull so resistors would only drain unnecessary current.

[nickey 0]

Thanks for both of your answers, that helps a lot. I will implement your suggested changes.

[nickey 0]

Ok, that just created a new question. I measured 178ma through the relay coil when activated and so I came up with a calculated 68 ohms for a collector resistor.. but when I put it in the circuit, the relay wouldn't trip any more.. I had to drop down to a 22 ohm resistor before it would engage, and after only a few seconds the resistor (1/4 watt) gets very hot.. don't think that will work well when the pump is on for 5 or 10 minutes at a time.

[RobG 1,892]

I measured 178ma through the relay coil when activated and so I came up with a calculated 68 ohms for a collector resistor...

That is holding current.

...but when I put it in the circuit, the relay wouldn't trip any more.. I had to drop down to a 22 ohm resistor before it would engage,...

Actuating current, which is needed to move the mechanical parts, is much higher than holding current.

...and after only a few seconds the resistor (1/4 watt) gets very hot.. don't think that will work well when the pump is on for 5 or 10 minutes at a time.

I don't think resistor in collector will work well, I don't think I have ever seen it used this way. Limiting relay's current is needed only when you need to preserve energy and it should be activated after the relay engages. For example one of the relay's contacts could be used as a feedback to the transistor to lower the current, but then the transistor will get hot.

 

That said, you do not need any resistors at all, the only component that is needed is a diode.

[nickey 0]

Sounds good to me.. thats the way I will keep it, with no resistors.. Thanks Rob

[zeke 693]

Take a look at this relay driving basics article.

 

It's a great summary on the topic.