#### Hurdy

Joined Feb 27, 2006
137
Thank you. Though I have the circuit working fine now once I had put a 70 ohm resistor on the emmiter.

#### Papabravo

Joined Feb 24, 2006
13,735
Originally posted by Hurdy@Mar 16 2006, 11:45 PM
I calculate this.

V = 5.5
I = 40ma

5.5 / (40x10^-3) = 137.5 ohms

I take it I have to take away my coil resistance from this?

137.5 - 125.7 = 11.8 ohms

Is this how you do it?
[post=15082]Quoted post[/post]​
I thougth I had posted a response, but I can't seem to locate it. Your calculatins are correct.

In a previous answer I tried to show the power in the 11.8 ohm dropping resistor.
It was (40mA)*(40mA)*12 = 19.2 milliwatts. At this power consumption a quarter watt resistor is more than sufficient.

I also suggested that the aapropriate value of the base resistor would be based on having a transistor with a minimum beta of 40. If the PIC output is at 5.5Volts and the base is at 0.7 Volts, then you need 1mA of base current to control 40mA of collector current. (5.5 - 0.7)/.001 = 4.8KOhms.

If you make the resistor smaller or the transistor beta is larger than 40 then the transistor is driven harder into saturation, but the external circuit will limit the collector current to the design value of 40 mA.