I read this question's answer on calculating the base resistor value for a 2N3904 transistor. I'm in a similar situation however I'm using a TIP3055 to switch a \$V_{CE}\$ of 12V and a \$I_{CE}\$ of 250mA, with a base voltage of 5V.

I'm having trouble following the above answers calculations mainly because the datasheet for the TIP3055 doesn't have any values listed for \$I_{CE}\$ currents of 1A or less.

Can I just use the smallest values in that datasheet to calculate the resistor's value, or is there some other way to calculate the base resistor value when the listed datasheet values are way out of range?

EDIT: I just followed this post that suggested working it out by taking the base current as one tenth the value of the collector current. By following this I get the following values:

$$I_{CE} = 250\text{ mA}$$

$$V_{CE} = 12\text{ V}$$

$$I_{BE} = 250 / 10 = 25\text{ mA}$$

$$R_b = \frac{5V – V_{BE}}{I_{BE}} \\

= \frac{5 – 1.8}{0.025} \\

= \frac{3.2}{0.025} \\

= 128\text{ Ohm}$$

128 Ohms seems like a very small value. I could try it but I only have one TIP3055 on me at the moment and would rather not damage it.

## Best Answer

Why you would want to use a darlington transistor like the TIP 3055 on a small 250 mA load, but yes. Use the lowest hfe you see to calculate the base current needed. Any value resistor that would work on the Ice of 1 A will work just as well for 250 mA, putting the transistor in saturation. It will be inefficient on the base side though, so make sure your source can provide that current.