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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.

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  • \$\begingroup\$ Plugging the numbers into this transistor base resistor calculator gave a value of 94 Ω, not far off from what you calculated. \$\endgroup\$ – tcrosley Oct 16 '14 at 9:08
  • \$\begingroup\$ @tcrosley I swear I went through all 40000 pages of Google looking for one of those... thanks! \$\endgroup\$ – jduncanator Oct 16 '14 at 10:08
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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.

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  • \$\begingroup\$ The only other transistor I have are 2N3904's which can't switch a current of 250mA sadly. The source is an Arduino which can provide a max of 40mA per pin, so I'm getting the picture that it may not be enough current. \$\endgroup\$ – jduncanator Oct 16 '14 at 1:24
  • \$\begingroup\$ You can try over driving the 3904. At worst It might blow the transistor, or shorten its life, but the Ice max isn't a hard limit, its a design limit. 50mA over isn't too much. The other option is using two in parallel, with the bases tied together. \$\endgroup\$ – Passerby Oct 16 '14 at 3:46
  • \$\begingroup\$ @jduncanator see electronics.stackexchange.com/questions/77045/… for more info on that. Two 3904s with a shared base resistor, and a small series resistor on each emitter will help. \$\endgroup\$ – Passerby Oct 16 '14 at 3:49
  • \$\begingroup\$ Thanks for that, seems quite interesting, didn't even cross my mind about using multiple transistors in parallel! I did manage to get the TIP3055 working with a 120ohm resistor in the end! Whilst overkill it works nicely! Thanks for your help! \$\endgroup\$ – jduncanator Oct 16 '14 at 5:21

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