# Help with figuring out cause of current drop on TIP 31 NPN transistor in project

1. Dec 1, 2009

### SF49ER

Im working on a project that involves me making a electrical circuit that will supply initially around .52 amps of current for and then drop the current to at least .40 amps of current to a actuating wire that uses current to cause wire to contract. The way i have my circuit made is that i have a 6v supply that sends this voltage to a 556 timer and four resistors connected to the collector of a TIP 31 NPN transistor when the supply is on. The timer circuit is made to have a time period of 1.47 with a sine wave that is high for .77s and low for .7 s.The output of the 556 timer is connected to the base of the TIP 31 transistor with an resistance to supply current to the base. The emitter of TIP 31 is connected to ground.The collector has four resistors that supply about .72 amps of current to my actuating wire that has a resistance of about 4.5 ohms resulting in about .48 amps across actuating wire.
So when the timer is on, and the current drops across the collector of the transistor causing a drop of current to my wire to about .38 amps across actuating wire but goes back to .48 when the timer is off.
Another i noticed is that as the current decreases across the base of transistor, more current dropped to my wire when the timer is on. ive attached the circuit from multisim so u can see what it looks like.
My questions are
1.why does the current drop when the timer is on
2.why does the current drop increase as the base current decreases
3.How can i calculate the cause of this drop so that i can find the right current across the base to give me the exact drop of current i want when the timer is on (.40 or .41 amps)
Thanks in advance if anyone has any insight on this matter

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2. Dec 1, 2009

### rolerbe

Assuming this is not a homework project.

Some quick observations from static analysis with simplifying assumptions. Assume a perfect transistor, perfect battery, etc. With the transistor OFF, you have a basic voltage divider. 12.75 total load ohms, and therefore 480ma current. That's what you are getting. So far so good.

Now, turn the TIP ON. From the spec sheet, the VCE SAT is 1.2 volts max. So, not a perfect ground (if it was, you would get NO current through the 4.5 ohm load). A 1 volt VCE ON (you can directly measure this) would give you 222ma through the load. You are getting about 380ma which would imply VCE in your circuit of 1.7 volts. So, a little more current to the base and you should be able to reduce the TIP ON load current some, but probably never less than about 200ma.

Make sure the 556 you are using can source more current (and if not reverse the whole load topology and reverse the logic), and reduce the base resistor. Try half the base resistance value, say 2.2Kohms (a standard value).