![]() TIP120 transistor is known for its high current gain (hfe = 1000) and high collector current (IC =5A) hence it is normally used to control loads with high current or in applications where high amplification is required. This increases the current gain and current rating of this transistor. The Darlington pair inside this transistor is shown clearly as its internal circuit schematic belowĪs you can see, there are two transistors inside this TO-220 package in which the emitter of first transistor is connected with the base of second transistor and the collector of both transistors are connected together to form a Darlington pair. If you need more voltage then you can check other transistor in the same family lit the TIP122. It can also withstand about 60V across its collector- Emitter hence can be used to drive heavy loads. It functions like a normal NPN transistor, but since it has a Darlington pair inside it has good collector current rating of about 5A and a gain of about 1000. The TIP122 is a Darlington pair NPN transistor. Note: Complete Technical Details can be found at the TIP120 datasheet given at the end of this page. Continuous Collector current (IC) is 5A.High DC Current Gain (hFE), typically 1000.This makes it suitable for medium and high power electronics like controlling motors, solenoids or high power LEDs.Ĭontrols the biasing of transistor, Used to turn ON or OFF the transistorĬurrent flows in through collector, normally connected to loadĬurrent Drains out through emitter, normally connected to ground It can switch loads upto 60V with a peak current of 8A and continuous current of 5A. 7 SO it does the amplification, But as I'm reading the output AC voltage from my alternator i notice a lot of noise and not a clean sine wave, could it be because of the amplification that there is noise introduced to the field coils.The TIP120 is a NPN Darlington Power Transistor. The behavior I noticed when i apply the control voltage to the op-amp now is that, the voltage seen at the field coils is greater than Vc. Op-amp -> BJT -> Darlington -> Darlington -> Field Coils. However, before i tried your circuit, I placed a BJT and ANOTHER Darlington in series with the Darlington of the original circuit. Hey CrossRoads i was about to implement the circuit you were going to use, I see how you state that it is easier to pull the current into the coils rather than push the current into it by placing it at the emitter. Would a Heat-sink be needed for this BJT also, since it is passing a big gain in current? Should i include a priming BJT to the Darlington and then retest the circuit?Īdditional thoughts: The Additional BJT will be powered by 15 volts at the collector. 1956/.02 = 9.78 A/A I'm assuming this is when the gain starts failing because of the limitations of the op-amp to output a max of 20 mA, because we expect about 600 to be the gain when operating correctly. Re = 23 ohms, So this is when the darling ton begins to fail. I started with 44 ohms and worked my way down to 23 ohms. I applied a control voltage to the op-amp non-inverting terminal, then i measured the Voltage at the Emitter while varying this resistance. Hey So i followed Rugged Circuits Method of dropping the resistance until i see a a drop in Voltage at the emitter Is there any method to increase this voltage up to about 12 volts? I have supplied the datasheet of the transistor, and a copy of the image I believe that there must be something in the data sheet of the Darlington i must not understand that is giving me such a huge voltage drop. The voltage drops from 12 volts all the way down to 1.4 volts. However, when i connect the VE to alternator coils, which i measured to be about 4 ohms. When there is no Load attached to the emitter, this is exactly what happens. What the expected behavior of the circuit is suppose to be is that when i apply a control voltage(Vc) to pin 12, the collector to base voltage(Vcb) will drop the required amount of voltage so that the voltage at the emitter(Ve) will be the same as the control voltage.įor example if i applied 12 volts to VC, i should expect that the Vcb will drop 3 volts so that VE will equal 12 volts. Seen in the bottom picture is the schematic I am testing.The op amp is powered with +/- 15 volts and the collector of the Darlington is powered by 15 volts also. I'm trying to figure out an issue I have with a Darlington Transistor.
0 Comments
Leave a Reply. |