Any such current in the PNP base region would be multiplied by the transistor's gain (perhaps as high as 100 or more at elevated temperatures), and that would slow down switching speeds and be undesirable. The purpose of R1 is two fold it bleeds off any leakage current in the PNP transistor's collector-base junction, as well as charge stored in the base junction capacitance. For analog amplifiers, that statement is not true. As said before, for switching purposes, the values are not critical. So calculate the resistor value as 11.4 (that is 12 VCC - 0.6 VBE) divided by 0.02 amp = 5,700 ohms, and then pick a lower value such as 1K or 2.2K to ensure sufficient current in case of lower gains or lower temperatures (gain drops significantly with lower temperatures). For a 1 Amp collector current, and assuming a saturated current gain (gain varies with current) of 50, that suggests a base current of 20 mA. It's value is determined by choosing the base current which is dependent on the required collector current. The purpose of R2 in Ken's drawing is to provide base current to switch the PNP transistor. No one is wanting to "slap you back", we're just trying to answer legitimate questions. This means that at 1 Amp, a darlington would need to dissipate 1 Amp x 0.7 volts = 0.7 watts, whereas the non-darlington would dissipate 1 Amp x 0.1 volts = 0.1 watts the former requiring a heat sink at that current, and the latter probably not. ![]() Non darlingtons can saturate to better than 0.1 volts. ![]() Depending on load current, make R1 somewhere between 22K to 47K, and R2 from 1K to 2.2K (neither are critical).Īlso, it is best NOT to use a darlington transistor for the PNP power transistor as by their internal configuration they cannot saturate (switch on hard) to better than about 0.7 volts. If you follow Ken's second drawing, you will get the desired results. This is best accomplished with a second small NPN transistor with a grounded emitter configuration. In my post I said that the PNP emitter should be connected to 12 volts, and the load between the collector and ground.Īs further stated, this does require you to pull the PNP base to over 11.4 volts for shutting the transistor OFF, and below that to turn it ON. You have the emitter and collector reversed. You have the emitter of the PNP transistor at a lower voltage (through the grounded load) than its collector. The drawing in your schematic is incorrect. If not, then you should use a PNP transistor with its emitter connected to 12 volts, and the load between the collector and ground, and switch the base voltage (current limited by a resistor of course) from above 11.4 volts (12-0.6) for OFF, to below 11.4 volts for ON. ![]() Can you not rewire the load to be connected from the 12 volt source to the collector, and ground the emitter? ![]() Likely you are just switching it to only 12 volts, hence you are not saturating the transistor, and therefore develop heat.įinding a source for switching the base to higher than the collector voltage may be more trouble than its worth. In order to saturate (fully switch-on) the NPN transistor in this high side configuration, the base needs current supplied trhough a limiting resistor from a supply more than 0.6 volts higher than the collector voltage 12+.6=12.6 volts. You are using the NPN transistor as a "high side" switch (abnormal configuration) rather than as a "low side" switch, which as you say is normal, and I suspect therein lies the answer. You have answered the heat question yourself, but perhaps don't realize it.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |