Cmos and Tip 42c

[zotdoc]

I am using a cmos logic circuit running on 5 volts to trigger 12 volt automotive relays. I have a bunch of Tip 42c PNP transistors that I got off ebay for a song. I know they are overkill for this application, but since I already have em, is there any harm in using them and are there any special tips I need to know about?

[dyarker]

Depends on relay coil current. With a min DC gain of 15 to 30, CMOS logic may not sink enough current for the base of a TIP-42.<p>If you already knew that, and already planned for a 2N2222 or 2N44xx or other between the CMOS and TIP-42, then you got good deal; use them.<p>Cheers,<p>[ April 26, 2005: Message edited by: Dale Y ]</p>

[zotdoc]

The relays I have are seimens 12 volt coil, 35 amp automotive relays. I can't find a data sheet anywhere, including the seimens site but similar relays have a coil resistance of 80 ohms or so. I'm about ready to try to make the logic pc boards. If anyone knows a good way to select the proper transistor or if a small signal transistor would work please let me know. The longest time that a relay would have to be on is 45 minutes if that is a factor. Thanks.

[philba]

Here's a page on controlling relays with logic signals. http://ourworld.compuserve.com/homepage ... r_ctrl.htm<p>Take a DMM and measure the coil resistance. Just about any NPN transistor will do the job. You just have to figure out what kind of current gain you will need. Basically figure out how much current your logic can handle comfortably (i.e. well below the limits) and then divide that into the current draw of the relay coil (using ohms law to figure the coil draw). Any transistor that has an Hfe above that will work. <p>I always use 1 mA as my logic current and thus the coil draw in mA is the Hfe needed. In general, 2N3904 is my driver of choice because I get them in surface mount for $.04 each. A logic mosfet would be a better choice but they cost a lot more.<p>For long holding times, calculate the wattage through the driver and check against it specs. 80 mA * 12V =~ 1 watt so you better have a beefy transistor with a heatsink. I suspect your coil is much lower than that, though. To reduce holding power (and thus heat), you could use PWM for the holding current as drop out voltage (V below which the coil opens) is usually quite a bit lower than the rated voltage of the coil.

[zotdoc]

Thanks to all. The resistance of the coil on these relays is 90 ohms. I am going to use a npn signal transistor and connect it's emmiter to the base of the TIP42c, which is a PNP transistor. Then I'll hook the emmiter of the TIP to +12 volts and the collector to the coil of the relay, then the other side of the coil to ground. This should turn on the relay with a positive output from the cmos right?

[k7elp60]

I don't think that will work as the NPN signal transistor will not turn on from the 5V logic.
I suggest you put a 1k resistor from +12V to the collector of the NPN signal transistor. Connect the collector of the NPN signal transistor directly to the base of the TIP42 transistor. Ground the emitter of the NPN signal transistor. Put a 3.3K resistor from the CMOS logic to the base of the NPN transistor. When the CMOS logic goes high the relay will energize.

[zotdoc]

When I hook the collector of the 2n3904 to 12 V and then hook this to the base of the TIP 42C(PNP) wont this energize the TIP regardless of the logic signal to the base of the NPN? Thanks.

[dyarker]

Almost like you said, try: TIP emitter to +12V, TIP collector to relay coil, other end of coil to common. (Don't forget reverse biased diode across coil!) TIP base to a resistor to 2N3904 collector, another resistor from 2N3904 collector to +12V (stops TIP from being a little bit on due to leakage when 2N3904 is off). 2N3904 emitter to common. A resistor from base of 2N3904 to CMOS logic output (a bare base/emitter has low resistance and would overload the CMOS).<p>TIP collector current = 12V / 90 Ohms = 0.14A
min TIP-42 low current gain is 30 (spec sheet)
TIP base current = 0.14A / 30 = 0.0045A<p>TIP base/emitter volt drop plus 2N3904 Vsat maybe almost 2V. 12V - 2V = 10V across TIP base resistor.<p>TIP base resistor = 10V / 0.0045A = 2222 Ohms,
2.2K Ohms standard value.<p>2N3904 collector to +12V resistor is not critical, 10K Ohms should be okay.
It's current = 11V / 10000 Ohms = 0.0011A<p>2N3904 collector current = 0.0045A + 0.0011A = 0.0056A<p>2N3904 min gain is 100 (memory, recalc if different).<p>2N3904 base current = 0.0056A / 100 = 0.000056A<p>2N3904 base resistor = (4.5V - 0.7V) / 0.000056A = 67857 Ohms. But since 56uA is a really small current for a switching circuit, use 10K to 22K Ohm resistor; it'll work fine, last long time. (Or even 3.3K as mentioned by k7elp60.)<p>When on, the TIP will dissipate less than 0.2W, so minimal heatsinking needed. The most likely cause of failure will be coil kick-back voltage if you forget the diode.<p>Cheers,<p>[ April 29, 2005: Message edited by: Dale Y ]</p>

[zotdoc]

Thanks to everyone. I went home and breadboarded the circuit just like Dale and K7elP60 said. It works lime a charm. I used a single aa battery as my "logic source and it turned on the relay perfectly. I grounded the battery to the common of the circuit. I won't forget the diode reversed across the coil! Thanks again everyone for helping me understand this part of my project!
Zotdoc