Need help adding a power transistor to my 555

[adown]

Hi everyone, I have a timer circuit that is going to power my arduino micro and a relay. I am going to run the 555 output to a 7805, or 7806 reg. The nominal relay current is 120ma, and the Arduino is 22ma running code. Do I need to need to run a transistor between the 555 and the 7805, Or will the 555 output be sufficient for the voltage regulator. I have some TIP32AJP PNP power transistors 3A TO_220, but in my case, the smaller footprint is better. Not sacrificing reliability, but not over kill either. Here is the circuit, I'm sure someone will recognize it. Thank you

 

[crutschow]

If you are running the 555 from 12V then you could use a 2N2222 or similar NPN as an emitter follower at the 555 output to power the regulator and it's load. No resistors needed but you need a diode, such as a 1N4001, in series with the transistor emitter to prevent reverse base-emitter breakdown when the transistor turns off from the voltage on the (required) input filter capacitor at the regulators input.

 

[adown]

I guess the TIP is a little over kill. Now that you mention it, I have a couple dozen 2n4401's that spec real similar to the 2n2222. Is this what you had in mind? Thanks

 

[ChrisP58]

I guess the TIP is a little over kill. Now that you mention it, I have a couple dozen 2n4401's that spec real similar to the 2n2222. Is this what you had in mind? Thanks

View attachment 91293

No. As you have it shown the transistor will act as a dead short across the power supply, which will kill either the transistor or shut down the power supply.

For an emitter follower, the supply connects to the collector, and the load (your 7805) will connect to the emitter.

 

[Tony Stewart]

(EDIT)
The NE555 has a push pull driver capable of 200mA but with about 20 Ohms +/-5 or a 3V drop at 150mA or 450mW and overheat.

The Itty Bitty 1555 is the same 25 Ohm, Rce

So all you need is any N Ch Fet with an RdsOn of a few Ohms and a 1~10K Pullup resistor with reverse diode across the coil.
https://www.digikey.com/product-detail/en/2N7002P,215/568-5818-1-ND/2531105

$0.15 (SOT-23) 1.5 Ohms.

This will invert the output of course for low side driving. If not good, use P Ch and drive Coil on high side. Then 555 active low will pull gate on P FET low and pull coil ON on high side.

 

[koolguy]

its very simple just connect mosfet or darlington like TIP122 i.e. npn when base is +ve It will turn!

 

[Reloadron]

I would go with Tony's post #5. Since you have a pile of 2n4401s I would likely just use one of them to drive the small relay and let the relay power the LM7805.

Ron

 

[MikeMl]

Another idea. There are regulators that have a fourth pin called an enable or power-down input. Use a CMOS 555 (for lowest standby power) to control the enable input on the regulator....

 

[cowboybob]

Hi everyone, I have a timer circuit that is going to power my arduino micro and a relay. ...

Given your stated total current requirements (<200mA) and the Micro's on-board regulator, I don't see why you need to complicate the circuit with an additional external regulator, etc. The 555 alone should handle both loads.

Unless, of course, you're driving additional loads with the Micro (not stated in your post).

 

[adown]

Cowboybob,, thats what I want. An uncomplicated SMALL circuit, but I want it to work. All these posts are starting to confuse me. I have the arduino and 1 relay. I was going to try to drive it off the 555, but I've been accused of trying to write my own electrical laws before, so I want to get it right, not just what makes sense to me. lol

 

[cowboybob]

Trust your math.

The only concern of mine is as I said: any output of the Micro (to greater and lesser degree and the USB port, if used to power something) will, of course, draw additional juice.

Might help us to know what the total Micro system you've envisioned is expected to do.

 

[adown]

This circuit is only a delay off circuit. the arduino will normally be run on a 7805 with vehicle ignition on. on ignition off this power off circuit will power the arduino for one final code energizing one relay for 16 seconds then shut down.

 

[cowboybob]

This circuit is only a delay off circuit. the arduino will normally be run on a 7805 with vehicle ignition on. on ignition off "this power off circuit" will power the arduino for one final code energizing one relay for 16 seconds then shut down.

(My quote marks)

What is powering "this power off circuit" providing power to the Arduino, once ignition power is removed? And from your description, the Arduino is powering the relay you originally referenced? [On that note, the Arduino I/O ports can only drive about 40mA]. And are there two (2) relays? Where does the 555 circuit come into play?

Please excuse my confusion.

 

[adown]

The power off circuit will be directly connected to the battery, and the 555 switched by the ignition. There are 2 relays, but only one will be powered at time of power off delay. I had the circuit wrong, this is what I meant to do.

 

[cowboybob]

Your circuit above (post #14):

What is its purpose?

 

[adown]

The purpose of the transistor is to provide a voltage source with enough current to drive the 7806 with out pushing the limits of the 555 output. I will be around 140ma running at 14.3 volts. The 555 will synch or source 200ma at 15 volts, I just want some headroom, and I've been told it will provide impedance matching. The 100 µf cap is just a filter. I was too lazy to use the correct sign, but I can see that could cause problems in electronics.

I didn't mean to put resistor looking squiggleys on that cap and the 7805. I got in a hurry.

 

[cowboybob]

Try this instead:

R1 and R2 will need to be adjusted to accommodate the output level of the 555.

 

[upand_at_them]

Why is R2 needed? The 12V from the 555 ouput is fine driving the NPN. Also, don't you need some protection so the 100uF cap doesn't destroy the transistor?

EDIT: With almost 12V at the emitter, will the transistor even turn on at all? Don't you need to drop some voltage at the collector with several diodes to make it work?

 

[adown]

I honestly don't know why either resistor is needed at the base. The 100 uf cap should be a .33uf. It's just to level off transients. I have a couple circuits on my board, and I just can seem to keep this thread straight. A diode could be added to prevent damage on turn off, but I don't think thats what you were thinking.

 

[upand_at_them]

I'm not as experienced as many on here, but here's what I'm thinking...

Yes, a diode does need to be added to protect the transistor on shutoff. But the diode drops I was referring to at the collector are for something else.

An NPN can only turn on if the base is at a higher voltage than the emitter. Higher by the amount of B-E voltage drop (about 0.7V). This is why NPN's are used for low-side switching, not high-side switching like being done here. If the NPN is "full on" then the emitter voltage is only slightly below the collector voltage. So almost 12V. This means the base has to be at least 12.7V, which won't happen in your setup. And definitely won't happen with R1 and R2 creating a voltage divider.

So my thinking was to drop some voltage at the collector with diodes so that the emitter is low enough for the base to work.

Or, use a logic-level MOSFET?