Power to 555 pulse gen

[con-Fused]

First post - thanks for being here!

Trying to use a signal transistor as a switch to power a 555, collector to vcc. The 555 output will feedback to the trans base to bias it into saturation and keep the device turned on once its intially powered on with a momentary switch. The goal is that the device will turn off with zero power consumption once the 555 output pulse expires (few minutes after the switch is pushed). I don't want the 555 to counsume battery power once the pulse is completed. The transistor will run the 555 when turned on by the battery, but not when the 555 output tries to bias the base - which nearly the same as vcc. Is it negative feedback? 555 output is proportional to Vcc, collector output depends on the base input current even in saturation, and the whole thing turns off? Its instantaneous as far as I can tell.

I know I can solve this with a small relay replacing the transistor to decouple (but I dont have the right one), but Im just curious if my feedback theory is correct. Or more importantly, what am I overlooking. I'm a beginner.

con-Fused.

 

[KrisBlueNZ]

It's a kind of latching action so I would describe it as positive feedback.

But there is a problem with using an bipolar junction transistor as an emitter follower when you're driving the base from a circuit that is powered from the emitter. The base needs to be around 0.7V more positive than the emitter. If you drive the base from the 555's output, and the 555 is powered from the emitter, the transistor will turn OFF because there won't be enough voltage on the base to sustain conduction.

The answer is to use a PNP with its emitter connected to the positive rail from the battery, and its collector connected to the 555's power pin. The base needs to be pulled down towards 0V to make it conduct. The simple way is to connect a resistor from the PNP's base to the collector of an NPN whose emitter is grounded and whose base is forward-biased by a positive voltage when the 555's output is high. Both of these transistors should also have a resistor from base to emitter, to ensure that they turn fully OFF. Make the base-emitter resistor around about 20% of the base feed resistor.

This arrangement is also better because the transistor that switches the power rail will saturate, so the voltage drop across it will be pretty low. An emitter follower does not saturate, and will waste more voltage.

You could also use a P-channel MOSFET instead of the PNP, to achieve a lower voltage drop, especially if the 555 is driving something that draws a lot of current. Also, in either case, you must have some decent decoupling across the 555, especially if it's the old type.

 

[con-Fused]

Thanks so much!

I see the problem with the 0.7v - I knew I was overlooking something.
I think I have your recomendation diagramed below. The LED load should assure the NPN turns off. I assume decoupling with the 555 is a capacitor on the 555 output before the NPN base.

Its simple, but I would not have come up with this. Could see using this general concept in many places.

Thanks again KrisBluesNZ
Con-Fused

 

[KrisBlueNZ]

That's the right idea, but that capacitor is not what I meant. You need to remove the capacitor from the path from the 555 output to the NPN's base. The decoupling capacitor(s) I'm talking about are across the power and GND pins (8 to 1) of the 555. They're needed to ensure it works properly, especially now that there's resistance between the battery and the 555's power input.

You should also have a base-emitter resistor on the NPN; the LED load is not enough to guarantee <0.7V at the 555 output when it turns OFF. But that's not the only reason for the base-emitter resistor.

I had a think about whether the falling VCC voltage could cause the 555 to raise its output as it is powering down. I don't think it will, because its threshold voltages will be falling while the timing capacitor voltage is high and this causes the 555 to drive its output low. But this is another reason for the base-emitter resistor.

When the 555 turns OFF and the transistors turn OFF, the 555's power supply rail will start to fall. Eventually it will fall so low that the 555 will not be able to operate properly; in this case it might drive its output high, and that would power up the circuit again. We need to make sure that when the 555's power rail reaches that voltage, that the NPN can't be turned ON by it. Adding a base-emitter resistor turns the base drive circuit into a voltage divider; you choose the resistor values so you need at least 6V (for example) at the 555's output to produce the needed 0.7V base-emitter voltage to turn the NPN ON. Once the 555's supply voltage goes below 6V, even if the 555 goes haywire, there's nothing it can do to turn the NPN ON. I hope that makes sense!

 

[con-Fused]

Yeah realized that capacitor wasn't going to allow base current to flow. Thought about putting one from the NPN base to ground to keep the device on long enough to allow the timing capacitor to discharge, but it turns out is does anyway as the device goes off. I also tied the RC timing leg off the PNP collector and the capacitor discharges rapidly to 1v as the device goes off then slower through the 555. Even without the enhancements, its working as intended and I'm excited about that! It goes off smoothly.

Its a push switch that lights a starring night scene in my son's school project. Stars stay on for 20s. Runs on batt thus the interest in no basal power use.

Thanks again.

 

[KrisBlueNZ]

Cool