1
\$\begingroup\$

The following project is built:

  • A 9 V battery is used – no socket required. This is important to gain experience for future projects.
  • There is an under-voltage lockout, which switches off when the battery becomes weak.
  • A 555 timer, wired as an oscillator and with a 50% duty cycle, makes a red LED flash.
  • The circuit is designed to save power.
  • Unfortunately, the circuit is not yet power saving enough.

When I do the math: I buy this 9 V battery with 1200 mAh, and the circuit consumes 20 mA, then it runs for 60 h – so less than 3 days. I have to change that. The point is that this is the first milestone. Later, the circuit will be extended. I want to connect a microprocessor to the same battery later.

I only work with PSpice for TI 2022; this is a cut-down version of PSpice; in other words, I only have models from PSpice and Texas Instruments available.

By the way: the operational amplifier in the UVLO gets its own 9 V battery. This way the voltage is always the same, as it consumes very little current.

Circuit

Current consume enter image description here

\$\endgroup\$
6
  • 2
    \$\begingroup\$ You may find that "working in SPICE" is, in this instance, rather counter-productive. If you want to save power, you'll need some sort of switching power conversion, so that the resistive energy losses will be minimized. You'll want to use a CMOS 555, if a 555 is even appropriate for this application (I doubt it is). \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Aug 22, 2022 at 16:29
  • 2
    \$\begingroup\$ I see you are using a 375 ohm resistor to represent an LED drawing 20mA. An LED will give substantial brightness even when drawing 2 or 3 mA. Couldn't you increase the value of the 375 ohm resistor so it represents an LED being driven with a smaller current. \$\endgroup\$
    – user173271
    Aug 22, 2022 at 16:48
  • \$\begingroup\$ Why do you need the Q2? The 555 can source or sink 200 mA. \$\endgroup\$
    – RussellH
    Aug 22, 2022 at 19:42
  • \$\begingroup\$ First thing to look at is - obviously - where is the power going?? how much of that 20mA is the LED, how much is U1, how much is U2, etc. Don't just guess - find out! \$\endgroup\$
    – user253751
    Aug 22, 2022 at 22:24
  • \$\begingroup\$ Ra seems much lower than it needs to be, especially compared to Rb, especially if you don't need a close-to-50% duty cycle. \$\endgroup\$
    – user253751
    Aug 22, 2022 at 22:25

1 Answer 1

Reset to default
7
\$\begingroup\$

At 50% duty cycle about an average 8mA is going into the LED and another 4.5mA is going into Ra.

The TLC555 itself draws maybe another 0.6mA.

That's 13mA * 60h = 780mAh. not so far from your 1200mAh rating depending on the cutoff voltage. You could easily increase Ra to 50k or 100k and hardly affect the duty cycle. Or connect the 1M to the 555 output and don't use the discharge pin at all, since you have a driver transistor for the LED.

You are also wasting power in both Q1 and Q2 because they're emitter followers. Using a saturated switch is better (MOSFET or BJT).

So after you get rid of the wasted power in Ra you can play with the LED current, duty cycle and the cutoff voltage to get longer life. Note that you should have a large electrolytic capacitor across the battery to get more effective life from the battery when you have a relatively high pulsed current. Figure a capacitance that won't drop more than a few hundred mV during the pulse even if the battery wasn't there.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.