Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are there power consumption benefits between a discrete timer: astable multivibrator or some other; and an integrated circuit?

I have a time dependent circuit in a project and I went looking for the simplest, as in component complexity, discrete circuit I could utilise to allow my project to self regulate its timing and I came across the astable multivibrator astable multivibrator

I built it into the project but now that the whole thing is up and running I am starting to wonder if my interest in simple components is hurting my greater interest in lowering power consumption.

Would an IC like the 555 consume less power than a discrete astable multivibrator designed to meet my circuits specific needs?

..also, suggestions on lower power consumption discrete timers other than the AM, or discrete timers with less components than the AM, as well as any direction on how to accurately measure power consumption of the circuit would be greatly appreciated.

share|improve this question
6  
Hi, Hugh, and welcome to StackExchange. Please edit your question to show us your 'simplest discrete circuit'. There's a schematic button in the editor toolbar. I don't know if you're aware that your shift and punctuation keys don't seem to be working so if you can get them going you could fix the typos too. – transistor Jan 4 at 19:57
    
it says in the question it is an astable multivibrator, a google search would show what my discrete circuit looks like, but i'll add a picture for anyone who is unfamiliar.. and in keeping with the tone set by you i'd ask why you require capital letters to answer a question about power consumption of integrated versus discrete circuits? what typos? – Hugh Jan 4 at 20:03
    
Your circuit works essentially, by powering some LEDs via current-limiting resistors, and then turning them off by shorting around them with transistors!!! This means even more power is used when a LED is off than when it is on. First drive the LED's properly, then worry about the consumption in the oscillator. – Kaz Jan 4 at 20:16
    
i was better off leaving any picture out of it, this is just some random am from google with an i.stack.imgur url, i am uninterested in literal suggestions about improving this circuit, the question is clear in asking for benefits of discrete versus integrated circuits in regard to power consumption, that is all i want an answer to – Hugh Jan 4 at 20:40
2  
Hi Hugh, @transistor is a nice guy (or gal) and was gently prodding you :) You have a good vocabulary. You are correctly using commas, paragraph breaks, semicolons, etc. You're spelling words correctly. Why ignore proper capitalization? It's all part of written English. By the way, you ask a good question. I'm looking forward to seeing the answers. – bitsmack Jan 4 at 21:11
up vote 7 down vote accepted

555 timers consume on the order of single milliamps of current (the lowest is around .5 mA as far as I know). That's far from the lower limit as far as timing-capable ICs go, though, as even a small microcontroller can easily work with consumption around 200 μA and some (if not most) RTCs consume mere nanoamps (not that those will help in this particular case, since you'll require a microcontroller to extract the time from the IC).

Now let's look at the symmetric multivibrator. At any given moment exactly one of the transistors is conducting while the other is not, so there's current going through that transistor's B-E junction and R2/R3(depending on which of the two is conducting), and also through R4/R1 and the C-E junction. Additionally, in your particular schematic there's constant current through R1/R4 and the respective diodes. All in all we have current going through both 470Ω resistors and one 10K resistor (additionally, the capacitor on the opposite side is being charged via the other 10K resistor, but we'll ignore that for now), giving us a current on the order of 9V*(2/470Ω + 1/10KΩ) ≈ 40 mA. This is several times higher than what the ICs can reach.

Note though, that, barring possible minor nonlinear effects, this is mostly dependent on the resistors used; scaling all the resistances up several times dramatically reduces the current consumed by the multivibrator itself, with the only drawback being that the usable output current is reduced accordingly, however in that case you could just use another transistor to switch your load.

Another obvious thing to do would be to move the LEDs to not bypass the transistors, cutting the current used almost in half. Similarly, if you only use one of the halves for your switching, putting a large resistor in place of the load will cut the losses in the idle state.

To summarize: discrete timing circuits, including symmetric multivibrators, can be more efficient than IC-based ones, but they can also be very inefficient, and the particular configuration posted by you is one of the latter.

share|improve this answer
    
thank you for the concise answer – Hugh Jan 4 at 21:09
    
ics have that literal black box magic quality to them, and i was wondering if the semiconductor scale components had any physics based efficiencies to them, of course the solution to any unfamiliarity is do the math, i should have started there, but again, i appreciate your way of expressing that fact more.. also your suggestion of a third transistor to overcome the output current reduction is excellent brain fodder – Hugh Jan 4 at 21:49
    
@Hugh: Yes, there is a direct physics based component: the smaller your component the less current that flows through them. Now, this is normally stated as a disadvantage: "the smaller the component the less current it can handle" but if you're concerned more about power consumption rather than robustness then it can be an advantage. – slebetman Jan 5 at 1:39
    
In case of the comparators, ABSOLUTE TRANSISTOR MATCHING. Where as discrete transistors are always mis-matched. – ammar.cma Jan 5 at 12:54
    
@ammar.cma: How can this be achieved under mass-production conditions? If I understand correctly, transistor parameters are randomized even when they are on the same die. – FlashCactus Jan 5 at 18:38

The differences in power drawn between your discrete circuit and IC solutions are going to be due to the differences in the internal circuitry and how it operates. For example, here is the internal circuitry of the 555 timer you mentioned, which uses bipolar transistors like your circuit. It draws several mA.

enter image description here

Note the three resistors in blue, each of them 5K -- that's where the number 555 comes from.

The CMOS version is here. It's designed to draw under 150 µA (except for the load).

enter image description here

Exact same functionality, different circuit.

share|improve this answer

your circuit wastes almost 20mA on the unlit side. but if you move the LEDS up to be in series with R1 and R4, and add a 10K bypass resistor across the LED you will find that it uses about half the power.

a 555 will be in the same ballpark.

share|improve this answer
    
thanks for the response, interesting suggestion with moving the leds, if you could i would appreciate a description of the reasoning that led you to the suggestion.. also, i should have stipulated that my question arose because of the size of the discrete timer versus the ic, i wondered if i was losing power to connection resistance between the individual components, should i assume that to be negligible? – Hugh Jan 4 at 20:25

You can use a CMOS 555 and low power LEDs and get very low power consumption. Or use a small 8-pin microcontroller with a low frequency (eg. 32kHz) clock.

Either will draw less than 1mW, so the power consumption will likely be dominated by the LEDs.

share|improve this answer
    
thanks for the suggestions, but the question is about comparing the power consumption of the integrated circuits versus a discrete circuit made with 'jelly bean' components – Hugh Jan 4 at 20:31
    
The posted circuit draws more than 300mW, so hundreds of times worse. – Spehro Pefhany Jan 4 at 20:39
    
If you use a micro you may as well use a PWM signal to drive the LEDs, which will achieve full brightness at a duty cycle of around 40%, cutting the power they use as well. Using a controller with an internal RC oscillator may be a very good choice for power reduction, but may be limited in the accuracy of the timing it can provide. – Mike of SST Jan 5 at 9:51

This is really a reply to your follow up questions, "is the resistance of discrete connections negligable" and "are there any physics-based efficiencies of IC devices".

The resistance of the wires is negligable. However, when trying to build extremely power-efficient circuits, down into the nano-and femto-amperes, you might start to have a different problem: the resistance of the PCB isn't really infinite. You may find a tiny amount of current leaking through a fingerprint on its surface, for example. See http://m.electronicdesign.com/test-amp-measurement/whats-all-femtoampere-stuff-anyhow

A more serious problem if you build a circuit out of visible components is that the power required to keep a bipolar transistor "on" is proportionate to its size. FETs are much nicer, they only leak a tiny amount through the gate, but act like little capacitors which must be charged up.

ICs will always win the power consumption competition because they can be made much smaller, in neat insulating packages, with very repeatable properties. The ultimate expression of this is the digital watch, which can run a timer and display off a tiny battery for years.

share|improve this answer
    
Does anyone really care about nA or fA supply current? Obviously, Pease is talking about measuring bias currents, which is quite different. What device can work with nA supply current, apart (perhaps) from an LCD display? Who really cares about PCB leakage as a contributor to device power consumption? – Oleksandr R. Jan 4 at 23:06
    
Digital watches, like I said, have nanoamp supply currents - and are timer circuits like the question asked about. – pjc50 Jan 4 at 23:34
    
Really? An LR44 cell has a capacity of 150mAh. If a typical watch employing such a battery has a useful life of 10 years between battery replacements (which is on the generous side), then its current draw must be about 2µA. Not low nA. My point is that such extremely small currents are far below self-discharge currents, so there is no practical reason (that I can think of) to reduce the power consumption of any circuit to such an extent. Maybe if it's powered by energy harvesting? – Oleksandr R. Jan 5 at 0:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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