See all 'Engineer In Wonderland' postsI have always been a great fan of the 4000 CMOS series, and its 74HC derivatives.
So when an engineer friend of mine was mulling over options for a five minute delay circuit, I chipped in that I thought the 4060 14-stage ripple counter and oscillator should be able to do it.
But try as I might, I couldn't think of a way to wire it up to act as any kind of monostable without additional components.
Can you do it?
Or with only a few external components?
The circuit is battery powered, so has to have minimal power consumption when timed out, although it can draw a milliamp or so when active.
There are many other circuits that could do it, but the damp application environment rules out using a single long time constant RC, and a dislike of testing code means no microcontroller - even if it has only got a few pins.
Somewhat later.............
Paul replied (below) with a cunning one diode answer
I did mess around with his values a bit because, I think, the equation on that website should have RC in the denominator.
I chose a biggish capacitor to keep the timing resistor low and minimise the effect of damp-induced leakage.
As Paul points out, the only static consumption is through the series combination of 39k and 470k - about 8µA at 5V.
Neat
'Alice''
Respond below, or to alice@electronicsweekly.com
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Comments (1)
Configure the gates around pins 9,10,11 to be a classic CMOS oscialltor (not my drawing but there's one here: http://www.elecfree.com/electronic/wp-content/uploads/2007/07/cmos-logic-clock-oscillator-by-ic-4011.jpg
Using C=100n, R=470k gives a clock speed of 25Hz so that Q14 goes high in 8000 counts or about 5 minutes.
Connect a 1N4148 from Q14 (anode) to pin 11. The start button is connected to the reset input.
With Q14 low, the diode is reverse biassed and the oscillator runs free. When Q14 goes high the diode conducts and this stops the oscillator and so Q14 remains high indefinitely.
Use a NPN or PNP common-emitter on the output to give the desired logic output from Q14.
The only leakage current will be through the 470k resistor. The actual value here will have to be a trade off between quiescent current and potential damp leakage across the capacitor.
What do you think?
regards
paul
Posted by Paul Harding | September 26, 2008 1:49 AM
Posted on September 26, 2008 01:49