Design Help 2 - Please NO PIC

The relay driver transistor should be an npn too.

John

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Isn't that going to reverse the on off?

The pnp, as drawn, will never turn on, because it needs negative base drive and there's none available. If you want the relay on when pin 3 is high, the driver should be an npn.

And you could eliminate two more parts, the 10K and its diode. Just let the first transistor discharge the cap.

Or replace the 555 with a simple schmitt trigger inverter.

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No. If you really had a conventional PNP output transistor in the circuit as your drawing shows, the collector base diode would be forward biased and pull the base high enough to reverse bias the base- emitter junction into avalanche breakdown, which would probably have destroyed the transistor.

For the long delay time that you want, a CMOS version of the 555 might be attractive.

I hope your 100uF electrolytic capacitor is a tantalum device - back in the 1970's, when I would with these sorts of time constants, tantalum electrolytic capacitors offered much lowr and more stable leakage currents than aluminium-based devices.

-- Bill Sloman, Nijmegen

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These might help:

Have fun,

Michael

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Guess I'd better check my parts again because I have something on the breadboard that works. I thought I had a pnp there but maybe I forgot what I put there. I'd better look at the part number on the transistor in question :-)

I'll try it without the 10k and the diode. Easy enough to pull them off. I'm assuming you mean on the output of the first transistor.

Working with what I have in the parts bin at the moment, 555s,

2n2222s, and 2n2907s :-)

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Not with the 1K in the way. There might be a very slow beta degradation, which is irrelevant in a circuit that doesn't work anyhow.

Common aluminums have self-discharge time constants measured in hours or days, sometimes weeks. Try it.

John

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This guy integrates 10nA on a 4,700uF electrolytic to drive a motor:

Cheers, James Arthur

For the long delay time that you want, a CMOS version of the 555 might be attractive.

But it does not have the drive capacity of the bipolar device, so you need a sensitive relay or a transistor driver.

I hope your 100uF electrolytic capacitor is a tantalum device - back in the 1970's, when I would with these sorts of time constants, tantalum electrolytic capacitors offered much lowr and more stable leakage currents than aluminium-based devices.

You can get aluminum electrolytics with very low leakage and high stability for much lower cost than tantalum.

Paul

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Yup, note his capacitor quality check: no significant self-discharge after "a few days."

John

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No, now I see what you mean. The c-b-e path could conduct through the relay coil in addition to the base drive through the 1K.

That might zap the transistor, but pnp's seem to often have higher b-e zener voltages than npn's, so the current may be low. Again, it doesn't matter much if you damage a transistor that doesn't work anyhow.

John

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"Common aluminums have self-discharge time constants measured in hours or days, sometimes weeks. Try it."

I will, I always thought that the aluminums had a high leakage current. I've got a Nichicon HE series 1000uF and 50V. The leakage current is listed as 500uA. This would imply a 'resistance' of 100 k ohm and a time constant of 100 seconds.

OK in 20 minutes the voltage has dropped by less than 10%! (And typically there is a 'fast droop in voltage after the initial charge.) Will this get worse with age? Why such a conservative spec if they are much better than this?

George Herold

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I suggest you don't need a transistor to invert the trigger pulse. Just connect the switch to V+ and 2 resistors to create a voltage divider to ground. Thy can be quite large as the trigger current of the 555 is very low. The midpoint of the divider should be at the specified trigger level (about 5 volts I think). When he switch opens the resistor to ground pulls the trigger line low. The NE555 should be able to run a 150 mA relay directly (200 mA drive capability).

Cheers

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I have a constraint on the switch. It's a pressure switch that is normally on and off is "on" for my application. It's a single lead and it grounded because it is screwed into an engine. So I can't change how I use the switch.

My bad on the second transistor. It's an npn too. It had no markings and I thought I took it out of my pnp bag but must have taken out of the npn's. Corrected drawing.

Removing the 10k and the diode results in the circuoit being on constaintly no matter what position the switch is in. So it must take more modifications to get rid of those two components.

I have a constraint on the switch. It's a pressure switch that is normally on and off is "on" for my application. It's a single lead and it grounded because it is screwed into an engine. So I can't change how I use the switch.

My bad on the second transistor. It's an npn too. It had no markings and I thought I took it out of my pnp bag but must have taken out of the npn's. Corrected drawing.

Removing the 10k and the diode results in the circuoit being on constaintly no matter what position the switch is in. So it must take more modifications to get rid of those two components.

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The diode discharges the capacitor through Q1. The 10k resistor is needed as a pullup for the trigger input. It may also be necessary to tie pin 4 (Reset) to pin 8.

It looks like the opening of the NC switch will discharge the capacitor and provide a low trigger signal, which should drive the output high and energize the relay, but you might get a high current through Q1. A 100 ohm resistor in series with the 100 uF capacitor might be a good idea.

When the switch opens, the trigger goes high, and the output should stay on until the threshold rises to 2/3 Vcc, at which point the output should turn off, and the transistor on pin 7 should discharge the capacitor. If that is true, then you might not need the diode.

Paul

You evidently missed the part where the switch has one terminal accessible, and "ground" is bolted to an engine block.

Hope This Helps!

I've seen his latest 555 rendition, and I think I like it. ;-)

Hey, OP! Does it work?

Cheers! Rich

Because they're NOT much better than this at their maximum rated temperature, which is sometimes what they're run at. If it's rated to

105C (I checked below, it is), put it in boiling water (put it in a baggie meant for boiling food) and see what the leakage current is. It will surely be a lot higher than at room temperature.

Here it is:

Leakage current is .01CV = .01 * .0001F * 50V = 500uA. Leakage current is guaranteed to be no more than that over the temperature range of

-40C to +105C.

I heard Bob Pease speak a few years ago, he said there are engineers who call him (or National) to solve problems - they were getting bit because they designed based on what's in the "typical values" column of data sheets. He said you should always design based on the worst case figures, as they're the only ones that are guaranteed. I was shocked, shocked I tell you! to learn that engineers would design based on anything other than worst-case specifications.

Remember this leakage current thing when this thing is in the engine compartment of your idling car in a traffic jam on a hot summer day.

I took some red and some blue paint and marked all my transistors now I can tell at a glance the polarity of the transistor,

I used cheap spray paint (because I had it) applied with a small brush. a paint marker would be a tidier solution,

spray is good for small jobs because it doesn't dry up if not used.

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OkI I see the problem. Is it possible to have a floating PS so that the motor ground is actually connected to V+???

Gottta think lateral!

Cheers