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Beginner question : 555 timer pin 4

Morning all - newbie to the forum. Not my first forum though and I have searched through old threads before asking my Q.

I'm a complete beginner as far as SS electronics is concerned I'm afraid. When I started into DIY audio 3 years ago I decided teaching myself High Voltage valve amp building would be a great idea. Really enjoying the results, but in hindsight starting with low voltage would have been more sensible - especially as I didn't know the difference between volts and amps!

Anyway, my latest valve amp is using SS electronics to provide DC heating for the valves and it's going in a cupboard, so I am putting a small and quiet fan at the back. I've designed a control circuit with a thermistor/trimmer setting the output voltage from an LM317 - that sets the fan speed nicely if the heat increases. I also wanted to add a warning light to alternately flash a couple of red leds if the temperature on a different thermistor exceeds a set level - should be OFF when cold.

The 555 timer bit is fine, plenty of material and sample circuits around and I've played with it on a breadboard to see how stuff changes with different values and configurations. I've got this working in 2 ways and wanted opinions about them.

1) 555 pin 3 to resistor & led - but use a BC547 (because it's to hand) transistor to control the current flow. Collector to LED cathode, Emitter to 0v, Base to junction of thermistor & trimmer allowing me to set the trip point. This works ok but the led flashes faintly as the voltage approaches 0.6v. No problem, just bugs me a tiny bit. Also with this method I can't alternately flash the LEDs - I can put them in series and flash them together and that's ok.

2) As above, but remove the BC547 and connect the thermistor/trimmer junction to pin 4 (reset) of the 555 and configure it so that when cold the voltage at the junction is below 0.8v. This also works, uses less components and the switching seems to be binary. Also I can alternately flash as per original design.

Any reason not to use pin 4 this way, to hold the 555 DOWN while cold? Bearing in mind that this will be the normal state, so it's always going to be this way unless the back of the amp overheats the cupboard.
 
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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
in hindsight starting with low voltage would have been more sensible - especially as I didn't know the difference between volts and amps!

Hahaha. Glad you're still with us.

Any reason not to use pin 4 this way, to hold the 555 DOWN while cold?

No, that's what it's for.

You will find that as the voltage transitions between about 0.4V and 1V that the timer starts somewhere in there, at a point that might even change a little with temperature. You can't guarantee it will be 0.8V.
 
Many thanks Steve. I'll go with option 2 but experiment further with the trip point. I can adjust it with the trimmer, but it will be a balance between false positives and ensuring it trips if it gets hot. It's a circuit I don't ever want to activate ...

Your point is well taken about temperature drift. Given that this circuit needs to activate under temperature increase I should plan for (and test under) those circumstances.

With a 9v (regulated) supply I plan to use a 10k thermistor (at 25C). Say I set the alarm point at 35C, the thermistor should be around 6k7. To put 0.8v on pin 4 the Thermistor needs 1.22mA through it, so I need the trimmer at 655R. I plan to use a 1K 1/2 watt Bourns trimmer and adjust in situ. The trimmer (used as a rheostat) needs to be derated to 0.3W, but with only 1.22mA through it the dissipation is tiny.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Yeah, that 1k 0.5W trimmer can pass 22mA without dissipating too much power (presumably at 25C). So 1.2mA will be fine, even at quite elevated temperatures.
 
Hmmm how odd

I put a 6k8 in place of the thermistor to test, with the 1k rheostat to ground. Pin 4 connected to junction of 6k8 and rheostat, (6k8 and TRIM create a voltage divider, between 8.8v and 0v).

As I increase the trim (and hence the voltage on pin 4) it goes through 3 stages...

OFF
ON
BLINKING

Can anyone help me understand why the ON stage is there ? I'll measure the voltage boundaries later.
 
Another question - if I base the circuit on the attached diagram (intention being to flash the LEDs alternatively) it works fine, but when I ground pin 4 the top LED stays on (steady, not blinking).

When pin 3 is LOW, is it connected to ground ? I assume so.

Can't paste image - at this link (railway lights)

http://solderpad.com/psd/railway-crossing-lights/


upload_2018-11-8_7-59-42.png


[Mod edited to add image]
 
Last edited by a moderator:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Pin 4 is generally used with a signal which rapidly transitions. In my earlier answer regarding the voltage at which it transitions I omitted a warning about it acting odd during the transition (I've not seen this behaviour, but I'm not surprised by it either).

Another thing is that after being disabled, the first time interval will be significantly longer due to the fact that the timing capacitor has discharged below 1/3 Vcc.

The circuit you're using effectively uses the 555 to switch off one of the LEDs by shorting the junction to one supply rail or the other. If the 555 were removed (or the connection to pin 3 removed) both LEDs would likely illuminate dimly.

So, when the 555 is held in reset, the output is at one supply rail, and one of these LEDs is on.

In your case, I would remove one LED completely, and have the existing one go from off, to on, to flashing.

The strange behaviour you see as the 555 transitions from off to on is generally outside the specifications, and may not be consistent. Different manufacturers, and different types of 555 (a CMOS version for example) may operate differently.
 
Thanks Steve. I'll ditch the alternating LED idea and keep it simple.

So I guess I'm back to my initial question. Is it better to use a separate transistor in the cathode of the LED to switch off the flashing when cold, or better to use the 555 as switch by holding pin 4 at or near ground until the temperature rises to the warning level.

Most (hopefully all except during testing) of the time this circuit will be off. When the temperature rises I guess I don't really care if it illuminates steady ON and then flashing as the temperature rises further - I just feel that I should be able to understand the behaviour so I am better able to design it to requirements.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
I would go for the use of pin 4.

It would be better to switch pin 4 using something that has a fast (faster) rise and fall time, and one where the transition point (and hysteresis) can be controlled.

You could use a comparator with a little bit of positive feedback to ensure a definite switching on and off at two slightly different temperatures, without that having to occur at 0.8V.
 
Many thanks Steve. I had the thought of how I could switch it faster but didn't know how to approach it. Comparator was the word I needed, I'll go research.
 
Ok brief reading has told me to use a comparator and not an op-amp.

The TI LM239P is available on RS for very low cost - it's a dual device but no issue. It will handle my 9v supply and I figure I'll have 2 voltage dividers as input, one with the thermistor and the other with the trimmer to set the alert level.

My only concern with it is the low output is specified in the datasheet as typically 400mV but up to 700mV. That could result in false alerts. Is this typical of comparators or do some of them guarantee low being nearer to 0v?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
That value will be defined for a particular pull up resistor value, possibly 2k.

If you use a larger value resistor the comparator will probably pull the output lower.

You could also use a rail to rail op amp in this application as you're not concerned with speed.
 
More research, learning as I go here but posting for verification. I could use a pull down resistor on the comparator output? Datasheet indicates low output current is 6mA. A 62R to ground would put the voltage at just under 0.4v. would this work ?

When the comparator output is low I think that resistor would pull the voltage down ok. However when it's high (normal in my application) it will be at around 7v. A 62R to ground would draw 112mA? I must be getting something wrong here.
 
Last edited:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
If you look at the output of a comparator like that, a pull up resistor is required. I would probably use a pull-up of 10k.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Something with an open collector output cannot go high at all unless there is a pull up resistor -- the output can only pull down.

The lower the value of the pull-up, the harder it is for the comparator to pull down, so the higher the "low" output is. So, a higher value allows it to pull lower, and that's what you want.
 
Thanks Steve

The datasheet for the LM293 shows that the output stage is an NPN transistor. This explains why the output will be a diode drop above ground.

upload_2018-11-9_10-4-59.png

I accept that the pull-up resistor will help me ensure it drives close to 0v and thanks for your help. As always I like to experiment for myself so it locks into my brain better.

Out of interest, if I selected a comparator with a PNP output stage would that solve the problem ? How would I track down comparators which use PNP outputs ?
 
Follow on question. I want to use 2 thermistors to monitor the temp of the back of my amp (1 per channel). The LM293P has 2 comparators on board - piece of cake I thought.

Each comparator works individually. With only one or other of the outputs connected to the 10k pullup the voltage swings from 0.04v when thermistor is COLD and 8.4v when its HOT. This connects to the 555 reset pin 4, which switches on/off and all is well.

But if I connect both outputs to the bottom of the 10k pullup resistor the voltage swings between 0.04v and 0.06v. So the 555 doesn't trip.

I figured this would work like an OR combination - if EITHER or BOTH of the comparator outputs is LOW then the voltage at the bottom of the 10k pullup should be 0v ?

I've got something wrong in my head again. Someone put me on the right track please ?
 
Just to check I'm not wrapping myself up in logic..

9v supply
to 680r
to 1.6v 10mA led
2 bc547 npn transistors, both collectors to the led cathode , emitters to ground
Floating wire from the base of each npn

Measure voltage at the base of the led.

Result - voltage is high (7.3v) if neither npn is conducting, otherwise it's 0.24v.

That works as I expected, ie an OR gate. That should give me what I want right? Pin 4 will stay high only if BOTH comparator outs are high (neither npn in the comparator outputs) are conducting, otherwise it will go low.

What am I missing?
 
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