# Max voltage /heat problem

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#### Dallas90

##### New Member
O.K. I'm getting closer to understanding what I need to do to create a timing delay using a 555 ic and 24VAC. I know I've got to rectify the 24VAC and that I will probably lose a volt or so there. Then I need to drop it down to where a 555 can handle it. I've seen some data that indicates there are 555's that can handle more than 5 VDC but 5 seems to be the common value.
My current (pardon the pun) questions. Is there a good source for 555's that can handle more power (making it easier/cooler for me to use a step down regulator)?
How much heat am I going to generate if I have to step down from 24 to 5 and power a 555 delay circuit and a relay coil (maybe 30 mAmp)?

#### user_88

##### Member
Consider using a step down transformer at the output of your 24 VAC source .... get the AC voltage magnitude down to something that you can work with .... Take the final AC voltage and rectify it ... maybe use two 2:1 transformers ... so that your resulting AC voltage is 24/2/2 = 6 VAC. Your problem description doesn't seem to require any significant power.

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#### kchriste

##### New Member
Forum Supporter
When you rectify & filter the 24Vac you'll get more voltage because the peak of 24Vac is apx 34V.
A 555 can run up to about 15V safely. If you use a CMOS 555, you can cut your power consumption, and thus the wasted heat in the regulator as well.
EDIT: Or use some other circuit, such as an comparator & cap as a time delay, which can operate at a higher voltage. It all depends on the timing requirements.

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#### unclejed613

##### Well-Known Member
it's late at night, so if i goofed in this drawing, somebody please let me know.....

if you're feeling a little bit ambitious, and you have some parts, here's a regulator that's a bit more flexible than the standard 3 terminal types. this is basically very similar to what's inside the little 3 terminal ones. it's also a good opportunity to learn how they work. Q1 is a series pass transistor, and does most of the "heavy lifting" in the circuit. it's an emitter follower amplifier, which means it has current gain, but no voltage gain, and the emitter is the output element. Q3 and 4 are a differential amplifier. they respond to any difference between the voltages on the bases. the output of the diff amp is in the form of a current, which drives Q2, which is a voltage amplifier driven by the output current of the diff amp. Q2 and R4 form a voltage divider, which supplies drive to Q1. the zener is a 5V voltage reference at the noninverting input of the diff amp, and the feedback from the regulator output comes from the voltage divider of R5, VR1, and R6. basically the diff amp compares the voltage on the wiper of VR1 (connected to the inverting input of the diff amp)with the zener voltage, and changes the drive to Q1 to maintain a constant voltage. the load (whatever circuit you are feeding with this) is the load resistor for Q1 (the voltage divider isn't much of a load, and is only there to provide feedback for the diff amp.
there is a voltage drop of 0.7V between the base and emitter of Q1, but since there's feedback, this is compensated for.

the diff amp and voltage amplifier section is basically an op amp (with up to about 800ma output current, which you can't do with most op amps). Q1 is good up to about 3A, and should be heat sinked for any loads above maybe 100-200ma. Q3 and Q4 can use any pnp (they have to be the same type) as long as the BVceo is 20% or more above the input voltage. the resistor feeding the zener may have to be a higher resistance with voltages above 40V.

hope this helps.running an LM7805 or 7812 without a heat sink with 32V input is a very bad idea. and the 78xx regulators have a max input voltage of 35V, so 32V is pushing the envelope.

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#### Chippie

##### Member
If you cant use a transformer with a lower turns ratio then why not use a switching regulator to reduce your 34 something volts down to something more manageable for the 555 ?

#### Dallas90

##### New Member
Thanks for all the responses, This is just the last piece in a bigger system, and the whole thing has taken me a lot of time (because I was very close to zero knowledge about six months ago).
All I need to do is delay power to a relay for 40 seconds (+/- 5), after that the power to the relay coil should come on and stay on until the whole system is powered down (anywhere from 12 minutes to 120 minutes later).
The power coming into the system is 24 volt AC. I need it to be as small as possible, so the less number of components the better. It will be in a small, waterproof container. If I could cause a delay without conditioning the electricity (taking up space and causing heat) I would do that. So kcristie's idea appeals to me. How does it work and can I get 40 seconds of delay out of it?

#### user_88

##### Member
ThomasNet® - Product Catalog

Take a look at one of these ... 24V .... @ your specified delay range

possibly :
Item # 601-M2 Series 601 - ON Delay

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#### Dallas90

##### New Member
That might work. I forgot to mention I'm trying to keep the cost down around \$2.00. I'll send off for info. I'm very curious as to how they did it.

#### kchriste

##### New Member
Forum Supporter
Something like this should work:

For the OpAmp, you need to choose one with that will operate up to apx 40V or higher just to be safe. Something like the MC33171 or equivalent will work. The timing tolerance will be dominated by the tolerance of the 100uF capacitor. You can change the 180K resistor to compensate or sub a 200K pot and a 100K series resistor.

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#### Dallas90

##### New Member
The part that user 88 sent me to is great except it's too large. The delay that kcristie put up is great, and I like keeping everything at 24vac. I'm a bit intimidated by it at this point, but I have Sunday off, so I'll probably have to give it a try if I can get all of the parts together.
I just did some testing to see if I could make the relay coil work (and not create too much heat ) with just a bridge rectifier and a 7815 regulator. I shut it off when the regulator got to 155º F.
I've been looking for small transformers, but I can't find PC mount that will do the job and are small enough for me ( on the order of 1/2 X 1/2 X 1/2)

#### Dallas90

##### New Member
That's a bit of a problem as 32VDC is not a common rating on the relays I'm familiar with. I'll keep looking.
So if I want to ramp up the delay to say 60 sec, I'd have to change the capacitor by 50%? Sorry for the nubie questions, but I'm really trying to understand this.

Thanks,
David

#### kchriste

##### New Member
Forum Supporter
If you can't find a 32V relay, you could use a 24V one in a pinch. For example, if the 24V relay has a 2000Ω coil then you could add a 680Ω resistor in series with it. In a circuit like that, the resistor would only dissipate 0.1 watt which is negligible. Generally speaking, the higher the coil voltage of a given type of relay, the lower the required coil current is going to be which is why I suggested a 32V relay.
Yes, to increase the timing period you can increase the capacitor value (220uf --> 330uF) or the resistor value (180K --> 270K). I just realized I goofed on my timing calculations earlier. The original 100uF should have been 220uF for apx 43 seconds of delay.

#### Mr RB

##### Well-Known Member
Why not just use one darlington transistor, and a resistor and cap to give the start up delay?

#### kchriste

##### New Member
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While that would work, the timing wouldn't be as consistent with voltage variations and the relay wouldn't be as "snappy". But since you mentioned it:

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#### Chippie

##### Member
why not change the 2 2N222's for a Darlo.....?

#### Dallas90

##### New Member
Well, if the timing was within say +/- 8 I could probably live with it by jacking up the total delay. It appears at this point I just don't want to be less than 40 seconds. So if I could be close on timing without generating a lot of heat, taking up a lot of space, or costing much money I'd be happy.

#### kchriste

##### New Member
Forum Supporter
Dallas90,
The timing will vary with changes to the 24Vac if you use the transistor only method. So as the line voltage changes, the 24Vac will change, and thus your timing will change. How much is difficult to say at this point, since things like relay pull in voltages are unknown. You'd have to build the circuit to be sure.

why not change the 2 2N222's for a Darlo.....?
Yes, you could do that. Most people are more likely to have a pair of 2n2222's or other NPN transistor on hand rather than a darlington.

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#### Chippie

##### Member
Dallas90,

Yes, you could do that. Most people are more likely to have a pair of 2n2222's or other NPN transistor on hand rather than a darlington.

I must have a bigger spares box than many then....

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#### Dallas90

##### New Member
Could someone help me learn here, by walking me through the operation of the circuit? It appears that the delay may be caused by the charge time of the capacitors and the switching by the transistors , but it's not totally clear to me.

#### Chippie

##### Member
To keep things simple here's my take on it....

Without going into all the maths behind it the time taken to charge a capacitor to a given voltage is defined as the time constant... usually 63ish % of the supply voltage...ergo if you change the voltage supply the time will change although the constant doesnt....because it is fixed...Look at the drawing....and then look at the graph....

If you keep the value of the capacitor fixed and the same resistor, changing the voltage changes the shape of the curve...Fixing the voltage and changing the resistor has the same effect if the voltage remains the same...Got it?

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