ANDYZ28

I am installing a set of OEM dual electric fans on my '82Z28. Dpes anone know how many amps each fan draws? What is the circuit protection level on cars that have these fans? I am thinking the both fans can be operated with one 30amp relay,and one 30amp breaker?

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82Z 305 w/comp 270 magnum cam,Edelbrock Performer RPM,Holly 1850-4,serpentine belt drive.TH700R4 w/B&M holeshot 2000 converter,& megashifter.1LE front brakes, 9bolt 3.27 w/1LE rear brakes.Aluminum driveshaft,boxed rear susp.poly everything,IROC swaybar+wonderbar. 70mph@2200rpm ASE Master Tech plus L2
also recently obtained a
'69 chevelle SS396 w/Turbo 400,3.31 posi,11.0 to 1, headers,etc. Latest project car,'86 IROC stock 305TPI hit on left side,but not too bad

Vader

Andy,

The primary fan is rated at 150W, and the secondary fan is rated at 100W.

You can calculate the current:

P = I x E. Or in more common terms, Power (in Watts) = Current (I) times Electromotive force (voltage). If you rearrange the equation, current (I) = Power (Watts) ÷ Electromotive force (voltage). Got it?

I would suggest a 20A circuit breaker for each fan instead of a 30A breaker for both. If one starts to overload, the other can still run, preventing overheating. You should really use two relays and stage them at different temperatures so that the electrical load of the fans doesn't overwhelm the system (and alternator) at idle speeds when that much air flow isn't necessary. My little 305 stays plenty cool on only one fan. You should have enough extra openings in the cooling system to accomodate two sensors, and if not, you can get creative:



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Later,
Vader
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"I'm'a do Things My Way - It's My way or the Highway."
Adobe Acrobat Reader

ANDYZ28

Thanx VADER, I will run two seperate relays,and breakers. I will have to run both fans simutaiously though temporarliy. As I am running out of places to put sensors. The right bank cyl head plug has had the allen socket stripped. Thanx,ANDYZ28

JP84Z430HP

I tried it out on just a single 30 AMP fuse, and it blew it. The fans will RUN on a 30 amp circuit, but the startup current is too high. I split it in to 2 circuits, 1 that runs constant (when it's warm out) and the other on the 190* (or somewhere around there) fan switch I got.

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Working on:
'84 Z28 LG4 305 with 200,000 original miles!
Added dual elec fans.
145 MPH IROC Speedo
Building 430 HP 350 (ZZ430)
using primarily GMPP parts.
Short block sitting on a stand. (Man, those Fast-Burn heads sitting on it look good!)

Starting to look like the Kicker poster child!

ASE Certified Master Tech

ANDYZ28

Thanx,I am going to run them both at 205 degrees on-185 off. Thankx

Stuart Moss

EDIT = deleted my first post, superseded by the following:

The right (secondary or passenger) and left (primary or driver) fan current draw was measured from my ’91 Z28.

DRIVERS SIDE FAN

36 amperes @ 12.6V [startup/inrush current] = 454 watts
13.2 amperes @ 12.6V [run current] = 166 watts

49 amperes @ 15.0V [startup/inrush current] = 735 watts
18 amperes @ 15.0V [run current] = 270 watts


PASSENGER SIDE FAN

40 amperes @ 12.8V [startup/inrush current] = 512 watts
13 amperes @ 12.8V [run current] = 166 watts

46 amperes @ 14.9V [startup/inrush current] = 685 watts
18 amperes @ 14.9V [run current] = 268 watts

As you can see, both fans draw essentially the same current (18 amperes run condition @ 15V). Only the inrush current is a little different. Because it happens for only a fraction of a second, it was difficult for me to obtain an accurate reading (my DMM does not have a “peak hold” function - maybe the next one).

When figuring your circuit load, use the 14/15 volt figure because that is from an engine running condition. 14.9/15.0 volts is considered somewhat high for an automobile, but still within the normal operating range. It was cold outside when I took the measurements, and the alternator output (from the regulator) will be higher because it is temperature compensated. I had just started the car, so it (the alternator case/regulator) was cold and didn’t have time to warm up and reduce its output, which it would have done after ~15 minutes to 13.9 - 14.3 volts, depending upon ambient temperature (~14.2 volts during the winter, ~13.9 volts during the summer).

The post above mine stated that he could not use a single 30-ampere fuse. I wonder if that was to power both fans from this one fuse.

I tried a 20 ampere fuse for one fan, and it worked fine when tested (@ 12.6V, or ~13 amperes). Even though the inrush current exceeded the rating of this fuse, it was only 200% of the rating of the fuse for a fraction of a second (40 amperes). A slow-blow fuse would be a good choice in this case. But that is fine for an “engine off” (or < 12.6 volts) condition. We must consider an engine running condition, or > 14 volts. Now we’ll be drawing ~18 amperes, and that is pretty close to its 20 ampere rating. It might work, but it would probably be better to use a 25-ampere fuse (30-ampere worse case) for a larger safety margin.

Since the inrush current lasts only for a fraction of a second at about 196% of the rating of a 25 ampere fuse (49 amperes @ 15 volts), it should be fine. Just be sure that you use one fuse/circuit breaker per fan when calculating this way.

FYI, fuses work faster than circuit breakers, and permanently remove current when blown. Many (most?) circuit breakers will cycle on/off during an overcurrent condition because of the heating/cooling of the thermal element inside the breaker which simply makes/breaks a contact until the excessive load or supply current is removed. Personally, I prefer a fuse.

One 30-ampere relay per fan would also be fine.

As an aside, I do not like both fans to come on simultaneously because of the large load (sometimes called “shock load”) that will be present to the alternator. Vader almost touched on this, although his concern was for continuous load, a valid point nevertheless. This can potentially cause damage to anything connected to the electrical system. It’s very possible that no harm will come, but I wouldn’t want to chance it on my car. If I was going to modify the fans to come on simultaneously, I’d delay one to come on about 2-3 seconds later. As you can see from the figures I measured above, both fans coming on simultaneously will present a load of ~100 amperes to the battery/alternator, albeit for only a fraction of a second. But that could be long enough to cause the alternators’ regulator to produce a spike powerful enough to damage something.

Like Vader, my 305 has always remained <200°F with only one fan operating during the hotest of days in Virginia in summer (> 90°F, A/C off), so I have no reason for two to always operate simultaneously to just cool the engine.

If both fans came on simultaneously from the factory, and one was not sufficient to keep the engine cool, I would modify the circuit to delay the turn-on of one fan - a simple circuit I’d build costing < $5 in parts.

[This message has been edited by Stuart Moss (edited March 27, 2001).]

84FTA

I'm curious why everyone here has such high draw on their fans. I used a non-stock set-up, two large fans (actually a tad too large!!) and I pulled out the multimeter and it displayed 9.8 amps running for both. I wish I knew where these fans came from but I recieved them second hand from Delphi-Packard (place that tests 90% of domestic car parts).

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1984 WS6 Hardtop Trans Am
Former L69 Car under restoration
1984 T-top Trans Am
4-bolt main 350, performer intake, headers, Holley 650, T-5, hayes clutch, dual elec. fans and 3.23's.
Daily driver and restoration
13.98 @ 101

ANDYZ28

Stuart Moss, Please tell me how to make the second fan come on with a delay. Thanx,ANDYZ28

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82Z 305 w/comp 270 magnum cam,Edelbrock Performer RPM,Holly 1850-4,serpentine belt drive.TH700R4 w/B&M holeshot 2000 converter,& megashifter.1LE front brakes, 9bolt 3.27 w/1LE rear brakes.Aluminum driveshaft,boxed rear susp.poly everything,IROC swaybar+wonderbar. 70mph@2200rpm ASE Master Tech plus L2
also recently obtained a
'69 chevelle SS396 w/Turbo 400,3.31 posi,11.0 to 1, headers,etc. Latest project car,'86 IROC stock 305TPI hit on left side,but not too bad

Stuart Moss

In order to best explain how to delay one relay, I would need to post an image. Unfortunately, I do not have a scanner, so I can’t draw a schematic and convert it to an image. Even if I did, I do not have a site to post it to so that I can post it here. You will need basic electronic knowledge and soldering/construction abilities to make it. Since you asked for a circuit, I presume that you are not knowledgable with electronics.

There are many ways to make a delayed turn-on circuit. I’ll describe a favorite of mine that is simple (uses only seven components) and inexpensive. I have made many, some have been in use since 1985, so I know it works reliably. All parts should be available from Radio Shack (or www.digikey.com). All values are not critical; ±20% is fine. The only thing that will change is the delay time. A 1N914 or 1N4148 can be substituted for the 1N400X diode.

The heart of the circuit uses a 555 timer, which is an 8-pin IC. Pin 1 is closest to the notch molded into the IC. Looking at the top of the IC, pin 1 will be to the left of the notch. The pins are counted top to bottom, then bottom to top, so pin 8 will be opposite pin 1, and pin 5 is opposite pin 4. A top view would look like this, with a notch between the 1 and 8 pins:

1 8
2 7
3 6
4 5

On a perf (perferated) board (also available from Radio Shack), connect (solder) the IC pins with hookup or wirewrap wire as follows.

1 - ground
2 - pin 6 (see note)
3 - positive side of relay and cathode of a 1N400X diode (see below)
4 - pin 8 (see note)
5 - no connection
6 - pin 2 (see note)
7 - no connection
8 - pin 4 (see note)

PINS 2 AND 6 are connected together. To these two pins, connect the following:
A) an 18K 1/4 watt resistor to ground.
B) a 220uF capacitor to pins 4/8 (16 volts or higher - 25 volts better), with the positive (+) to pins 4/8.
C) a diode (1N914 or 1N4148) to ground, with the cathode (banded side) to pins 2/6.

PINS 4 AND 8 are connected together. To these two pins, connect the following:
A) a 220uF capacitor to pins 2/6 (16 volts or higher - 25 volts better), positive (+) lead to pins 4/8.
B) a 4.7uF (16 volts or higher - 25 volts better) capacitor positive (+) lead to pins 4/8, other side to ground.
C) +12VDC. +12 applied here will begin the delay - connect to the switched +12 of the output of the other relay

PIN 3 NOTE: The 555 can source/sink quite a bit of current. I believe 300 mA (0.3 amps), so it can drive many different types of relays directly. If you have a large relay (> 200mA coil), you should use a transistor to dirve it (see below). Most relays draw only 100-200mA, so it should be able to drive the relay directly. If you’re getting a generic relay (e.g. an ice cube relay), it probably is not polarized (the coil is not polarity sensitive), so you can connect the output of the IC (pin 3) directly to any side of the relay coil. The other side of the relay will be ground. The 555 is only 42¢ from DigiKey (web address above), and I’ve seen it cheaper elsewhere (29¢ at Holsfelt), so it’s an expendable part to test it.

Connect the cathode (banded) side of a 1N400X diode (1N4001, 1N4002, etc.) to pin 3. Connect the other side (anode) to ground. This will keep any voltage spikes that the relay coil produces from entering the IC.

If you need a transistor to drive the IC, two additional components would be needed; an NPN small signal transistor such as a 2N2222 and a 6.8K 1/4 watt resistor. Advise and I’ll describe how to connect it.

NOTE: If you want to begin the delay with switched ground (vice switched positive), that will require a PNP small signal transistor and two resistors. Or simply switch the ground (pin 1) wire.

The 18K resistor and 220uF capacitor form the RC timing network (RC = Resistor Capacitor). If you change these values, you’ll change the timing. Timing can be determined by multiplying resistance by capacitance and 1.1.

In other words (resistance in meghoms)(capacitance)(1.1) = time in seconds. So, using the values in this circuit;

(18K) (220uF) (1.1)
= (0.018) (220) (1.1) [18K = 0.018M ohm]
= 4.4 seconds

Substituting a 12K resistor for the 18K would provide a delay of about 2.9 seconds.

This simple circuit operates as follows. When power is first applied (+12 applied to pins 4/8), the output (pin 3) will remain low for the time determined by the values selected for the timing components of this circuit, ~4.4 seconds, at which time it will go high and remain there until power is removed from the IC. The delay will begin again when power is re-applied.

You can put the circuit in a small box, encapsulate the circuit in epoxy or body filler, or if you were careful constructing it, it’ll fit inside a plastic 35mm film container. If you encapsulate it, you won’t be able to fix it if something goes wrong. But with a cost of <$5 to make, it shouldn’t be too much of a concern. It may be a good way to not only get an enclosure for the circuit, but also waterproof it.

REGARDING THE POST ABOUT HIS FAN MOTORS DRAWING A LOT LESS than the values presented here. You stated you measured 9.8 amperes (I presume for each of your two fans). You didn’t state at what voltage that was measured at. As you can see from my measurements, I saw ~13 amperes with the engine off (12.8 volts). With the fans running at 11.5 volts, it’d probably be around 11-12 amperes. Either way, 10 amperes is not much of a difference than 13 amperes. Even if you had GM motors I wouldn’t be concerned with only a 3 ampere difference - if your measurement was taken at a voltage less than what I did, then the difference would be even less once you calculate the current draw at the same voltage I did. If there was a big difference, I would attribute it to the variations between manufacturers such as motor strength or hp, rpm, load (resistance of the fan), etc.

[This message has been edited by Stuart Moss (edited March 28, 2001).]

84FTA

It was at 14.2 volts if I recall correctly. It was, infact, for both fans and not just one fan. Which means my fans draw 1/2 half the amperage of normal (OEM). These can possibly not even be GM fans, as they test most domestic models and some foreign parts.

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1984 WS6 Hardtop Trans Am
Former L69 Car under restoration
1984 T-top Trans Am
4-bolt main 350, performer intake, headers, Holley 650, T-5, hayes clutch, dual elec. fans and 3.23's.
Daily driver and restoration
13.98 @ 101

TunedPort350

[QUOTE]Originally posted by Stuart Moss
NOTE: If you want to begin the delay with switched ground (vice switched positive), that will require a PNP small signal transistor and two resistors. Or simply switch the ground (pin 1) wire.

Stuart,

I have built the circut as you explained, and it works great for switching the postive end, can you please explain in detail how to switch the negative.

You mention two ways can you please explain this.....


Thanks

ANDYZ28

"TunedPort350", If you get this figured out. Can you make me one. I want to be able to switch BOTH fans on with a delay. This way in extreme cicustances I could switch both fans on to cool with a single switch.
I am in the process of installing a 406" SBC in my '82 Z28. So I am probably going to need all the cooling help I can get.

Thanx,ANDYZ28

TunedPort350

Andy,

Actually I did create this circut and it works very good. I have it so that one signal (from the ecm) goes to the two relays for the fans and the time delay circut delays (about 5 sec.) one of the relays from turning on.

I was curious about how Stuart was saying that you can switch the negative (like stock or GM does) because right now I am switching the positive for the delayed relay (which does work fine) but I was just curious.

I'll post a picture of what it looks like if you are interested to see it.

Do you have a dual fan setup right now (with two relays) where one fan is switched by the ECM and the other by the thermal switch? If so it's real easy to integrate this circut, otherwise if you only have one fan you'll have to add another relay, which isn't too difficult either.

ANDYZ28

I no longer have an ECM. My fans are wired in the following manner: Left fan on with A/C high pressure (switch closes and grounds relay). Right fan with engine high temp (switch closes and grouns relay).

I also want do switch both manually.

Thanx,ANDYZ28

Stuart Moss

TunedPort350,

Regarding the two methods for applying a switched ground to trigger the switched delay, I'll try and explain below.

There are many ways to do this, and the easiest way is to simply switch pin 1 of the 555 IC to ground when you want to start the delayed output.

If you can breadboard this circuit (which I highly recommend doing first with any circuit - good breadboard available at Radio Shack) first, simply connect a wire to pin 1 of the 555 and connect it to ground to test it.

The second method I referred to was to use a PNP transistor (2N2907A or TIP30A, TIP32A etc.....) to supply (switch) power to the 555 IC. It will be wired to supply +12 to the 555 by switching ground to the base of the transistor. Wire it as follows:

The transistor has three leads - base, emitter and collector. On a 2N2907A transistor in a TO18 package (metal can), the lead closest to the tab is the emitter, the middle lead is the base and the other (farthest from the tab) is the collector. For a TIP30 or 32 PNP transistor, I believe the pinouts are BEC (Base, Emitter and Collector) as you look at it with the mount of the transistor facing down and the pins toward you.

1. Connect the switched ground (from your switch or whatever) to one side of a 3K 1/4 watt resistor (value is not very important - a 2K or 4K will work equally well).

2. From the same switch ground lead that is connected to one side of the above mentioned resistor, connect a 10K ohm 1/4 watt resistor (again, the value is not that important), with the other side going to +12. So now, the wire on switched side of the switch sould be connected to three places - (1) the 3K ohm resistor, (2) the 10K ohm resistor and the switch terminal itself.

3. Connect the other (free) side of the 3K ohm resistor to the base of the PNP transistor.

4. Connect the Emitter of the PNP transistor to +12 - so now you have both the emitter and one side of the 10K resistor (see step 2) connected to +12.

5. Connect the collector to pins 4 and 8 of the 555 IC.

The transistor will now switch +12 (minus ~0.7 volts) to the 555 whenever a ground is applied to its base via the 3K resistor.

This will be one of my last, if not my last, post to this forum as I have been assigned to Islamabad, Pakistan. I will be departing in about a month for a two year tour. After that I will be assigned to other locations throughout the world for the next 15-20 years.

Unfortunately, it will not be practical to keep my '91 Z28 which has been my "life" for the past three-odd years. I like it so much I thought about storing it for 10-20 years, but at $1,000 per year, well, you can do the math. Plus, who knows if we'll even be using gasoline in 20 years. I could always have it shipped to me (at U.S. Government expense) and "store" it whever I live, but it's better to "let go of the past".

I think I've sold it (I'm just waiting for the balance due) - but if the balance doesn't come through, I'll let it go for whatever price (hopefully at least a reasonable price) I can get. I'll know for sure in a couple weeks. Someone's going to get a very nice car.

This site has increased my knowledge base of cars and mechanical things tremendously, and am very fortunate to have found such a great place to hang out.

Cheers

//Stu

ANDYZ28

Stu, You will be missed!

Thanx a ton, ANDYZ28

TunedPort350

Stu,

I have this circut on a perf. board. The reason I posted the question is because on the 555 timer ic lists pin 1 as ground and pin 3 as output. I wasn't sure if I could just switch the two to go from pnp to npn (and of course the diode from the banded end going to #3 to the banded end going to +v now, ..... right?).

I, as well as many others I am sure, will truely miss you and your expertise.

Thanks for all of your help.

Bill

Stuart Moss

Bill,

You cannot do as you propose because pin 1 is ground for the IC (and pin 8 is VCC [or positive] along with the "reset" pin 4 which must be high [positive voltage > ~ 3 volts ?]).

Pin 3 is the switched output of the 555 IC. The only reason I put a diode across pin 3 and ground is to supress the voltage spikes that occur whenever a relay coil (or any coil for that matter - even applying or removing power to a transformer will produce a voltage spike because it's a coil just like a relay coil) is made or broken (voltage applied or removed). This occurs whenever a magnetic field collapses.

Anyway, the purpose of the diode is to supress this voltage spike. While the IC may work without it, chances are that sooner or later a spike will destroy it. This diode is often placed directly at the relay coil (cathode, or banded, side to the positive side and anode to the ground side of the coil). Since you're switching the relay coil with positive voltage (via the 555 IC) and the other side of the coil is connected to ground, it doesn't matter if the cathode (banded) side of this diode is connected directly to the relay coil or to pin 3 of the IC - it's all the same circuit (wire).

If you were to reverse this diode (anode to pin 3 and cathode [banded side] to ground, you would probably destroy the 555 IC and the diode or both. This is because the diode would conduct current from pin 3 to ground as soon as the IC turn's it output high (as in when it switches on your relay).

For the sake of simplicity (there is truth to the KISS (Keep It Simple Stupid) principle), I would suggest that you simply switch the ground (pin 1) to operate this circuit. Although it is common practice in the electronic community is to switch the positive power to a circuit, not power ground.

However, for this simple circuit, controlling the circuit by switching power ground should be okay.

BTW, I'm impressed that you actually assembled the circuit (and on perfboard). Most people I know would have no clue what a perfboard is let alone how to get a working circuit from it. Congrats!

//Stu

johnsjj2

I just ran each fan on a separate relay, and then ran them into a 90amp breaker. It has worked great for almost a year.

ANDYZ28

My fans are wired up exaxtly as the factory had them on the donor car. Exept for the ECM part.

TunedPort350

Josh,

We also have one relay for each fan, what we are talking about here is the control portion (what turns it on) of the relay. If you have one switch turning on both relays (fans) then they are both turning on at the same time. The two fans coming on at the same time puts quite a load on the alternator, voltage regulator, battery etc...

The circut we are talking about delays one of the relays (fans) from turning on (for 4-5 sec) even though there is only one signal (or switch). So one fan turns on (drawing about 700 watts to start up) then 4-5 sec. later the other fan turns on (drawing about 700 watts to start up), instead of both fans turning on simultaneously and drawing about 1400 watts.




Stu, (if your still there)

I think I confused you a bit. My previous post was inaccurate, sorry. What I was trying to ask is how to switch the ouput from pos. to neg. Not the trigger.

Matt87GTA

G0DSpeed Bro, and keep your head down, Hooah.

Stuart Moss

Bill,

Yea, you confused me but now I guess you know how to switch it with either applying a positive or negative voltage!

You were 100% correct then. If you want the output to switch ground instead of a postive voltage, just do as you stated - reverse the diode so it's pointing toward the positive lead on the coil, or in other words, the anode to pin 3 and the cathode (banded side) to positive. This can also be done by simply soldering the diode across the relay coil leads.

One side of the relay coil (the cathode side of the diode) will connect to +12VDC and the other side will connect directly to pin 3 of the 555.

Remember the 555 can sink ~ 200mA (0.2 amperes), which should cover just about all general purpose 12VDC relays which typically draw ~100mA. If you use a relay that draws more than 150-175mA, you'll should use a transistor (or another smaller relay to drive the larger relay) to increase the current rating.

//Stu

EDIT:If you don't have access to an ampmeter to test the current of the relay, a good test would be to simply touch the top of the 555 while it is operating (energizing) the relay. If it is too hot to touch, then it is probably providing too much current to whatever it is switching (via pin 3).

This is a good general purpose test I often do to any electronic circuit to see if it's operating too hot, and is a good test to see if a transistor (or?) needs a heatsink, or a larger heatsink (or fan, etc........).

8Mike9

You could build the circuit above, as described, or cuse a timed relay, or if you already have your existing setup installed and working, use a timer (looks about like a relay in size/shape) and install it to feed your current setup. Some are fixed timers, some are adjustable.

Aw, relay logic...my first job in the "real word". Heck, now days the only time I use test equipment is on my car

8Mike9

Oh, and FWIW, my '89 has 30amp fuses (or relays??, don't recall), and both fans come on at the same time, only Alt problem I had was with the original one losing the rear bearing...it was still charging though. Latest alt has maybe 15-20K miles on it.

Guess it could be an issue if you have a big amp, or other large draws.