I'm having trouble understanding this schematic for a quadrajet relay powered choke. Specifically the #85 and #86 terminals. As far as I can figure, when the ignition switch is turned to the "on" position power is sent to the #2 terminal on the alternator which then powers the relay coil and switches it from terminal #87A to terminal #87 which powers the choke. However, as soon as the engine fires, the #2 connection at the alternator breaks, then the relay coil loses power to it allowing the relay to switch back from terminal #87 to # 87A which shuts off power to the choke, thus disabling it form functioning. This makes no sense to me. What am I missing here?
If it's tied to gen light feed that's what it will do. Your choke relay would also work if the gen/alt took a dump threw the belt etc. anytime the gem light goes on the choke will work. Why a relay anyway?
This style of powering a quadrajet choke is something GM did from the factory. The other way I found is using the oil pressure switch to do it, but I will be running a gauge and not an oil light, so wanted to do it this way. The idea is to have the choke powered only when the engine is running. Oil pressure switch method:
GM three wire alternator, right? You don't have all of the diagram showing, that's why it's confusing. The wire off the alternator to the relay/idiot light is coming off the T1 terminal (or should be). The idiot light is fed with 12 volt power from the ignition switch. The light comes on because without the alternator charging, the T1 terminal supplies the ground for the light. No charging, no power to the relay. Start charging, and the T1 terminal now shows 12 volts. Because the light is now getting 12 volts from both sides (no ground or difference of potential), it goes out. But the relay is now getting 12 volts, and has it's own ground path, so the relay closes. Power to the choke, yer good to go....
Because there's no difference of potential. All wires on the power side of the relay coil have the same voltage in them when the alternator is charging. That's the same reason the relay doesn't pull in when the light is on.
The following post is a good explanation of the two types of circuits used by GM. I've only ever used the circuit that feeds through the oil pressure switch, so I can't speak to the relay circuit. Per the explanation below, the relay shouldn't see 12V until the alternator is turning; then the field wire output energizes the relay coil. This was posted by Joe Padavano on another forum: This question must come up monthly, so I figured it was finally time to write a little tech article. Many folks use the wiper power feed, but this has problems. Simply using a switched power source, especially one that is hot in the ACC position, will cause the choke to open prematurely if the key is on but the engine is not running. Also, the combination of the always-on choke heater and the wiper load may blow the fuse. GM used two different methods to power the choke, depending on the vehicle. Most used an oil pressure switch in the power line to the choke. This way, the choke doesn't start to heat up until the engine is running and there is oil pressure. On my 62 F-85, I ran a fused wire (20 A fuse) from the BATT terminal on the alternator to the switch, and from the switch to the choke. Very simple to wire and very clean. I like to use the Standard P/N PS64 switch. The middle terminal works the OIL light, the other two are a normally open switch that closes with oil pressure. Use these to control the choke. If you have an oil pressure gauge instead of a light, simply use a brass tee fitting to connect both this switch and your oil pressure sender. Here's the sender: Here's the wiring diagram: Report this image The other way is to use a relay that is triggered by the field wire on the alternator (brown #1 wire on SI-family alternators, for example). The field wire is only hot when the alternator is turning. This is how GM wired the stock G-body cars with electric choke when the gauge package was installed. Of course the down side with that is that the choke will close if you break the alternator belt. A typical wiring diagram for the second method is here:
what is the resistance of the relay coil, vs the resistance of the bulb? it's a voltage divider circuit, do the math.
Getting one to work on a Qjet is a pain, also you have to remember to push the knob in, which is my main problem. It's not that difficult to get the choke to work right, if you're patient.
There's still some confusion on how this works..... I'll take another shot. Here's a couple of diagrams showing more of the circuit (or at least the parts we're interested in). The top diagram is when the alternator ISN'T charging. Close the ignition switch and the 12 volts will flow through the idiot light to the T1 terminal and the light comes on. Now, T1 technically isn't a 'ground', but because it's potential (voltage) is zero, it acts as one. The power is simply flowing through the 'dead' circuit and using it's ground. Because T1, wire junction '2', and 85 on the relay are all common (the latter two in effect connected to T1) and 'grounded', the grounded 86 on the relay has the same zero potential and no current flows. The bottom diagram represents the alternator charging. The regulator applies 12 volts to T1 once the alternator starts charging, and now you have 12 volts feeding into both sides of the idiot light; in effect, this would be the same as connecting both lamp leads to the positive battery post. So the lamp goes out, no difference of potential between the leads. But you now have 12 volts at 85, it flows through the relay coil to ground and the relay operates. It's possible to have current flowing through both the light and relay, but only if the voltage present at T1 is different from the voltage available at the switch side of the light. How much current flows will depend on the difference, but if the light goes out, the circuit will work as-designed. If the alternator partially fails and voltage at T1 drops enough for the light to glow intermittently or dimly, voltage to the relay may drop below the relay's 'pull-in' voltage and the relay may open. As an example, if voltage at T1 is 11.5 volts and it's 12 volts at the ignition side of the light, you have a difference of .5 volt going through the light. So with a lamp resistance of about 36 ohms, current through the light would only be about .013 amp, or about .07 watt (about 2% of rated output). What the resistances of the lamp and the relay coil are would have no effect on the voltage delivered, it would just change the amp numbers in the individual branches. I'm sure the reason for the relay rather than a direct connection to the alternator is the required current for the choke heater is beyond the regulator capability.
Shouldn't those drawings show the ignition switch as closed? Which is how the 12V goes to the alternator to excite it? And from what I've read, the choke is wired through the relay so that it only gets voltage when the engine is running(alternator charging)
Yeah, I could have showed the ignition switch closed, but to make it more obvious I showed it open. And the relay is only there to protect the voltage regulator from too much current draw; the circuit would still work if the choke heater was directly connected to T1 for the same reasons outlined above if the regulator could deliver enough amps.
OK, I see what you are saying now. I kind of find it confusing because it appears as though there would be 12V to the choke as soon as you turn on the ignition when wired like this. The ignition sends 12V to the alt terminal one, but it seems to also send it to the choke at the same time. Then once the alternator is charging, it sends the 12V back which shuts off the alt light, but it is now sending 12V to the same relay coil terminal, just from a different direction. I can't figure out why the 12V from the ignition switch through the alt light doesn't power the choke relay. I understand that it works, just don't know why??
Because electricity always takes the path of least resistance and given the 'choice' of going directly to ground though a very small resistance versus going through a larger one to get to ground, it will take the small one every time. There is no doubt a very, very small amount of current going to the relay even when the light is lit, but it's so small you won't be able to read it on a hand-held meter; you'd need ultra-sensitive precision lab equipment to measure it. For practical purposes, that amount is zero.
How do you wire this using a 1 wire alternator? I ran fused power from alternator to #30 on a relay then from # 87 to choke Checked power and relay doesn't close. it seems like I always have power to #30 (13volts when switched off) I toook # 86 and went to ground still doesn't close circuit. Do I need to wire differently? Just a little more info, I have alternator wire going from amp meter to alternator so I picked up my alternator power from the amp meter. Does that work?
To save time and a headache, just wire the relay from the HEI Batt terminal if you have HEI, or the coil wire coming from the OEM wire harness. The choke will only be active when the Ignition is on. The relay will switch with reduced voltage from the resistor wire if you are still using one. Simple relay information; 85 is the input and 86 is the ground terminal of the relay, they control the on and off ( switching ) of the relay. #30 and #87 are connected together inside. 87A is non-functioning when relay is OFF. Once you switch the relay ON, #30 and #87A are now connected and #87 is non-functioning. #85 and #86 are independent of the other three, they do not touch or use the same power source in this case. 85 or 86 must go to ground, the other must have a power source that can be turned on and off ( ignition feed ).
Since the wire from the alternator is always connected to the battery, the thing will never turn off. Connect the relay relay ground wire to a normally open, two terminal oil pressure switch, and the other side of the pressure switch to ground.
You have no switched power going to terminal 85 to control the relay, only 3 out of your 4 needed wires are hooked up.
Hot rods are fire hazards...you an use an ammeter with an alternator, if you make sure the wiring and ammeter are sized for the current that they will encounter.
This. Now, just run a wire from the + terminal on your coil to terminal 85 on the relay, whenever the key it turned to the Run position the relay will be activated and current will flow to the choke coil. Normally it's not wise to connect anything to the + terminal of the coil, in this case the current draw is so low it will not negatively affect the operation of the coil. For peace of mind you could fuse protect the wire from coil to relay.
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