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Dear Lazyweb...

I tinkered with some of my electronics projects last night; one of them is the toaster whose pop-up mechanism I plan to transform into a tea-bag popper. I've completely disconnected the heating elements and other extraneous stuff, so I have a pretty good view of the circuit board now. Here are pictures. When you push down the toaster's handle, it closes a circuit (which presumably includes an electromagnet and a potentiometer, a.k.a. the light–dark dial) by holding a couple copper tabs against metal contact points on the circuit board. I don't completely understand what keeps the handle down (and thus the circuit closed), but I'm not about to plug it into a 120V wall socket without the heating elements attached in an attempt to understand.

That brings me to what I believe is the next step: converting it to draw power from a 9V battery instead of 120V AC. In the lightly-documented example I'm following, their toaster's circuit board includes a bridge rectifier, which is not a component with which I'm familiar. Can I assume mine has a bridge rectifier as well? Can any of you tell by looking at the pictures whether it has one?

Comments

( 6 comments — Leave a comment )
clahey
Jun. 30th, 2006 05:17 pm (UTC)
I'm assuming that the little mini circuit board has an IC on it?

So, the copper tab that connects when you push down the button is connected directly to the AC, so that's gonna have to be rewired somehow to take power from the battery. It's not clear what all the AC is connected to. It looks like one side is connected through a diode to the IC and the other side is connected through a capacitor to the IC.

It looks like one side of the electo magnet is connected to the circuit board and the other side is connected directly to the AC. This might mean the electro magnet takes AC power.

My first guess was that that was a transformer instead of an electromagnet, but there's only 2 leads going to it instead of 4, so it's probably an electromagnet.
barawn
Jul. 6th, 2006 01:24 am (UTC)
Yeah, that's an electromagnet.

But there's no bridge rectifier there. It looks like a half-wave rectifier is built out of the cap and diode on the side near the IC with the two resistors setting the DC voltage (I think). If you trace from the AC on the left (by the dial), you'll see it go through the diode, then the cap goes in parallel back to the AC through a pair of resistors. That's a half-wave rectifier, with the IC acting as the load. There could be a resistor on that circuit board as well, but I can't tell (it looks like it's just epoxied on there, in case someone attempts to steal their precious toaster technology - and maybe because of the heat).

My guess for the way the whole thing works: the light-dark dial is a voltage divider on the DC generated by the half-wave rectifier. The IC gets both the DC (for power) and the divided-down DC (to determine how long to stay down). When the time's up, the IC connects the DC to the electromagnet, which ... does something to make it pop up.

Unfortunately to check if I'm right, you'll probably have to plug it in. If you remember how to reassemble it, you could always solder wires to either side of the capacitor, reassemble it, plug it in, push it down, and measure the voltage across the cap.

I don't have access to Make to see the way the other one was built, but if I'm right, you should be able to convert this to DC pretty easily.
barawn
Jul. 6th, 2006 05:07 pm (UTC)
comment 2, "I'm a dipwad" edition: most likely when the DC comes on (when it's pushed down) the IC connects the DC to the electromagnet, which holds the switch down. When the time's up, the IC probably cuts the voltage, and it pops up.

But the main reason I posted this is: why are you worried about connecting it without the heating element? How's the heating element connected up? It's probably just connected up in parallel with the rest of the circuit, which means that plugging it in without the element in won't do anything bad at all, though you'll need a way to close the switches.
radhardened
Jul. 8th, 2006 01:17 am (UTC)
why are you worried about connecting it without the heating element?
The article (wish I had an easy way to reproduce it for you) warns against plugging it into the wall because the circuit board's counting on having most of the voltage dropped by the heaters.
barawn
Jul. 8th, 2006 03:30 pm (UTC)
Where does the heating element plug in? I thought it looked like it plugged in in parallel: if it plugs in in series, then that could be a concern.

If it plugs in in parallel, though, it doesn't matter (voltage drop on one half of a parallel circuit is the same as the other half - Kirchoff's first law), and I'd bet they're in parallel, because those two resistors aren't going to be able to handle the current needed to make the heating element red hot, and the resistors are *definitely* in series with the rest of the circuit.

What are the color codes on those two blue resistors? Digital camera pictures: not good for decoding resistor bands.
barawn
Jul. 8th, 2006 03:50 pm (UTC)
Oh, I forgot: here's what I think the circuit traces out as. The IC connections were guessed as I couldn't see it in the pictures. Obviously, don't have any of the resistor/capacitor values as I couldn't see them either. I guessed on the directionality of the diode in parallel with the electromagnet, as well.
( 6 comments — Leave a comment )