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Topic: Doubling PWM voltage (For LED Driver) (Read 2 times) previous topic - next topic

retrolefty

So you got two different posting threads going on the same project at the same time? The mods usually stomp on that kind of thing.
;)
Lefty

Longarms

Mike, ahh okay, yeah I see the difference. Hey, I did something right!

Lefty, I wasn't sure what to do 'cause this post is about the project, but I had another question about a piece of hardware not working how I expected it to. So... project here, hardware in hardware.

A mod should probably just delete both of these and i'll start over... =\
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

Longarms

I picked up some IN4003 diodes, in series between Vin and the resistor, but they're only dropping the voltage by 0.3V each. Don't really want to put 5 diodes on each of these.

Anyone have any ideas? (Either about why the diodes aren't working properly, or other options to drop from 11.6 to 10)
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

Grumpy_Mike

Quote
but they're only dropping the voltage by 0.3V each.

No they don't. Have you put any sort of load on it?
Try a 4K7 resistor.

Longarms

I did a 1k resistor, I'll try a 4.7K tonight.

To double check that I'm doing this right, the "load" resistor goes where my driver would go, right? And I measure voltage between the resistor and ground?
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

Grumpy_Mike

Using 1K is more of a load than 4K7 so I would expect no difference. You simply must be doing something wrong. What sort of diodes are you using? It is physics a silicon diode drops 0.7V across it.

Longarms

#21
Apr 05, 2013, 03:06 am Last Edit: Apr 05, 2013, 03:08 am by Longarms Reason: 1
Sorry for the long delay, I was sick all of last week, and I'm just now getting everything back in order... So lets get into this so we can figure out how I'm screwing everything up =)

I ripped my breadboard apart, and started over. First thing first, getting Vin down from 11.9V to 10V

Here's a gallery showing my set-up. Vin->IN4003->IN4003->Resistor (A 4K7 is pictured, but I also used a 1K and 10K with the same results)
Pictures 4 and 5 are swapped around in the gallery, but they're meant to show that the load doesn't change voltage.
http://imgur.com/a/Awx8u

I'm not sure if I'm measuring right, I dono, I'll let the pictures speak for themselves...

Thanks! I'll do my best to reply to any questions as quickly as possible.
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

Grumpy_Mike

Sorry to hear you have been ill, it's no joke.

Right. When I say put a resistor to load it, I mean connect the resistor from the output, that is the cathode of the last diode to ground. In that way it pulls current through the diodes and gives you the correct voltage drop.
The way you have it the resistor is in seriese and as you rightly say will not make any difference.

Longarms

#23
Apr 05, 2013, 03:33 pm Last Edit: Apr 05, 2013, 03:46 pm by Longarms Reason: 1
Ahhhhh, okay!
I'll try it when I get home! Hopefully I can get this working this weekend.

Thanks!

EDIT: To be clear, is this what you mean? http://imgur.com/Xf9W8wo Of course, the driver won't be hooked up yet ('til I know this is correct)

Actually, I don't understand the difference between what I did on the breadboard, and this drawing. Picture 4 (should be 5) showed this...
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

Grumpy_Mike

Quote
To be clear, is this what you mean?

Yes that's right.

Quote
I don't understand the difference between what I did on the breadboard, and this drawing. Picture 4 (should be 5) showed this

If you look at the picture you will see that you just had the other end of the resistor not connected to anything. Therefore it was not in the circuit. That is it is not affecting anything in the circuit. When you connect your meter to the other end so it is in the circuit it is "only" in series with the meter. As the meter is a very high impedance the resistor is insignificant in comparison and so does not (quite rightly ) affect the reading.
By putting the resistor to ground you have a circuit for the current to flow round, from the power +ve, through each diode, through the resistor to ground and back into the power supply. The current flowing is mainly determined by the resistor. It is this resistor that is pulling current round the circuit and will allow the diodes to drop their normal forward voltage drop.

Longarms

Ahh, of course, absolutely. I knew I understood you at first, then i made the drawing, then i guess I forgot what I understood in the first place. Got it, will report back in a few hours (Maybe... I think the girlfriend wants to go on a date tonight...)
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

Longarms

Hey! It worked!
                                                                                                                                                                                                   
I'm at 9.82 with 3 diodes, I think that should be good, no? My other option is 10.44

I was kind of surprised that I could measure the -.62 across the diode... I think I have no idea how electricity works.

Now onto hardware design. This is so exciting!
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

Grumpy_Mike

Well done.
Have you tried a heavier load like a 1K resistor? Also check it in the circuit you are using as three may be a touch too much.

Longarms

I tried a few different loads, 1K 4K7 and 10K I think.

Should I start a new thread for my hardware (Drivers, connectors, wires, housing, etc.) stuff? I'm putting together some drawings now, but I want a critique before I go buying stuff.

Thanks!
We auctioned off our memories In the absence of a breeze. Scatter what remains, scatter what remains.

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