Maker Pro
Maker Pro

1N4007 question

john2k

Jun 13, 2012
188
Joined
Jun 13, 2012
Messages
188
I have 2 LED drivers that I want to drive my LED unit with. One driver is of higher constant current than the other. Both drivers will never be on at the same time. The output of driver 1 and the output of driver 2 will be joined together in parallel and then connected to the input of the LED unit. As a pre-caution I want to add a 1N4007 signal rectifier diode on the positive output of each driver so that current only goes out. Wiring it up without the diodes work fine, but when i put the diodes, I noticed that the diodes do get a little warm. the one linked to the higher current driver gets much hotter. The diodes are supposed to be able to handle 1A. Would it be better to maybe add the 1N5408 which are supposed to be able to handle 3A. Will the heat be much less on a 3A one as apposed to a 1A one?

Just wondering what would happen to a 1N4007 if it overheated or burnt out and damaged? what would it do to the current flowing through it? Would completely leaving out the 1N4007 be a better option? without the 1N4007, each driver can send current through the output of the other driver, will that effect the other driver? do drivers reject current coming in from the output?

Thanks
 
Last edited:

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
What is a little warm? Warm to touch, but you can leave your finger on it without burning it? If so, that is perfectly okay. If you want to dissapate less power in the diode, using a higher current diode will not help, though if it is phyically larger it might get less warm. To reduce the power, use a shottky diode.

Bob
 

john2k

Jun 13, 2012
188
Joined
Jun 13, 2012
Messages
188
The diode that is connected to the positive line of driver 1 gets very hot i can only hold my finger on it for about 5 or 6 seconds. I believe driver 1 may require a higher rated diode because i think it is actually trying to push 1.8A through whereas driver 2 is pushing only 0.7A through hence why driver 2 is not causing the diode to get hot.

I am not using the diodes to reduce or restrict power in any way, just want to make sure when one driver is on and the other is off that the power from the on driver output doesn't end up going into the off driver output. can a driver that is turned off actually get damaged if power is pushed through the output by the other driver?
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
You are correct to isolate the two drivers with diodes. Do use a diode that is rated for more than the current you expect to go though it. If you want less heating, use shottky diodes, they have a lower voltage drop and will therefore not heat up as much, or waste as much power.

Bob
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Perhaps you need to calibrate your finger?

It seems you're between 4 and 5. This is probably acceptable, but I really don't like things getting that warm.

Leaving longer leads (to help radiate heat), having adequate ventilation, and soldering the device to a board with generously sized pads and tracks can all help.

A higher rated diode won't get less warm (see note below), but will likely be physically larger, providing more area to radiate heat and thicker leads to draw it away from the junction.

As noted by previous posters, a schottky diode has a lower forward voltage drop, and this will reduce the heat dissipated at a given current. Watch out because these have (generally) lower voltage ratings too.
 

john2k

Jun 13, 2012
188
Joined
Jun 13, 2012
Messages
188
thanks guys,

i bought a 3A Schottky Barrier Rectifier - 1N5820.

It definitely works much better and hardly produces any heat that I can feel.

I have a few questions about the Schottky Diode 1N5820, hope you guys can help. On the regular signal diodes, if I do a continuity test on it with a multi-meter, it will only beep if wired the correct way. However, with the Schottky Diode it beeps both ways. At first I thought oh no it doesn't seem to block current from flowing wrong way. So then I wired up a DC power source to measure the voltage coming out of the Schottky Diode wired both correct and incorrect way, I also was able to read the voltage with a multimeter with the Schottky Diode wired incorrectly. Once again, I thought, oh no it's not blocking. Then final test I wired up LED light incorrectly and correctly, and it only lit up when wired correctly.

Just wondering is this all normal for the Schottky Diode? because a regular signal diodes doesn't behave this way.

And finally, will the Schottky Diode work for my particular application to isolate the two drivers with the diodes?

many thanks
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Schottky diodes are generally quite leaky.

In this application they will isolate the two sources, but there will be some leakage. If you could tell us the voltages we could figure it out from the datasheet. (you could measure it across the diode).

Because of their high leakage, you can find they get heated more, not from the forward current, but from the reverse!
 

john2k

Jun 13, 2012
188
Joined
Jun 13, 2012
Messages
188
Thanks for your reply Steve,

Driver 1 delivers 3.2V at around 1.3A. whereas driver 2 delivers 3.2V at around 0.6A. The regular 1N4007 on the positive output line of driver 2 seems fine in terms of letting current flow through and does not heat up much. Since Driver 1 is delivering a higher current, I want to put the 3A Schottky diode on there. So i guess my question is, will a regular 1N4007 on driver 2 be able to block current from driver 1 and will the Schottky diode be able to block current from the lower current driver 2?

I've quickly sketched up a diagram of the diode application below, do you think the diode arrangement will be ok?

Thanks

diodelayout.jpg
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,700
Joined
Nov 17, 2011
Messages
13,700
The 1N4007 should be fine in this application.
The Schottky diode, too. The leak current should be too small to bother driver 1.
 

CocaCola

Apr 7, 2012
3,635
Joined
Apr 7, 2012
Messages
3,635
Note that you will no longer have 3.2V, due to the forward drop of the diodes...
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
Unless it is a constant current supply, which he did mention in the OP.

Bob
 

john2k

Jun 13, 2012
188
Joined
Jun 13, 2012
Messages
188
Unless it is a constant current supply, which he did mention in the OP.

Bob


Yes you are correct it is a constant current supply. And the voltage doesnt change even though I have the diode. I am guessing the constant current driver automatically adjusts to allow the small extra voltage for the diodes?
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,700
Joined
Nov 17, 2011
Messages
13,700
Yes, it does as long as the load (LED) is within the compliance range of the driver. Which it obviously is.
 

john2k

Jun 13, 2012
188
Joined
Jun 13, 2012
Messages
188
I've just got a few 1N5822 SCHOTTKY BARRIER DIODES. The specs are:

Vfmax = 0.525 @ 3.0 A
PIV = 40VDC (28VAC RMS)
If = 3.0A Continuous (80A PEAK)

Just wondering what the 0.525 forward voltage is about? Is that the amount of voltage that the actual diodes uses? or is that the maximum voltage the diode can let through?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
It is the amount of voltage that is dropped across the diode at the given current. If I used a water analogy, imagine the diode is a one-way valve. The valve is not perfect and if the input pressure is x psi, the output pressure is a little less -- the difference is the pressure dropped across the valve. Diodes are also like valves in that the amount of voltage (pressure) dropped across them grows as the current (flow rate) increases.

0.525V is a fairly small amount for a diode at that current (3A).
 
Top