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Diodes in Series misbehaving - need electrical help!

dethlore

dethlore

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I have been trying to build a fairly elaborate circuit for the past month or so and kept running into an underpowered problem, so last night I did some actual testing with the bare essential components. Here's what I have:

Two 18650 in series (7.4V)
LM1185 regulator set to 7V and 700mA
Two Red Diodes rated for 700mA

Tests:
Powering just one diode(tried both separately), I get the 700mA no problem, with diode Vf of 2.7V
Powering dummy load for 6V, I get 700mA
Powering two diodes in series and I can only get 450mA out of each with a total Vf of 6V

Problem:
Obviously... these should be at 700mA, not 450. What in the world could cause this? I have done hours of research and I am stumped. I need help!

Thanks!
 





Sigurthr

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Draw a schematic of your actual circuit with the two diodes in place, from there we can help.

Btw two LiPo/LiIon in series is 8.4V, not 7.4V.
 
dethlore

dethlore

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Sigurthr said:
Draw a schematic of your actual circuit with the two diodes in place, from there we can help.

Btw two LiPo/LiIon in series is 8.4V, not 7.4V.
Click to expand...

Yeah sorry. Here's a diag. I didn't write on there (sorry) but the output current should be 700mA (and it is for one), but is only 450 when two diodes are in series. oh... and the un-valued resistor is 20k. The specs for the LT1185 makes this ~7V , 700mA

I know the voltage is 8.4 when freshly charged, I just said 7.4 because they are 'nominal' at 3.7 per cell.

Thanks again

1185diodes.jpg
 
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BShanahan14rulz

BShanahan14rulz

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Your schematic may not match your actual circuit. If it does, I would say you need new batteries, if two cells sag down to 2.7V when pulling 700mA from them :whistle:

But you might also consider taking a photo of your setup for us.
 
dethlore

dethlore

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BShanahan14rulz said:
Your schematic may not match your actual circuit. If it does, I would say you need new batteries, if two cells sag down to 2.7V when pulling 700mA from them :whistle:

But you might also consider taking a photo of your setup for us.
Click to expand...

Sorry, thought I mentioned... the volt drop of each diode when in series (measured on the anode and cathode of the diode) is about 3V. So when it's in series, there is actually 6V across the 2 diodes which leaves another 1V or so to be burned off by the regulator (although I didn't actually measure what its Vd is).

You may be right on the cells though. I may indeed have a bad cell... I'll have to figure out each cell's health to see if it is a bad one. Setting the regulator to a max of 3V and 700mA gives me no problems. Single LED has 2.6Vd and 700mA.
 
KrowBar

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Maybe I'm missing something, but why isn't Vout connected to your load? It looks like the diodes are connected directly across the battery in parallel with what is effectively a 20micF cap and your LT1185. This will load the batteries in some way as to drop the voltage (the undrawn series resistor due to internal battery resistance) With the current at that voltage split between the 1185 and the diodes (and the caps during the transient). The behavior you reported suggests that the series diode load with its higher effective resistance (than a single diode) is pulling less current. Double check your circuit/schematic and get back to us.

[edit] What stumps me is how you could be lucky enough to get the behavior you expect from the dummy load
 
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dethlore

dethlore

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KrowBar said:
Maybe I'm missing something, but why isn't Vout connected to your load? It looks like the diodes are connected directly across the battery in parallel with what is effectively a 20micF cap and your LT1185. This will load the batteries in some was as to drop the voltage (the undrawn series resistor due to internal battery resistance) With the current at that voltage split between the 1185 and the diodes (and the caps during the transient). The behavior you reported suggests that the series diode load with its higher effective resistance (than a single diode) is pulling less current. Double check your circuit/schematic and get back to us.

[edit] What stumps me is how you could be lucky enough to get the behavior you expect from the dummy load
Click to expand...

derp... yes. I drew it wrong. The load neg connects to pin 4 (not the resistor, but to pin 4) my bad. It's driving me up a wall that I get the right load on the dummy, but not the diodes. I know diodes differ between each other, but that much of a difference doesn't make any sense.
 
KrowBar

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What is your dummy load? How are you measuring the current both for dummy load and series LD?
 
dethlore

dethlore

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KrowBar said:
What is your dummy load? How are you measuring the current both for dummy load and series LD?
Click to expand...

dummy load is just like the oooold popular one here on the forums but it's with .7V 3A diodes and a 1 ohm 5 watt resistor. I measure current two ways. 1, I measure mV across the resistor which makes 1mV = 1mA. 2, with the Multimeter in 'current' mode up to 10A, I connect the driver out pos to mmr pos, and mmr neg to load pos. load neg goes to capacitor neg and then to pin 4.

measuring current on the diodes, I use method 2 (connect mmr in current mode in series with the load).

instead of re-drawing it.. just mentally disconnect the line that goes from cap to pin 3 on the right side, and then connect that cap to pin 4. :) (I'll work on getting it uploaded... gimme a few minutes)
EDIT: Updated the diagram (see above)
 
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Sigurthr

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I haven't looked at the datasheet to confirm it so take this with grain of salt; but either your pinout on the drawing or your wiring has pins 1 and 3 swapped (shown: 1 to Vcc and 3 to gnd).
 
The Lightning Stalker

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This isn't a super high draw circuit. I'd say eliminate the batteries and substitute a good 2-3A 12V power
supply for testing purposes. If it still doesn't work right, then you know the problem is somewhere other
than the batteries. Recheck the component layout, and the actual voltage drop of each diode running by
itself at 700mA. Each diode is unique and there will be variance from one to the next.
 
Cyparagon

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I ran those resistors through the formulas, and you've got it set at 7.87V with a 750mA current limit. I don't know that the regulator will operate happily since the minimum dropout is ~.4V and your supply is 7.6V. Furthermore, minimum dropout voltage might be higher when current limiting.
 
The Lightning Stalker

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I think Cy is on to something. The voltage should be set much higher than the drop of the diodes, as high
as it can go. I think pin 2 should be tied directly to pin 1 and leave out the two resistors attached to pins
1, 2,and 4. Verify operation with a test load though just in case I'm wrong. The other resistor should be
closer to 21.4k to get 700mA. A multiturn pot might be better there than a fixed resistance.
 
dethlore

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I think you guys have it. ..
1. I remember previously building a driver for an m140, and I completely removed the voltage setting. That seemed to work very well for that. . I'll try that in the morning.

2. I tested the batteries with drlava's method and found out my batteries are in really bad condition! Haha. I'll have to hook up the power supply for testing.

I'll keep you guys posted. Thanks a ton for helping me see the things I overlooked! !
 
dethlore

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Interesting... my batteries have huge internal resistance. Some different batteries did a better job at 580 mA, but their Ri was high ish as well. Power Supply gives 680mA and odd enough, these diodes were running 3.4V for that current. They're rated for 2.6v@700mA

Used power supply on a full circuit through a transistor and logic mosfet and I'm back to 450ma, but I think it's because I wasn't expecting 3.4v Drop at desired current.

Anyways.. couple more tests to run... but I think their higher than expected forward voltage is the reason I'm having issues.
 
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