Author Topic: Switching a load with no diode drop. Is it possible?  (Read 9014 times)

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Online IanB

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Re: Switching a load with no diode drop. Is it possible?
« Reply #25 on: December 11, 2016, 07:09:56 pm »
Quote
Just FYI, I have 2.8 V from a pair of NiMH in series. I would not like to connect them to an LED without any current limiting.

Please read reply #16. It's an LED with internal control.

That information wan't available at first.

It's also helpful to know for anyone else reading that NiMH cells do not have an unvarying output voltage of 1.2 V.
 

Offline hggTopic starter

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Re: Switching a load with no diode drop. Is it possible?
« Reply #26 on: December 11, 2016, 07:14:27 pm »
I know, but the flickering IC LED works fine from 2.0V up to 3.0V without any current
limiting resistor.
 

Offline hggTopic starter

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Re: Switching a load with no diode drop. Is it possible?
« Reply #27 on: December 11, 2016, 07:16:23 pm »
@Benta
Quote
Nah, makes no sense. Your statement that the battery needs to last for years and the LED is a type with IC makes it moot.

I asked for a schematic because it looks like that your suggestion will not work.
I might be wrong.
 

Online mariush

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Re: Switching a load with no diode drop. Is it possible?
« Reply #28 on: December 11, 2016, 07:22:29 pm »
Not to mention you didn't say they're rechargeable.  Regular NiMh alkaline batteries can be up to 1.65v when new, so you'd have 3.3v in series. 

There's also lithium batteries in the AA format, like the Energizer Ultimate Lithium, which could have as much as 1.7v when fresh out the box (but they quickly drop down to about 1.45v) : https://www.amazon.com/Energizer-Ultimate-Lithium-Batteries-Longest-Lasting/dp/B00003IEME?th=1

They're more expensive at around 2$ a piece, but at low currents they last way longer than regular alkaline batteries. Here's a graph showing the discharge at 200mA and a comparison between a lot of batteries available commercially :  http://batteryshowdown.com/results-lo.html  (click on the graph picture in the bottom right to see larger)
As you can see, at 200mA constant drain, they last about 50% more compared to alkaline, at only 20mA the difference will be much higher.
 

Online IanB

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Re: Switching a load with no diode drop. Is it possible?
« Reply #29 on: December 11, 2016, 07:31:14 pm »
the Energizer Ultimate Lithium, which could have as much as 1.7v

1.82 V even

Quote
when fresh out the box (but they quickly drop down to about 1.45v)

Not that quickly, in my experience. Even after considerable use I have still seen 1.6 V.
 

Offline madires

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Re: Switching a load with no diode drop. Is it possible?
« Reply #30 on: December 11, 2016, 07:39:23 pm »
BSS138
 

Offline hggTopic starter

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Re: Switching a load with no diode drop. Is it possible?
« Reply #31 on: December 11, 2016, 07:42:29 pm »
I have used the Lithiums.   Great batteries. 
Now I am using the AAA Eneloops BK-4MCCE.

@madires  Are there any similar low signal mosfets in THT versions?
 

Offline madires

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Re: Switching a load with no diode drop. Is it possible?
« Reply #32 on: December 11, 2016, 07:46:26 pm »
I'd assume so. You could try Digikey's parametric search.
 

Online KL27x

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Re: Switching a load with no diode drop. Is it possible?
« Reply #33 on: December 11, 2016, 10:02:06 pm »
BSS138 doesn't even begin to switch until about 2V. At resistance of >4ohm. Why even throw that out there?
You don't need a "brand name" part. You don't need switching speed. Rise/fall times. Just do search of small signal FETs. Enter gate voltage SUB 1V. There are many with upper gate threshhold voltage (guaranteed max voltage where switching begins) of under 800mV! Another thing that will universally be the case with these guys is MAX +- gate voltage of 8V and MAX drain-source of 20V. This is because to make the gate more sensitive, dielectric must be made thinner. Already instant tipoff that BSS138 is garbage (in this context) is 50V rating and +-20V on the gate.

And sort it by price. Cheaper = better. :) Look at price, package, and the the other things I mentioned. Then to dig a little deeper before you pull the trigger, look at the little graph in the datasheet. The one that shows drain-source resistance as a function of gate voltage. That and the price/package are really the only things you need to know for your particular problem. Every other spec for any FET on the market will more than meet all your other requirements. But you can't sort by "graph." What you want to see in this graph is a low turnon point, significantly lower than 2.4V. And a low resistance at 1.8 to 2.4V. While BSS138 might work fine for one light, if you make 5 of them and want them to match, you don't want to cut it that close for no good reason. There is considerable variation in the start of saturation between individual FETs.
« Last Edit: December 11, 2016, 11:59:52 pm by KL27x »
 
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Offline Audioguru

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Re: Switching a load with no diode drop. Is it possible?
« Reply #34 on: December 11, 2016, 11:20:41 pm »
The Dollar Store sold cheap Chinese LED ornaments with a colors-changing-LED that has a controller IC inside the LED. The ornaments are powered with three 1.5V button cells in series and have a 51 ohm current-limiting resistor in series.

They also sell flickering LED tea lights but I have not looked inside one to see if it uses a current-limiting resistor.
 

Online David Hess

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Re: Switching a load with no diode drop. Is it possible?
« Reply #35 on: December 12, 2016, 08:26:30 am »
Thx. Trust me. Gate drive voltage > 2V and load driving voltage > 0.5V (approaching zero, even) and low current, there is no way BJT will compete.

*edit: And it's not even close.
At 300mOhm resistance, FET will drop only 0.003V at 10mA current. Since there's that little bit o drop, perhaps your LED only draws 9.5mA, though. Ya know.

They make low Vce saturation bipolar transistors which can easily meet or beat that like the ZTX618 (0.0007V at 10mA maybe?) but it is possible to do even better without saturated switches or even germanium transistors.  If current gain is not an issue, then swap the collector and emitter leads of a general purpose transistor for even lower Vce saturation.  They used to make special "chopper" bipolar transistors with symmetrical construction and low saturation but low gain for use where today JFETs and MOSFETs are used in chopper applications.

Using a bipolar transistor as a common emitter switch to drive an LED at low voltages allows another interesting possibility.  Use a pair of bipolar transistors, match them for Vbe, and configure them as a current mirror with gain using a resistor in series with the emitter of the lower current transistor.  Now the transistor which is switching the LED also operates as a rough constant current source.
 
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Offline orolo

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Re: Switching a load with no diode drop. Is it possible?
« Reply #36 on: December 12, 2016, 01:47:29 pm »
They make low Vce saturation bipolar transistors which can easily meet or beat that like the ZTX618 (0.0007V at 10mA maybe?) but it is possible to do even better without saturated switches or even germanium transistors.  If current gain is not an issue, then swap the collector and emitter leads of a general purpose transistor for even lower Vce saturation.  They used to make special "chopper" bipolar transistors with symmetrical construction and low saturation but low gain for use where today JFETs and MOSFETs are used in chopper applications.
Thank you for this pearl! The incredible versatility of a humble bjt connected in unconventional ways never ceases to amaze me. I'll put high on my experiment list to explore reversed transistor characteristics. They are often dismissed as low performing, without discussion of possible merits.
 

Online Zero999

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Re: Switching a load with no diode drop. Is it possible?
« Reply #37 on: December 12, 2016, 03:28:50 pm »
@benta
Quote
but the simplest solution would be to connect the LED with series resistor directly to the battery and short the LED with an NPN transistor during daylight. Eventually a resistor in the collector to limit IC.

Can you post a schematic of your suggestion ?
Can you please post a schematic of what you tried originally?
 

Online KL27x

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Re: Switching a load with no diode drop. Is it possible?
« Reply #38 on: December 13, 2016, 02:23:28 am »
@ David Hess
Mind blown. I ordered a few to test. I am sure I won't figure out a practical use for another 10 years.* Esp since they are kinda pricey. (Not because this isn't amazing; just cuz I'm dumb.) But yeah. Mind blown. I'm hardwired to think FET for efficient power switching; at lower frequencies, anyway. I guess at lower current, the script might flip!?

*Well, one thing. FET driver. Wonder where I got that idea? :)
« Last Edit: December 13, 2016, 02:57:34 am by KL27x »
 

Online David Hess

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Re: Switching a load with no diode drop. Is it possible?
« Reply #39 on: December 13, 2016, 03:14:32 am »
Thank you for this pearl! The incredible versatility of a humble bjt connected in unconventional ways never ceases to amaze me. I'll put high on my experiment list to explore reversed transistor characteristics. They are often dismissed as low performing, without discussion of possible merits.

The most common albeit still very rare place I see bipolar transistors used in reverse (Jim Williams calls it "inverted mode") is when an integrating capacitor needs to be discharged to as close to zero volts as possible.  This is not quite such a big deal now when we have low threshold voltage MOSFETs available but what if you want to build a high performance integrator which runs on the lowest possible supply voltage?  Linear Technology application note 15 has examples in figures 3 and 6 and discusses the problems with FETs in general on page 4.
 
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Offline rob77

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Re: Switching a load with no diode drop. Is it possible?
« Reply #40 on: December 13, 2016, 10:30:50 am »
If current gain is not an issue, then swap the collector and emitter leads of a general purpose transistor for even lower Vce saturation. 

interesting , definitely going to try it  :-+  but it will be good for small voltages only as the Vbe breakdown voltage is only few volts.
 

Online David Hess

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Re: Switching a load with no diode drop. Is it possible?
« Reply #41 on: December 13, 2016, 01:59:54 pm »
If current gain is not an issue, then swap the collector and emitter leads of a general purpose transistor for even lower Vce saturation.

interesting , definitely going to try it  :-+  but it will be good for small voltages only as the Vbe breakdown voltage is only few volts.

Low Vbe breakdown voltage is definitely a limitation.  On the other hand some transistors, usually PNPs but not always, have high Vbeo breakdown voltages.  In a few cases, their Vbeo breakdown voltage is higher than their Vceo breakdown voltage like the currently produced Toshiba 2SC3326 which is intended for inverted operation for muting in audio applications.  The CMPT404A which is a surface mount 2N404A is another good example designed for chopping applications.
 

Online KL27x

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Re: Switching a load with no diode drop. Is it possible?
« Reply #42 on: December 17, 2016, 03:37:33 am »
Color me a nerd. I got my transistors, a couple days ago. And since it is Friday night, I obviously have plenty of time to play with transistors, lol.

I have wired up the FET vs BJT challenge.

Red corner:
ZTX618
NPN transistor.

Blue corner
BSS816NWH6327XTSA1 (this just rolls off the tongue, don't it?)
This is the "PFET" which I suggested earlier, based on a 30 second Mouser search, sorted by price. And apparently, I can't tell the difference between a PFET and an NFET, cuz it's an NFET, lol

 
So I put 200R resistor on Collector/Drain. And for the NPN, I put a 2K resistor on the base. Gate, direct connection. Gate/base biased with Vsupply.


Red corner - 30 cent BJT
@ 2.4V, my PSU shows 9mA (experience tells me this reads low, but it verifies circuit working as intended) and my multimeter (Cheapo Uni-T pen meter) shows Vce 0.004V, Vsupply 2.396V
@ 1.8V, my PSU can't decide if it's 4mA or 6mA... and penmeter shows Vce of 0.004V


Blue corner - 8 cent FET
@2.4V, 8mA, Vds 0.001V, Vsupply 2.396V
@1.8V, 4ma, Vds 0.001V

The BJT is impressive to me, but the numbers don't lie. FET still wins, here. Well, THIS particular sample, anyway. Maybe next sample of same FET not as low. I'll have to try the BJT upside down, next. :)

Real beauty of FET for low frequency power switching, though... can run them in parallel ad infinitum to decrease voltage drop even further while increasing max heat dissipation. BJT seems to bottom out, as per datasheet, 4mV is the drop at even 100mA.

So, anyhow, OP has many options. Yes, you can switch without a diode drop using BJT or FET.

* Using 29.3 ohm load resistor, to give ZTX room to shine, Vsupply 2.31V (voltage drop from long leads!), and 200R base resistor, Vce is 0.009V. So it is not meeting spec at this voltage. (This is only 78.8ish mA, but I didn't have right resistor to get 100mA.)

Same load resistor and supply voltage, NFET has Vds of 0.009V. TIE!

At 2.3V,  BTS FET is right on par with graph in datasheet at this point. ~9.46mV drop is what would be expected per the graph. I'm not inclined to think this is particularly "good"  example of this FET.... even though it outperformed at 10mA, the graph doesn't scale down very far. But maybe the BJT is a bad one...
« Last Edit: December 17, 2016, 07:01:27 am by KL27x »
 
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