Switching a load with no diode drop. Is it possible?

[hgg]

Hi,

I have a supply of 2.4V from two NiMH batteries and I would like to switch on a (flickering IC) LED
at night with a transistor but I get a diode voltage drop so the LED will not light up.

I was wondering if it is possible to switch the LED without any diode drop.
(I want to avoid the Joule thief solution)

Thanks.

[rob77]

with a emitter-follower you'll always have 1 diode drop - at the B-E junction.... a common emitter topology (BJT as a low-side switch) will eliminate that 1 diode drop.
but anyways - you need a current limit for your LED - either a current limiting resistor or drive the diode with constant current.

[hgg]

I do not need any current limiting resistor, because at 2.4V the LED draws 10mA.

Don't we have a voltage drop from Collector to Emitter?

[amyk]

You can always use a MOSFET instead.

[hgg]

I've tried it with a P-Channel mosfet, high side switching, and I get the same voltage drop. 

[Gyro]

With a Mosfet you need enough gate drive voltage to turn it fully on. That might be a problem at 2.4V.

[hgg]

I used the IRFR9024N with a Vgs of -2V min and -4V max.

[orolo]

Try a PNP (2N3906) instead of a NPN driving the LED (maybe with a moderate base resistor to limit current). The C-E saturation voltage can be in the tens of millivolts. If your LED is white, it will not glow very much anyway with 2.4V, but the transistor will not be responsible for that.

[Benta]

Don't know how much capacity your battery has, 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.

[2N3055]

You have to switch on low side.. Bipolar transistor in saturation, with few mA over them will have 0.1-0.2 V drop... Mosfet can be less than that,but you will have trouble switching it on without 4-8V..

[kosine]

The circuit you have is almost there. You need to move the LED above Q2, add a small current limiting resistor (about 47R) and drop R1 to say 10k to give Q2 a bit more base current.

When the solar cell input drops, Q1 will switch off and Q2 will switch on hard. It's Vce will be about 100mV, so nothing to worry the LED.

The value of R1 will determine the quiescent current because Q1 is always on. It'll probably be 200uA or so with a 10k, but the higher you go the greater Q2's Vce. 47k might be OK.

R2 will determine the LED current. Note that the circuit you show will indeed work without it due to the Vbe drop across Q2, so the LED is only seeing 1.8V. But if you move the LED above Q2 it'll see 2.3V or so, and that'll need a current limiter (R2).

(Another way you could avoid R2 is to increase R1 to limit the base current of Q2. That'll also limit the collector current of Q2, but it'd dodgy since it depends on the exact gain of Q2.)

[retrolefty]

I used the IRFR9024N with a Vgs of -2V min and -4V max.

 Keep in mind that is the Gate Threshold Voltage, where the device just starts to conduct. So according
to that spec your specific device, worst case may only start to conduct (250 uA) when up to 4 volts gate/source voltage. So it really can't be used in your circuit using a 2.4v Vcc.

http://www.mouser.com/catalog/specsheets/international%20rectifier_irfr9024n.pdf

[hgg]

Low side switching indeed worked better.
Guys, thank you for your suggestions!

In the BJT, is the diode seen only between base and emitter?

[KL27x]

I do not need any current limiting resistor, because at 2.4V the LED draws 10mA.

This might be a problem. If the LED only draws 10mA with NO limiting, it is barely functioning, at all. This is bad design, to begin with. That said, there is still something to learn and possible borderline success.

+1 amyk. She is correct.

I've tried it with a P-Channel mosfet, high side switching, and I get the same voltage drop. 

You tried it with a 55A power FET with min Gate threshold higher than 2.4V, on average. You looking for a small signal FET.

With a Mosfet you need enough gate drive voltage to turn it fully on. That might be a problem at 2.4V.

You do not need to turn it on fully, in order to beat a BJT in this instance. And FET operate at 1.8V all the time. 50% of our modern computers have hundreds of thousands of FETs in the processor operating at 1.8V. You will easily beat the BJT. Any resistive load, no matter how high impedance, is going to beat the diode drop of a BJT in this instance. Since your LED voltge drop is approaching 2.4V, this means any resistive load will have voltage drop approaching zero, which is much, much less than minimum voltage drop of a diode junction in a BJT. (Also current will approach zero; but is enough to get your LED to light, I promise.... all these "approaching" things are why this is bad design. Running LED by voltage is like dividing by zero, lol).


Here is a small signal PFET, for example. For high side switching.
http://www.mouser.com/ProductDetail/Infineon-Technologies/BSS816NWH6327XTSA1/?qs=sGAEpiMZZMshyDBzk1%2fWixDt2IZOmxjILFS%2fF%252bTDmNA%3d
You can just as easily use NFET for low side switching.

At 2.4V, this has resistance of about 300 milliohm. This will not greatly reduce current thu your LED. Output impedance of your battery and wiring probably is greater than 300mohm.

So this is bad circuit, but given directly driving the LED at 2.4V works good enough for you, this will also work good enough for you. Drive it directly, not thru a BJT. To get fuller saturation. Guarantee this will work (as good as your original design intent, anyway. LED should never be driven by unlimited current source and controlled by voltage, only, unless by active control. Even the 300mOhm resistance of your FET is not enough resistance to make this a good circuit. Very small change in voltage will make significant change in LED output or even burn it out.) But bottom line, in this particular case, I guarantee an optimal (and cheap) FET will give your LED more current than any BJT in any configuration.

[Audioguru]

I used the IRFR9024N with a Vgs of -2V min and -4V max.

That is its "threshold voltage" when it is almost turned off conducting only 0.25mA. Maybe yours had a threshold voltage near 4V.

If your LED lighted without current limiting when the battery was 2.4V then how many milli-seconds will it last when the battery is fully charged at 2.9V?

[Audioguru]

Why not buy a solar garden light for one dollar? It comes with an LED, a 1.2V Ni-MH battery cell, a solar panel and a circuit that charges the battery when there is sunlight and boosts its voltage when it detects darkness.

[hgg]

I forgot to mention that the led I used was a flickering led...
It works down to 2.0V but with 2.4V or even with 3.0V it did not need any current limiting.
It was drawing no more than 11mA when it was directly connected to the 2.4V battery.
Anyway, I might use a resistor just to be on the safe side.

[hgg]

@audioguru   Because they do not use any kind of overcharging protection and I need
the batteries to last for many years.

[hgg]

@KL27x  Thanks for the MOSFET links.  I need to buy some of these.

[KL27x]

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.

[mariush]

People are gonna jump all over me, but here it goes.

Just buy a PicKit programmer and a PIC10F222 : http://www.digikey.com/product-detail/en/microchip-technology/PIC10F222T-I-OT/PIC10F222T-I-OTCT-ND/1015706
Tiny 6 pin micro, works from 2v, uses something like 100-200nA @2v when in stand-by, internal oscillator, each output pin can do up to 25mA (i think it depends on input voltage as well and you can probably connect 2 output pins in parallel for more current), built in adc to measure voltage from solar cell
 
You basically have a loop in the micro, measure voltage from solar cell, if it's high enough to think it's light outside turn off led, if it's too low, turn on led (connected directly to micro). sleep for 15s - 30s - 1m - 5m , repeat.

Pretty good for 50 cents, and it's a tiny sot 6pin part, compared to 2 transistors and resistors taking more space and potentially costing more money. 

[hgg]

@benta

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 ?

[Benta]

Can you post a schematic of your suggestion ?

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.

[IanB]

I do not need any current limiting resistor, because at 2.4V the LED draws 10mA.

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.

[Benta]

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.