# Can this be done?

#### mkp

Mar 30, 2013
7
I'm looking for a transistor device (I think), but instead of turning on when the gate gets above a certain voltage, it will turn on when the gate voltage drops below a certain voltage (ideally 3V or so).

Basically, I'm looking for a low cost, low power, no voltage overhead way to drop resistance when the voltage drops to 3V. The easiest way I can imagine this to be done is a switch that would turn on when the voltage drops below a certain point to bypass a resistor.

Any help would be greatly appreciated.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
One way to do this is to have a transistor with a voltage divider going to the base so that it turns on when the input voltage hits about 3V (I assume it's not critical that it changes state at exactly 3V or does so with a very sharp transition).

This transistor can then be used to turn off another transistor.

Of course, we need to know something about that resistor you want to short, and whether one end of it is ground with respect to the 3V signal.

#### mkp

Mar 30, 2013
7
Below is an image of a simplified circuit with this special transistor that turns on when the voltage drops below 3V effectively shorting R2 and reducing the resistance of the circuit.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
What is the aim of this circuit?

#### mkp

Mar 30, 2013
7
Unfortunately, I can't disclose any details as to why I need this but I think something like this could have many applications.

For instance, say my source is a 12V battery that drops voltage with use (typical alkaline). At some point, the current running through the circuit has become too low for practical use. So, in order to bump up the current, we need to drop resistance.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Have you considered a constant current source? Or a voltage regulator?

If your application is secret, it's really hard to help.

Placing resistors in series with batteries to power a generic "load" that requires a lower voltage is beginner thinking for the most part.

#### mkp

Mar 30, 2013
7
I understand. I guess the quick answer is no it can not be done easily. I didn't know if I was missing something obvious (like a special magic transistor I haven't heard of).

I could put a voltage regulator into the circuit but this bumps cost and complexity up an order of magnitude. If you disagree, please point me to one that can output a constant 4V from a source that will decrease from 6V to 2V. Current is about 120mA. Cost should be less than a $1 for parts. Thanks, and yes, I'm a novice at best. Kind regards, Mark #### (*steve*) ##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator Jan 21, 2010 25,505 Well, a resistor won't work when the input voltage falls below the desired output voltage. It sounds like a voltage regulator is what you need, but you have a very tight constraint on cost. If your budget stretched to$2 then you could use something like this.

It would be much more efficient than a resistor, and would allow you to use a higher voltage input source and retain efficiency.

You could replace the entire set of resistors with a transistor, a resistor and a zener diode -- this would require the input voltage be at least 0.7 volts higher than your desired output voltage, but would be very inefficient at both low currents and higher input voltages.

If you're using a 12V battery to power a 6V load, the regulator I pointed to you would be ideal. In addition to it maintaining the voltage at a constant value (until the input falls to less than 6V), your battery would last much longer -- possibly up to twice as long.

If we had some idea of the range of voltages your load can operate at and the range of currents it requires, we could make more recommendations.

#### mkp

Mar 30, 2013
7
Steve, I have actually tested that exact unit (from amazon). On amazon it says as low as 3V input and Ebay says 4V. It works quite nicely and I haven't seen it for that cheap so that now becomes an interesting option. I'm curious now how much power this consumes. I will need to do some more testing with it.

In regards to our application, we are making an LED light. We are considering 3 to 6 AA batteries. The load is several LED's in parallel (probably 12, 5mm LEDs or maybe 2 higher power smd). So, ideally we have them at 3V and about 10mA.

Here is a typical battery performance curve. So if we are using 3 AA batteries, the voltage will start at 4.5V but then quickly go to 3.3V with a significant amount of energy still in the battery. If we use a resistor like most LED lights use, then the light will be super bright for about 2 hours and then start the slow decline. So, my thought was I could start with say 100ohms and then reduce it to 50ohms at some later voltage bumping up the voltage to the LEDs.

Jan 5, 2010
7,682

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
One issue is that you really shouldn't place LEDs in parallel.

See https://www.electronicspoint.com/got-question-driving-leds-t256849.html

Since you seem not to need a particular number of LEDs, I would suggest a single high powered LED connected via a current source to your battery.

For multiple LEDs you may want to consider placing them in series and using a boost SMPS to raise the battery voltage..

It all really depends on whether you need 1 LED or many, and how bright it needs to be (i.e will it be 10mA, or 250mA)

Also, knowing the colour of the LEDs you plan on using (presumably a single colour) would be helpful.

#### mkp

Mar 30, 2013
7
Steve,

That is a good write up. Thank you for putting that together.

To simplify our discussion, let’s assume we are using this white light 5mm LED for the multiple, low power, LED light..

Replies
12
Views
3K
Replies
21
Views
3K
Replies
17
Views
636