Transistor base current cutoff-saturation ratio

[yhelothar]

I'm looking for a transistor whose base cutoff is greater than 0mA and the saturation is 2x the cutoff current. Is there such a transistor that exists?

Ideally, I would like a transistor with:
I(base) cutoff = .5mA
I(base) saturation 1mA

 

[yhelothar]

I just came across digital transistors. Are these transistors that are either fully off or fully on?
http://61.222.192.61/mccsemi/up_pdf/Digital Transistors.pdf

 

[William Gaatjes]

I have some questions before i can help you out ?
What are you going to use it for, do you need to switch a led or a high frequency signal ?

Because digital transistors are good to use, but they have a higher input to emitter voltage than a normal transistor. Can be as high as 2V because of the internal resistor divider.

I get the impression that you just want a bipolar transistor that you want to drive with 500uA ?
A BC847 (NPN)or a BC857(PNP) will work fine but with a BCE of 0,7V.
Basically those digital transistors are sort of BC847 en BC857 types with integrated base / base emitter resistor divider.

If speed is not an issue, a mosfet such as the bsn20 , FDN337 will also work but both have a higher threshold voltage. The advantage is that mosfets only draw current during the transition from 0V to for example 3V3 and vice versa : Switching on and off or off and on. This is of course because of the gate capacitance in combination with the miller effect.

EDIT
Also, usually a mosfet does not require a resistor to the gate if only using small signal mosfets.
Perhaps a 1M or 100K resistor between gate and source when connected to a floating output such as the i/o pin from a microcontroller during reset and startup. Some microcontrollers might even allow to fuse program pull up or pull down resistors to always be active even at reset. Then an external pull up or pull down resistor is not needed.

Had to look it up in my database :
More small signal N mosfets :
BS170.

The BSS138 has a minimal threshold voltage of 0,6 volts.

 

[William Gaatjes]

I misread it :

Off means 500uA ?
On means 1mA ?
Why do you need the 500uA ?

Ignoring the reason :
This can be achieved by a resistor between base and emitter of the transistor.
For example 1000 Ohm. at 500uA there will be 0.5 Volt over the resistor.
At 1mA there will be 1 Volt present over the resistor if there was no transistor.
Rule of thumb :
1V - 0,7V = 0,3V. I would think you would have 300uA to drive the transistor by using a simplified Kirchoff view.
Thus you have more then enough current to drive the transistor when using a BC847C or BC857C.


I leave it to you to see if it works and calculate it.

 

[yhelothar]

Thanks a lot for the info.

I am trying to build an ammeter for a current controlled source/load.
This ammeter will use LEDs to display the current. I plan to have 4-5 LEDs with each LED at 500uA intervals up to 2-2.5mA.

My plan to do this right now is by placing the ammeter in series after the load, which would take the current and put it through a current divider and run each current branch into a transistor that will switch on LEDs.

The transistors I've tried so far have a cutoff to saturation ratio of 1000x, from 10uA to 10mA. This means to drive the transistors in fully on/fully off mode, my LEDs will be a 1000x logarithmic scale, a far cry from the linear scale I'm looking for.

So while placing a resistor would allow me to control the operational range of the transistor, it seems to me that this won't work with the current divider principle I'm trying to employ.

 

[Modelworks]

The typical way to do an ammeter is to to use an opamp to convert current to voltage and then connect that to transistors using voltage dividers to light the selected LED, you could also reverse bias the transistors to control the scaling better.

good explanation of how to limit transistors to just switching on or off.
http://www.electronics-tutorials.ws/transistor/tran_4.html

 

[uclabachelor]

I'm looking for a transistor whose base cutoff is greater than 0mA and the saturation is 2x the cutoff current. Is there such a transistor that exists?

Ideally, I would like a transistor with:
I(base) cutoff = .5mA
I(base) saturation 1mA

You want to use a comparator (or a string of comparators) in this case, not a transistor.

 

[William Gaatjes]

You want to measure current ina range from 2 to 2,5mA at steps of 500uA.
You will need a resistor to change the current into a voltage.
But this is dependent on how much voltage drop you can have you have.
Then as the others have mentioned, an opamp is the best choice as amplifier or comparator.
Far more easy. An comparator such as the LM339 or 2 LM393 will do fine.

If you want an easy way, try the good old : LM3914. This is a led bar driver.
All you need to do is change the current into a voltage within the input range of the LM3914.

Just use a small resistor, an opamp to amplify the voltage drop over the resistor. It is easy if the current sense resistor is connected to ground. A TLV271 has a common mode range that included ground.

The AD8211 will also work when you need to measure high side voltages.
But a discrete solution with an opamp, a few resistors and a transistor will also work as current to voltage converter.


Voltage drop over your current sense resistor is key. At 100Ω (Ohm), it will be 0 to 0.25V. With 50mV steps per 500uA.

 

[William Gaatjes]

Here is the website to download the datasheet for LM3914.
An opamp circuit amplifying the voltage drop to the required input range is all that you need.
Just an idea :
Since the LM3914 has 10 outputs, you can have a more extended measurement range to include fault currents such as over current or under current.

http://www.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=lm3914&fileType=pdf


Good luck. ^_^

 

[yhelothar]

I will have to look into the comparator and transforming the current into voltage.
That LM3914 looks like it was made for what I'm trying to do!

I love you guys!