Alternative for 8 2N3904 transistors in a single chip?

[brokenbot]

Hi,

I am trying to build a project from instructables, http://bit.ly/1KqESIg , which needs 8 nos 2N3904 transistors. This would make my circuit bulky. Can I replace them with a simple IC, suppose which contains 10 of them in one block?

Also, I would need around 10nos 220Ω resistors. can it be packed in a single IC?

Helps are welcome

 

[Harald Kapp]

Welcome to electronicspoint.

Have a look at the ULN2003.
For the resistors look up resistor array or resistor network.

When size matters, use SMD components. These will require some advanced soldering skills, though.

 

[brokenbot]

Thanks Harald Kapp.
Will definitely help.

 

[AnalogKid]

ULN2803 has 8 sections. Also, the 2004/2804 have inputs designed for 6 to 15 V and can operate safely at 24 V, if you need that kind of range.

ak

 

[Minder]

The resistor arrays exist in dip or sip packages.
m.

 

[brokenbot]

I am loving the prompt reply by you! Thanks!

AK: Thanks for the pointer. I am looking at the datasheets.
M: Can you name a resistor array? It will be good if I get an array with different types of resistor in it, as opposed to having n nos of same values. May be a DIP.

 

[dorke]

Your best choice would be...
eliminate all transistors .
You may need a slight code modification

Use a 2ma 7 segment element instead of the 20ma you are using now,that simple.

 

[brokenbot]

Your best choice would be...
eliminate all transistors .
You may need a slight code modification

Use a 2ma 7 segment element instead of the 20ma you are using now,that simple.

Well, thats a great idea!
But, I wonder, would that be as bright as the 20ma?

 

[dorke]

Most likely not as bright.

 

[brokenbot]

That was what I thought. Thanks.

 

[brokenbot]

ULN2803 has 8 sections. Also, the 2004/2804 have inputs designed for 6 to 15 V and can operate safely at 24 V, if you need that kind of range.

ak

Considering my 7 segment display, the 2N3904 has hfe=300max and ULN2003a has hfe=10,000 max. It would ruin my display when i drive it from arduino output.

Edit1: As pointed by Herald Kapp.

 

[AnalogKid]

Considering my 7 segment display, the 2N3904 has hfe=300max and ULN2003a has hfe=10,000 max. It would ruin my display when i drive it from arduino output.
Edit1: As pointed by Herald Kapp.

No, it won't. Harold has only one post in this thread, and he does not say anything about that.

Of course, without a schematic it is impossible to say, but IF (note the big IF) you are using the 3904's as saturated switches, the 2803 probably will drop in with zero circuit changes. The higher gain is irrelevant because the display current, and hence its power dissipation, is set by external resistors. If anything, the 2803 will make ***less*** energy available to the display because it has a higher voltage drop across it when it is on. This leaves less voltage across the LEDs and their resistors.

And in fact, if your analysis is correct, you already are massively overdriving the display with a 3904. Let's assume an arduino output current of 1 mA, and you want 20 mA through each display LED. That would require a driver gain of 20. But if a 3904 has a gain of 300 (BTW, in this application it does not), then certainly that's enough whatever to "ruin" something. Except it isn't. Fortunately, Ohm's Law was developed in 1827, so for almost 190 years we've known how to limit the energy in a switched circuit.

ak

 

[brokenbot]

No, it won't. Harold has only one post in this thread, and he does not say anything about that.

Of course, without a schematic it is impossible to say, but IF (note the big IF) you are using the 3904's as saturated switches, the 2803 probably will drop in with zero circuit changes. The higher gain is irrelevant because the display current, and hence its power dissipation, is set by external resistors. If anything, the 2803 will make ***less*** energy available to the display because it has a higher voltage drop across it when it is on. This leaves less voltage across the LEDs and their resistors.

And in fact, if your analysis is correct, you already are massively overdriving the display with a 3904. Let's assume an arduino output current of 1 mA, and you want 20 mA through each display LED. That would require a driver gain of 20. But if a 3904 has a gain of 300 (BTW, in this application it does not), then certainly that's enough whatever to "ruin" something. Except it isn't. Fortunately, Ohm's Law was developed in 1827, so for almost 190 years we've known how to limit the energy in a switched circuit.

ak

Thanks for pointing out ohms law for me.
Let me clarify that I thought that this is a discussion and i will question if i see something intriguing.
As far as this thread is concerned, Harald pointed out 2003, and i was talking about facts obtained from its datasheet.
We can always have 20000 volts, and then make a drop and use only 2 volts from it. In todays world, this will not need Mr. Ohms himself for it. The question is efficiency.
Thanks to you for pointing out that for me.
Can you please suggest a design, where the display would be bright enough and there will not be a need to drop power? Of course considering what dorke suggested. We agreed that a 2ma display will never be as bright as a 20ma.

The real question here is finding a transistor array which has a gain near to 20.

 

[AnalogKid]

"
The real question here is finding a transistor array which has a gain near to 20.
"

There isn't one. At a collector current of 20 mA, all commonly available transistor arrays have a DC forward gain much greater than 20.

I don't understand why a transistor with a gain of 20 is the real question. A schematic would help greatly in clarifying your question. Also, can you share how driving an LED with a high-gain darlington transistor would ruin it? I can explain things more clearly if I know what your thinking is. It is clear from your statement in post #11 that your understanding of how transistor gain affects external circuits is incorrect. I brought up Ohm's Law because that is what determines the current in an LED, not a driver transistor's gain.

An LED is a "current mode" device. That is, its major characteristics (luminous flux, forward voltage, etc.) are defined by the current going through it. Because it is a diode, it has no inherent current limiting mechanism, and will burn up if the current through it is not limited externally. Since most circuits are powered by a constant voltage source, the usual way to limit the current through an LED is to put a resistor in series with it. Maxim and others make constant-current drivers for LEDs, but the power equation is the same. Even if you vary the LED brightness with a pulse-width modulated waveform, there still needs to be something to limit the current pulses to a value that is safe for the device. I'm not sure what you mean by efficiency, but for an individual LED the power equation always is the same:

(source voltage - LED Vf) x LED current = power dissipated in the current limiter.

For multiple LEDs you can use the persistence of the human retina to reduce the total circuit power by multiplexing the LEDs so they are not all on all the time. There still have to be current limiters in the circuits, but not as many.

If your goal is to have the drive transistor limit the LED current by not going into saturation, that is what Maxim does internally in their constant current LED driver chips. It is possible to do this with discrete components but not simple, and difficult to stabilize over temperature. Also, note that the system power dissipation will be exactly the same as if the transistor switch were saturated with a fixed resistor as the current limiter. Other than multiplexing, there is no way to reduce the total power requirement of a multi-digit display system (compared to an all direct-drive system).

ak