'Fraid so. It's a Darlington, intended for high voltage outputs.
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Phil Hobbs
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That was what I was afraid of. I should still have enough headroom, but that is wasting a bit of power...
Any suggestion on a replacement part? I need to enable/disable 8 lines with logic level inputs.
It can even be a 1-of-8 decoder I just need it to support up to 1A, and I'd like it to be relatively cheap. My load is going to 24 parallel LEDs (3.3v Vf for 16 of them, 1.8 Vf for 8), each with a constant current sink in line with them.
Cheap, for me, is < $1.25 for 8-in-8-out, or < $2.0 for 3-in-8-out (since I'd be able to replace two chips in that case).
You'd think that somebody would make a MOSFET version, but AFAIK they don't.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
It does not mean that Vce will be 1.3V at that current -- you could have
350mA at 50V, for a little while.
It DOES mean that the lowest that the transistor will pull the voltage down when 350mA is going through it is 1.3V.
"Saturation" in a BJT means that the Vce is as low as it's going to ever get at that collector current. Basically, the collector isn't acting like a pretty good current source any more -- instead, it's acting like a pretty good short circuit.
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Tim Wescott
Control system and signal processing consulting
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"Saturation" in a Darlington is equal to a Vbe + Vce-sat of one NPN, thus the odd-looking number. ...Jim Thompson
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I love to cook with wine. Sometimes I even put it in the food.
There are other chips in the same series that have pregain and a single output stage. I seem to have misplaced the number, but just threw a bunch of them away. Great chips.
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The main device... but the driver device saturates. ...Jim Thompson
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| James E.Thompson | mens |
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| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
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I love to cook with wine. Sometimes I even put it in the food.
Looking more carefully, I think I might need P-Channel MOSFETS, and then I'd need to invert my logic driver (replace the 74HC238 with a 74HC138). If I used an N-Channel, I don't think I could get the Vgs to be high enough.
I was thinking this part:
formatting link
Then I would have the source connected to +5v, the drain connected to to my load, and the gate connected to the output of the 1-of-8.
The load is some parallel LEDs (each in series with a constant-current-sink), so the drain voltage could vary between +1.8v up to +5.0v, The current can vary from 10mA to 1200mA. Depending on how many parallel LEDs are enabled.
The "current sink" is 3 8-bit shift-register constant-current-sink (TLC5196) which power between 0 to 24 LEDs in parallel (depending). The MOSFET is a "row enable" to choose between 1 of 8 sets of 24 LEDS. So in total, I'm powering 196 LEDs, just 24 at a time.
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