LED Driver: HEF4794B -- how much can it source?

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Hi Folks,

I'm not finding the information I need in the datasheet for this part... probably because I lack the mad skills needed to find it.

http://www.semiconductors.com/acrobat/datasheets/HEF4794B_2.pdf

Can you take a look and tell me if the HEF4794B can source, let's say 20mA, from all eight outputs at the same time?

The application diagram shows this chip sinking the current, but I want to use an NPN darlington array to do they sinking.

All help appreciated, thanks!

EDIT: I'm also wondering if my clock rise time is fast or slow? This will decide which pin to use for cascading. I'm running at 8MHz.

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Quote:
I'm not finding the information I need in the datasheet for this part

Because it is not there :?
Quote:
See “Family Specifications” except for: rating for DC
current into any open-drain output is 40 mA.

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

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Interestng data sheet.

I agree with JS that the key information one requires is not readily evident.

Regardless, the drivers are open drain, and are designed to sink current, as shown. You could always put in a string of resistors to V+, and essentially have a byte of control signals, but then you defeat the purpose of using a chip with built-in drive capability.

My read is that the implied current Sink is 20 mA per pin in normal operating conditions, and that one could run all pins at this mode, continuously. This ASSUMPTION is made based upon the "DC Characteristics" table, where it uses Iout < 20 mA for 5 or 10 or 15 Vsupply.

Perhaps Philips has some Application Notes on their web site you can search for?

JC

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Quote:
Wikipedia: An open drain terminal is connected to ground in the low voltage (logic 0) state, but has high impedance in the logic 1 state.

So I think you can't really source power from this chip... Not meant for that.

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I would suggest you this led driver:
http://www.st.com/stonline/products/literature/ds/13524.pdf

It has 8 (exist in 16 also) constant current output drivers, they can sink from 5 to 100ma per output (adjustable via external resistor) no need to place resistors in series with the led’s, as long as the voltage drop between the different led’s and voltage source is not too big. SPI interface, 30MHz clock frequency, output enable which you can drive with a PWM signal to control the led’s brightness, etc.

I’m using this chip and it works great, the advantage using constant current led drivers is that all led’s have the same brightness

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Well crap, I put in the order for that driver yesterday. Guess I'll have to make another order.

Thanks for the help all!

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Ok... still trying to work with what I have here. If I flip all of my LEDs around so that I'm sinking the current with the LED driver I just have to replace the transistor I was previously using on the ground side.

I have the potential need to source 400mA. If I use a PNP transistor will the 2n2907 work? This would mean supplying approximately 45mA to the base though... which I think is quite a bit too much for the AVR.

This is why I was using a darlington array before. Are there any PNP darlington arrays available? Any other types of suggestions?

EDIT: Wait, I think I did that math wrong. Do I only need 4.5mA to the base of a 2n2907 to shut it off?

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Maybe look at something like a UDN2891A for an octal source driver or ULN2803 for an octal sink driver.

ps. you apply current to the base of a transistor to turn it on.

Without looking at the specs for a 2n2907, at a guess the hfe is around 100, so 4.5mA might be closer to the mark, but I'm not too sure if this transistor is good for 400mA - I'd suggest a beefier device.

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Kartman wrote:
ps. you apply current to the base of a transistor to turn it on.

Yes, with an NPN transistor. But with a PNP transistor don't you apply current to the base to turn it off? Sink current from it to turn it on?

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sink or source - you need to apply a current! With a pnp you still need to apply current to turn it on, just that the emitter is usually at a positive voltage therefore the base current needs to flow towards the negative.