pull down resistors

Hello,

I'm using an AND gate IC, AHC08 is the chip... and I'm wondering something about inputting a logic low into this guy. basically I'm tying the emitter end of a transistor to the input gate, and sending these to ground through a pull down resistor. My collector is tied directly to a 5V source.

So, here's my question. Does it matter what the value od the pull down resistor is, and can it be too high? I imagine it can be too low, cause that will short the 5V to ground when the transistor turns on.... but if it's too high... would it essentially make it look like the input to the gate is floating?

On the datasheet to the AND gate it has an Ii (input current) value of

1uA. Is this current going to be running through my pulldown resistor to ground when the transistor is off?

One more question,

I have a D Flip-Flop, that says it has a IiL (current input low on the D input) of -1.6mA.... if I want to tie my input of the AND gate (and therefore the emitter of the transistor and the pulldown resistor) to the D input of the D latch, does this mean I'm going to have -1.6mA going through my pull down? In this case I would have to calculate a low input voltage with the pull down and figure out R.

The D Flip Flop:

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The AND gate:
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Much thanks!

Reply to
panfilero
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08.html

The CMOS gate has very high input impedance, so the pull-down resistor could be megohms and work. That's not good practice generally, because of possible leakage from dirt, humidity, EMI susceptibility, etc. Lower values are safer, and cost exactly the same.

The TTL flip flop input needs a pull-down resistor that can sink its input current and still meet its input-low voltage spec. That's going to be several hundred ohms, max.

The bigger question is what's the signal source, and why are you using an emitter follower to drive these inputs? And why the antique flip- flop?

-- Cheers, James Arthur

Reply to
dagmargoodboat

On a sunny day (Mon, 7 Feb 2011 06:46:46 -0800 (PST)) it happened panfilero wrote in :

When in the pull down situation, te emittor follower does basically nothing./ What counts then is tho RC time of the resistor and the input capacitance of the chip. This sets the maximum speed. So you discharge that input capacitance via the resistor.

1 uA source or sink?

Yes

Yes, and once you know R and the Cin, you can calculate the speed.

Reply to
Jan Panteltje

panfilero schrieb:

Hello,

a pull down resistor may be too high. You should think not only about a static low state, but also of a dynamic low state as a short pulse. All capacities at the signal line, the stray capacity, the input and output capacity must be discharged through the pulldown resistor. So calculate the total capacity and the time constant caused by the resistor and the capacity. Several time constants are needed to discharge the line from a high state to a good low state, lower than 0.4 V.

Bye

Reply to
Uwe Hercksen

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HC08.html

That's interesting, so these TTL chips suck? They require whatever hooks up to their inputs to be able to handle/sink a considerable amount of current... and the CMOS kind of logic don't.... interesting....

Reply to
panfilero

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Technically, the chip 'sources'. You have to suck.

(The other guys had a point about the speed too, using large pull-down resistors with CMOS is slow.)

-- Cheers, James Arthur

Reply to
dagmargoodboat

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You need enough to satisfy the D-FF, since that is the most demanding of all.

Just calculate the needed pull down for that. I'd use 1 to 2k.

Jamie

Reply to
Jamie

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HC08.html

1.6mA across a 1k resistor gives an input voltage of 1.6 volts. That's too high, he'll have to go lower than that.

-- Cheers, James Arthur

Reply to
dagmargoodboat

Thanks for all the responses, I think I get it now.

Reply to
panfilero

gate:

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it was a -1.6ma not a (+), which means in this case, the input most likely is a pnp and biasing it with a pull down of what I suggested should give you some where in the hood of 4 ma current drive to the D input of the FF.

From what I can understand from his post, He needs sufficient current to drive the D input of the D-FF (downward), the other circuit that is joined with it is CMOS and this does not require much..

Maybe I am going off the wrong end here and thinking of something else but I've always used between 1..1.2k to drive TTL inputs with no issues from a 5V source.

Jamie

Reply to
Jamie

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