current INTO output of 4093 Schmitt Trigger? - Page 1
Forum section: electronics / beginners

author: dentaku
author_rank: Frequent Contributor
timestamp: January 21, 2014, 03:48:05 am
content: I'm curious about the first circuit on this page.http://www.doctronics.co.uk/4013.htmThe first indicator section has an LED connected directly to pin 4 of the Schmitt Trigger. Is it OK to connect things like LEDs (even with a 680ohm resistor) directly to the output of a 4000 series Schmitt Trigger like this?What in the datasheet would tell me if it's OK to do this? You'd think the way to do it would be to use a BJT connected the same way as the second indicator LED section.I believe I've connected theCathodeof LEDs directly to the output of 7555 chips before and it worked, at least I've never killed one back when I was starting out and messing with 555s....Edited to change anode to cathode.... oops:)

author: leppie
author_rank: Frequent Contributor
timestamp: January 21, 2014, 05:05:14 am
content: That would be the output current. From the looks of it around 10ma (in total I assume).

author: lgbeno
author_rank: Frequent Contributor
timestamp: January 21, 2014, 05:37:09 am
content: Yup sinking current, many of these devices can actually sink more than they can source.

author: dentaku
author_rank: Frequent Contributor
timestamp: January 21, 2014, 03:36:58 pm
content: Quote from: lgbeno on January 21, 2014, 05:37:09 amOK, so then where in this datasheet, for example, would it show how much current is safe to sink into the output of the 4093?http://www.ti.com/lit/ds/symlink/cd4093bc.pdfThe datasheets say "All outputs have equal source and sink currents and conform to standard B-series output drive (see Static Electrical Characteristics)."Datasheets are quite technical for beginners and rarely have a "don't ever do this" section

author: PA0PBZ
author_rank: Super Contributor
timestamp: January 21, 2014, 03:45:25 pm
content: Quote from: dentaku on January 21, 2014, 03:36:58 pmPage 3, IOHand IOL, high and low level output current.

author: Kjelt
author_rank: Super Contributor
timestamp: January 21, 2014, 03:46:32 pm
content: Usually for these series of chips there is a seperate datasheet the so called family datasheet that specifies the common criteria.For instance for the HEF4000 series fromPhilipsNXPthey have their family datasheet here:http://www.nxp.com/products/logic/family/HEF4000B/And it states for instance on introduction the table:DC current into any input or output - max. 10 mAThis means that any of the in and outputs can sink or source max. 10mA BUT this does not mean that all the in and outputs can sink or source this together or for all voltages. For this you need to look at the maximum allowedPower dissipation per package and per gate which are 500mW and 100mW.So if your VCC is below 10Volts one output can sink/source max 10mA but above this is lower.And in total if your VCC is 10Volts you can have max. 5 outputs source 10mA.

author: dentaku
author_rank: Frequent Contributor
timestamp: January 21, 2014, 05:18:03 pm
content: OK, so that circuit probably won't kill a 4093 if you're just experimenting with it using one LED a 680ohm resistor and a 9V battery but you wouldn't do it that way in something that mattered.Would I be correct in assuming that a red LED has a Vf of around 2V, the battery is 9V and the resistor is 68Ohms so that's a current of around 10.3mA?I'm going to try this using a 74c14 instead of a 4093 (because I don't have one) but I'm going to use a transistor for the first indicator LED too, just in case, even though the 74C seems to be able to sink even more current than the 4093.Quote from: Kjelt on January 21, 2014, 03:46:32 pm

author: dentaku
author_rank: Frequent Contributor
timestamp: January 22, 2014, 03:51:06 am
content: One last question about this circuitThis time it's about the 4013 flip-flopI recently learned about not leaving the inputs open and the website even says "The inputs of the D-type must be connected, either to LOW or to HIGH, and must not be left open circuit. This includes the SET and RESET inputs which are connected to 0 V."I'm assuming this only means the side of the IC that's being used?There's nothing connected to pins 8-13 of the 4013 in the first example circuit.

author: granz
author_rank: Regular Contributor
timestamp: January 22, 2014, 04:29:01 am
content: The inputs to the unused flip-flop should also be tied high or low. The problem is that CMOS inputs are high impedance and will pickup noise and cause random state changes. This could potentially cause unpredictable power consumption and even cause the device to heat up from oscillation. In the real world, it probably won't do those things (in most situations) but just tie them to ground--it's the right thing to do.If you'd like to investigate further, you can connect an oscilloscope to the Q output of the flip flop and leave the inputs floating. Bring different random objects near the device and watch the output switch states. WARNING: a static discharge may destroy your chip if you're not careful (but wouldn't it be worth the experience?)Hope that helps.

author: Kjelt
author_rank: Super Contributor
timestamp: January 22, 2014, 12:27:14 pm
content: Quote from: dentaku on January 21, 2014, 05:18:03 pmYes that would be correct.Quote from: granz on January 22, 2014, 04:29:01 am+1 always connect the inputs to a fixed level be it ground or Vcc (depends per case).

author: dentaku
author_rank: Frequent Contributor
timestamp: January 22, 2014, 02:48:28 pm
content: Well, I didn't kill anything and it works. I didn't bother with the first indicator LED like in the original example because I just wanted to see if it works and I ran out of room.I built it on one of the tiny little breadboards I got yesterday from China (just to make it more challenging:) )I connected the unused SET2, RESET2, DATA2 and CLOCK2 inputs on the other side of the 4013 to ground.CLOCK2 is considered an input just like the others right?I used a 74c14 to create the pulses because it's so simple. At least that one doesn't need it's inputs connected to Ground considering how Schmitt Triggers aren't sensitive to noise.I guess if I would've used a 555 it would have given me the room for another transistor and the other LED.The LEDs aren't a big deal to me anyway because I connected it to my oscilloscope and the Flip-Flop was outputting exactly half the frequency as the Schmitt Trigger which was the whole point of this experiment for me.I wanted to build something that outputs a sound exactly one octave down from the signal you put into it.