Need help identifying datasheet - Page 1
Forum section: electronics / beginners

author: kresica
author_rank: Newbie
timestamp: June 12, 2015, 11:47:46 pm
content: As this is my first post here, I first wanna say hello to everyone here.I have a simple schematic that I need to describe for my college class project and I think I got it all, but I want to double check that I got the part on a photo in attachment correctly.I'd say that this is simple inverter connected to the Schmitt trigger. The part that confuses me is the cap in upper right corner. Since the input is from a keyboard, there will be some switch bouncing and that isn't solved before entering this part of a circuit.So I have two possible explanations:1) Cap is used so that input to Schmitt trigger can't jump around since cap (dis)charging takes time (especially since the R=1M and C=100uF)2) Cap is used for decouplingSo, anyone have any idea am I close?

author: kresica
author_rank: Newbie
timestamp: June 14, 2015, 10:36:49 am
content: Anyone at all?Please help, I still haven't found out what the cap is for...

author: AndyC_772
author_rank: Super Contributor
timestamp: June 14, 2015, 10:47:06 am
content: If it were a decoupling cap, it would be connected across the power supply pins of the logic gate.It's there to help eliminate contact bounce. Think about how it'll charge and discharge when the transistor below turns on and off, and what the effect will be on the voltage at the input to the Schmitt trigger.What if the transistor is switched on for, say, 250msec? What will the voltage across the cap be before, and after?Then, how about if the transistor is switched on for 2msec, then off for 2msec, then on for 246msec?In each case, how does the voltage at the input of the Schmitt trigger change, and what will its output do?

author: kresica
author_rank: Newbie
timestamp: June 14, 2015, 01:12:16 pm
content: Yes, of course it isn't decoupling cap, I don't know what i was thinkingI already put decoupling cap between Vcc and gnd of Schmitt trigger IC.When transistor switches on, cap is already charged, and because transistor switches on quickly, power supply can't handle big dI/dt spike so cap is a good option as a quick source of charge.As transistor switches off, remaining current flow from power supply continues to flow through cap, charging it once again.As for voltage at the input of Schmitt trigger, I'd say that when transistor is switched off, current flows through 1M res, and since transistor is off, Vc = 0 V. Therefore, input of Schmitt trigger is at approx. 0 V with small current flowing through and after Schmitt trigger, output is high which signals that key is depressed.When transistor is on, it has a Vce voltage drop of a few volts. 100 ohm res has insignificant voltage drop and it serves as a backup load when transistor is off and there is voltage drop between Schmitt trigger input and ground. Anyways, when transistor is on, input at Schmitt trigger is high and output is low indicating pressed key.EDIT: when key bounces and transistor switches on/off rapidly, cap can't discharge that quickly and voltage at the input of Schmitt trigger can't follow transistor switching on/off.Am I correct?

author: mikerj
author_rank: Super Contributor
timestamp: June 14, 2015, 01:31:21 pm
content: It's effectively a re-triggerable monostable which would be used to filter out contact bounce. When the transistor T1 is switched on, the cap C2 will quickly charge through the 100 ohm resistor. When the transistor switches off, the capacitor has a 1 megohm resistor to discharge through. This means a short pulse, or a number of short pulses into the base of T1 will be stretched out to a single clean pulse at the output.The schmitt trigger inverter is required to convert the slowly varying DC voltage on the cap to a clean logic level.

author: AndyC_772
author_rank: Super Contributor
timestamp: June 14, 2015, 02:41:21 pm
content: There's nothing here to do with big dI/dt spikes, or what the power supply can or cannot 'handle'. Nevertheless, I think you've got the essential operation of the circuit right.When T1 is turned off, no current flows through R4, and the voltage across C2 falls to zero because it discharges through R3. So both ends of C2 are at +5V.Now let T1 switch on. At the instant of switch-on, C2 still has 0V across it, so the input to the Schmitt trigger doesn't change. But current now flows out of C2, through R4 and T1 to ground, and over time, the voltage at the input to N1 will fall. The time constant is C2 * (R4//R3), which is almost the same as C2 * R4 alone.If T1 switches on and off a few times, with a period of just a few milliseconds, it doesn't really make much difference to what happens at the output of N1. C2 charges up whenever T1 is conducting, and it holds its voltage when T1 isn't. R3 is too large in value to significantly discharge C2 over that period of time.Net result: N1 output goes high when T1 has been switched on for long enough, in total. It doesn't, however, toggle many times just because T1 does.When T1 is finally turned off, then C2 slowly discharges through R3, and eventually N1 toggles back again. The circuit is now reset ready for the next keypress.

author: kresica
author_rank: Newbie
timestamp: June 14, 2015, 06:23:28 pm
content: Thank you both for detailed explanations. You've helped me a lot. Just one more question:When visualizing current flow, should I regard Schmitt trigger as an infinite resistance element?

author: dacman
author_rank: Frequent Contributor
timestamp: June 14, 2015, 07:22:18 pm
content: The print is blurry, but it seems to me that the asterisks for C2 shows 104, which would be a 0.1uF capacitor (or 100nF). Time constant for C2 and R4 is 0.00001 seconds (C2 charging), and 0.1 seconds for C2 and R3 (C2 discharging). When the switch is depressed, initially there will still be 5V at the input of N1 at the start of its charge, but with a time constant of 0.00001 seconds, within about 20 us the N1 gate will switch. (The circuit will trigger before C2 is fully charged.) If there is any switch bounce, the discharge path through R3 is much slower (by a factor of 10000), and C2 will not loose much charge. This, along with the Schmitt Trigger, keeps the output of N1 from bouncing. When the switch is released, it will take about 1 time constant (C2 and R3), or about 0.1 seconds for the Schmitt Trigger to switch back to off (if C2 were fully charged, which would require the switch to be held for 50 us). For an RC time constant, it takes about 5 time constants for a capacitor to fully charge from zero or discharge from fully charged, but the circuit will switch within one or two time constants for both charge and discharge.

author: Paul Moir
author_rank: Frequent Contributor
timestamp: June 14, 2015, 10:06:44 pm
content: Just under the schmitt trigger I think I see 4093. That seems to match the pinout too.https://www.fairchildsemi.com/datasheets/CD/CD4093BC.pdfIf you check the datasheet, bottom of page 2, the input current is + or - 0.1uA max. You have two paralleled so I guess that's 0.2uA max. I think you can ignore it as an element.Note that TTL logic (74xx series) does have some current going into and out of their inputs, so they can get significant.