[Solved]  Following + Inverting buffers for 555's output (transistor option) - Page 1
Forum section: electronics / projects

author: Bendba
author_rank: Regular Contributor
timestamp: March 21, 2017, 05:51:57 am
content: Hi,here is a issue I came across a few time when making electronic circuits:I want to buffer the output of a device with a follower and an inverting buffer (In the current case, a 555 timer to get one output high when the output of the 555 is high and get the other output high when the output of the 555 is low)I know I can just use a CD4009 to drive my outputs, that's how I normally get around it. But I would like to find a circuit to do it with transistors. I've tried making lots of different configurations of transistors, it always end up the same, the buffers work on their own but as soon as I connect the input of the two buffers together, they both turn on (either full on or partially depending on the circuit)Would someone have an example of circuit I can use to find what I am doing wrong?ThanksBen

author: StillTrying
author_rank: Super Contributor
timestamp: March 21, 2017, 11:16:14 pm
content: Quote from: Bendba on March 21, 2017, 05:51:57 amThe easiest way to do that is to use a 556, with the second 555 used as an inverter, trig and thresh connected to the 1st 555's output.If you need more drive then a single NPN or PNP will do depending on which direction you need most drive.

author: Zero999
author_rank: Super Contributor
timestamp: March 23, 2017, 11:03:49 am
content: Quote from: Bendba on March 21, 2017, 05:51:57 amWhat does the 555 do?What about replacing the entire 555 circuit with the CD4009? It's possible to a monostable and astable mulitivibrator with that IC.

author: Bendba
author_rank: Regular Contributor
timestamp: March 25, 2017, 03:47:53 am
content: Hi,Thanks for the answers.StillTrying,Indeed, I must have been brain dead, never thought about using another 555.I was trying go use transistors because I have plenty of them, not so many ic'sHero999,In this case, the 555 is a time base/pwm source. I can select the capacitor with a jumper (~10nF to 1000uF in increment of 10's factor, ~1nF by default if no jumper is inserted.) And a small potentiometer let me change the pulse width. It very crude but it works to do quick tests. The frequency roughly goes from 1/100 Hz to 120 kHz.I did try to make the whole thing around a 7404 (I don't have any 4009 anymore) but I can't seem to keep the oscillator stable at low frequencies or extreme duty cycles.I eventually worked out a circuit that sort of does what I want. In this case, the output of the 555 also drives a bicolor LED (common cathode, that's why I wanted an inverting buffer too) the color gives me an idea of the duty cycle.

author: tpowell1830
author_rank: Frequent Contributor
timestamp: March 25, 2017, 05:49:33 am
content: Quote from: Bendba on March 21, 2017, 05:51:57 amI'm not sure what you are doing, but could you clock a d flip-flop with the PWM and drive your transistors with the Q and QNOT outputs?

author: Bendba
author_rank: Regular Contributor
timestamp: March 25, 2017, 11:14:15 am
content: Hi tpowell1830,I didn't really try flip flops, as an inverter is easy to find. Unless there is a reliable transistor flip flop circuit out there.

author: Zero999
author_rank: Super Contributor
timestamp: March 25, 2017, 08:02:30 pm
content: Quote from: Bendba on March 25, 2017, 03:47:53 amThe old TTL 7404 won't work, unless the resistor value is very low <1k.You should use the CMOS CD4069 or 74HC04 instead. Another option is the 74HC14 which is much simpler to implement.https://www.eevblog.com/forum/projects/pwm-signal-from-555-timer

author: Bendba
author_rank: Regular Contributor
timestamp: March 26, 2017, 03:59:59 am
content: Hi.I still have a couple of 74HC04, no 74HC14 though. I'll try those circuits out.That said, I'm still trying to find a transistor circuit that does that follower + inverter function. That 3 pnp transistor circuit I posted earlier is the only way I found so far too do it. Does that curvy have any major flaw?Edit: the 2 LED's in the circuit is only to show the outputs, it doesn't necessarily have to be driving LED's, also, the input could be anything. In today's case, it is a 555 time but I've been making other circuits where I needed that buffer.

author: ebclr
author_rank: Super Contributor
timestamp: March 26, 2017, 06:42:14 am
content: One digital transistor will do that very easilyhttp://www.onsemi.com/PowerSolutions/parametrics.do?id=799

author: Bendba
author_rank: Regular Contributor
timestamp: March 26, 2017, 12:49:18 pm
content: Hi ebclr,Thanks for your answer. Sorry for being a pain, I'm still trying to work it out with common bjt's (I have over 500 S8050 and S8550)That said, those digital transistors are interesting if I plan something one day for which the part count is important.Regards

author: Ian.M
author_rank: Super Contributor
timestamp: March 26, 2017, 01:31:44 pm
content: Don't start with a bipolar 555. Its output cant get all the way up to the + supply voltage, so its a lot harder to buffer without significant losses. If you use a CMOS 555, an unbiassed complimentary pair of emitter followers will buffer the output and swing typically to within a volt of the rails.The least complex bipolar inverting push-pull buffer I know of is this:V1 would represent the 555 output and V2 its supply. Obviously voltages and some resistor values would need changing. I originally posted it here:https://www.eevblog.com/forum/projects/vacuum-fluorescent-display-multiplexing-problem/msg1032843/#msg1032843, complete with a LTSPICE simulation.Its *NOT* suitable for high current loads as the available output current is limited by the pullup resistor and transistor gain.

author: Bendba
author_rank: Regular Contributor
timestamp: March 26, 2017, 01:57:29 pm
content: Quote from: ebclr on March 26, 2017, 12:53:24 pmThat's actually what I started with. Those two circuits work well by themselves but as soon as I tie the two input together, both transistors start conducting unless I bring R1 and R3 higher but then, they don't respond to the input anymore. But then, Ian.M's comment might be pertinent. The output of my 555 might just not be going high enough to stop the pnp transistor's base from conducting through the npn's base.Ian.M,I never thought about putting a diode there. Looks like that's what I'm after. If I'm correct, with your circuit, I can still use the output of the 555 to drive another npn transistor for my non inverting buffer without interfering.I'll have a try tomorrow.

author: Ian.M
author_rank: Super Contributor
timestamp: March 26, 2017, 02:17:12 pm
content: If you are having problems with turn-off in your simple grounded emitter circuits, add a resistor from base to emitter. It forms a potential divider with the series base resistor, which increases the minimum voltage (with respect to the emitter) to turn the transistor on.The circuit I posted above is a way round the deficiencies of a simple collector pullup resistor. It acts to boost the available output-high current and, for lower voltages, works even better with a Schottky diode

author: ebclr
author_rank: Super Contributor
timestamp: March 27, 2017, 08:14:08 am
content: The idea is to use only one, not the two together, one is with NPN transistor another with PNP transistor , On the NPN one, the LOW will be strong and the high weak based only on the pull up resistor, On the PNP High will be strong and the low weak based, only on the pull down circuit.

author: Bendba
author_rank: Regular Contributor
timestamp: March 30, 2017, 12:24:11 pm
content: Hi,Thanks all for all your help. I ended up with three good working circuits.The base-emitter resistor did the trick, even my npn and pnp agree to work together now.I got a 3 transistor all pnp circuit working too.I still have to try Ian.M circuit but I can see some applications for it.Thanks again.