DIY Curve Tracer & Novice Questions - Page 1
Forum section: electronics / projects

author: jdutky
author_rank: Regular Contributor
timestamp: April 10, 2021, 12:31:45 pm
content: I'm building the DIY Curve Tracer from W2EAW's video #232 (), and I got frustrated with having two independent oscillators in the same system. My solution was to replace the long period oscillator (generated by an op amp) with a counter so that I could have a specific number of steps per long cycle (determined by the preset value of the counter). The master clock is still generated by a 555 timer, using basically the same layout from W2EAW's design (except I don't use the reset signal; the 555 is free running and drives all the rest of the system timing).I've been having all kinds of trouble both driving other circuits with the signals coming out of the TTL logic chips (the 74LS193 sync counter, and a 47LS04 hex inverter to change signal polarity as needed), and keeping the counter actually counting. I'm driving assorted 2N2222 transistors to charge or discharge capacitors, and when I connect the TTL signals to the bases of the transistors I occasionally get a much lower amplitude pulse than I expect (~1V rather than 5V), and either the transistor doesn't saturate, or the counter stops counting (sometimes both).I think that my problems come from a misunderstanding about how to drive the transistors. In W2EAW's schematic (https://www.qsl.net/w2aew/youtube/curvetracer.pdf) there are resistors in series with the base of transistors that are being used to drive a signal to ground (the 8.2K resistor going to the bases of Q1 and Q2, and the 3.9K resistor going to the base of the unlabeled resistor in the upper right). Am I correct in suspecting that these are current limiting resistors, and that I should have those in place any time I am driving a transistor with a limited current output signal (e.g. a TTL output)?Similarly, I know that I can't leave input signals to a TTL chip floating, that I must pull them up or down, but do I need to put a resistor in series with the input signals when I do this, or can I just tie them directly to Vcc or ground? Also, if I'm not using an output signal should I be pulling it up or down, or can I leave it floating? If I must pull it up or down, should I use a resistor, or can I tie it directly to the rail?I've been hacking all of this together on a solderless breadboard, which may also be responsible for some of my problems. I am in the process of drawing up a proper schematic of my design and will attach that to this post as soon as it's ready (very soon, because I also need it to keep track of what I'm hacking on the breadboard; without it I'm rapidly getting confused by my experimentation).Thanks for any advice or assistance.-- Jeff Dutky

author: ledtester
author_rank: Super Contributor
timestamp: April 10, 2021, 02:57:15 pm
content: QuoteShort answer: yes.QuoteTying a digital input directly to a rail is fine. Unused outputs may be left unconnected.Note there was a typo made on this topic in some published TI literature:https://electronics.stackexchange.com/questions/242683/what-to-do-with-unused-ttl-and-cmos-outputsUpdate:The "Designing With Logic"https://www.ti.com/lit/an/sdya009c/sdya009c.pdfdocument mentioned in the above SO question has this to say about certain 74-series logic devices:Quote

author: Kleinstein
author_rank: Super Contributor
timestamp: April 10, 2021, 09:28:39 pm
content: The old BJT based TTL chips recommend a resistor when connecting to VCC. Most of the times it works without the resistor, but there is a slighte chance to blow the chip without. The GND side can can go to GND directly. An open input is usually seen as high with the old TTLs, but not very robust to external interference.Modern 74HC and similar chips, which are CMOS internally can have the input directly tied to VCC or GND. An open CMOS input is bad an very sensitive to picup of all kind of noise and trouble.