Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!cmcl2!phri!roy From: roy@phri.UUCP (Roy Smith) Newsgroups: sci.electronics Subject: Re: Vaguely related to the light bulb problem Message-ID: <3887@phri.UUCP> Date: 22 Jul 89 03:02:00 GMT References: <4363@merlin.usc.edu> <1348@bnlux0.bnl.gov> Reply-To: roy@phri.UUCP (Roy Smith) Organization: Public Health Research Inst. (NY, NY) Lines: 30 In hobbit@topaz.rutgers.edu: > What's the correct way to drive TTL from an op-amp? [...] the op-amp seems > unable to pull *low* enough to make the Schmitt input see a low level. I > suppose I could go out and buy one of those fancy op-amps that drives all > the way to the - rail Not only is the question related to the light bulb problem, but the answer is too. In fact, at the risk of being pedantic, I'll just quote my answer to the light bulb question, since it applies here too. In article <3868@phri.UUCP> roy@phri.UUCP (Roy Smith) writes: > The problem is that TTL can sink current, but not source it. With >TTL you really have to think current sinks instead of voltage sources. I'll throw out a few possibilities off the top of my head. Try pulling the TTL input down with a resistor to ground. Or, use a CMOS gate as a buffer (much higher imput impedence and sufficient output to drive a single TTL input). Or, use a low-power TTL part for the first gate after the op-amp. Or, an op-amp which can sink more current. Or, a transconductance amplifier between the op-amp output and the TTL input. The bottom line is you don't really want to think about what the TTL input voltage is. What you want to think about is if you can sink enough current. A TTL input is a current source. You gotta do something with that current or the input drags itself high. -- Roy Smith, Public Health Research Institute 455 First Avenue, New York, NY 10016 {att,philabs,cmcl2,rutgers,hombre}!phri!roy -or- roy@alanine.phri.nyu.edu "The connector is the network"