{"id":106,"date":"2022-01-27T22:23:17","date_gmt":"2022-01-27T22:23:17","guid":{"rendered":"https:\/\/paulcarbone.com\/50days\/?p=106"},"modified":"2022-01-30T19:26:26","modified_gmt":"2022-01-30T19:26:26","slug":"day-2","status":"publish","type":"post","link":"https:\/\/paulcarbone.com\/50days\/2022\/01\/27\/day-2\/","title":{"rendered":"Day 2: More on the Multivibrator"},"content":{"rendered":"\n

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I wanted to look at this circuit in more detail to understand what’s going on.

I’m probing the collector of Q2 on the yellow trace, and the base of Q2 on the green trace.

Note that the time off and time on are different – I changed the value of the resistor on the base of Q1 to 10k, down from 22k. This means one side blinks faster than the other, and it just makes it a bit easier to tell one half from the other when looking at the trace below. <\/p>\n\n\n\n

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This mostly me just winging it, but:

When the yellow trace is low, Q2 is conducting, placing the cathode of the LED D2 at ground. (sorry, I didn’t label the parts, but you can imagine Q2 & D2 are on the right). When the yellow trace is high, Q2 is cut off, and D2 is off, without a path to ground.
When the green trace is at it’s highest ~.7v above 0, there’s current flowing through the 22k resistor, through the base of Q2, holding it high. But then how does the trace go negative, when we don’t have a negative supply? That’s when C1 is discharging through Q1 when it’s conducting. That negative swing also shuts off Q2, allowing C2 to start charging. C2 charges quicker, and gets higher (hence it looking more like a square wave) because it’s charging though a lower value resistor, and it’s being allowed to charge for longer, because the other side is slower.<\/p>\n\n\n\n

I don’t think I understand absolutely everything about this circuit, but I think I’ve mostly explained this correct. I’m sure I could read something that would explain it better, but I wanted to work through this on my own. <\/p>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

I wanted to look at this circuit in more detail to understand what’s going on. I’m probing the collector of Q2 on the yellow trace, and the base of Q2 on the green trace. Note that the time off and … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[5],"tags":[],"class_list":["post-106","post","type-post","status-publish","format-standard","hentry","category-blinkinglights"],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/posts\/106","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/comments?post=106"}],"version-history":[{"count":7,"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/posts\/106\/revisions"}],"predecessor-version":[{"id":116,"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/posts\/106\/revisions\/116"}],"wp:attachment":[{"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/media?parent=106"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/categories?post=106"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/paulcarbone.com\/50days\/wp-json\/wp\/v2\/tags?post=106"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}