Ripple filter through a transistor network? - Page 1
Forum section: electronics / beginners

author: de_light
author_rank: Regular Contributor
timestamp: July 24, 2020, 05:46:09 am
content: Please help me try to understand this part of a schematic. It is for a portable radio.This section is labelled 'ripple filter'. It is comprised of two NPN and one PNP transistors. I've never encountered a ripple filter made of purely transistors (I think I have only seen capacitors being used for this) so I'm trying to work it all out.Any kind souls able to walk me through the current flow?

author: Ian.M
author_rank: Super Contributor
timestamp: July 24, 2020, 09:16:27 am
content: You can start off with *ALL* the voltages marked in this section are wrong if external power is connected, as they are only applicable when its running from batteries.Q513 is a saturated switch - it disconnects R522 when its running on batteries, removing its approx 11mA load on B+ and also forces Q511 hard off so it doesn't draw current through the now forward biased C-B junction of Q512. When its on external power, the bottom of R522 is effectively grounded. You can also ignore D503, which is a 13V Zener, presumably to clamp transients from hot-plugging J4.R523 and R525 form a potential divider feeding the base of Q511 with 91% of the average DC_in (smoothed by C522 which with the divider forms a low pass filter with a cutoff frequency of 1.7Hz). D502 drops Q511's emitter about 0.7V below B+ when Q513 is on, approximately matching Q511's Vbe drop. Q511 thus acts as a comparator, turning on progressively as B+ drops below the smoothed voltage on its base, and drives the PNP pass transistor Q512, so B+ tracks the smoothed 91% of DC_in on C522.LTspice sim attached. I've disentangled the schematic so the 'flow' of the circuit is a lot clearer in the sim.

author: de_light
author_rank: Regular Contributor
timestamp: July 24, 2020, 10:21:51 am
content: I thank you for your reply and I'm in awe of your knowledge...but literally understood about ten percent of what you said. Can you translate into noddy language please?

author: Ian.M
author_rank: Super Contributor
timestamp: July 24, 2020, 11:21:39 am
content: 'Noddy language' doesn't help much when describing analog electronic circuits. The concepts aren't really expressible non-technically, so you end up having to build understanding one technical definition at a time till you 'grok' the particular application.However on the basis that a picture is worth a thousand words, and realising that you may not have downloaded and installedLTspiceyet, (you definitely should), here's the disentangled schematic I drew up.I suggest you start by figuring out where I got (i.e. calculated) the 91% of the DC in voltage and 1.7Hz cutoff frequency from as those are key concepts needed to understand the rest of it. Both are 'back-of-the-envelope' approximations ignoring the loading by Q511 base, not results from the sim.The following links *may* help:https://en.wikipedia.org/wiki/Voltage_dividerhttps://en.wikipedia.org/wiki/RC_time_constant

author: de_light
author_rank: Regular Contributor
timestamp: July 24, 2020, 12:25:17 pm
content: Okay, thank you.Using Electrodroid's voltage divider application, I get a Vout of 5.455V which I guess is near enough your 91% DC Input voltage. So this feeds the base of Q511.Q513's output I guess is around 5.5V. So this would lead to a current across R522 of around 11.9mA?In terms of the low-pass filter, the resistance going in by the two 4.7kΩ+ 47kΩ in parallel gives 4.273kΩ. Combined with a 22uF capacitance, this gives a frequency of 1.693hz.May I have the next clue to the puzzle of understanding this please?

author: Ian.M
author_rank: Super Contributor
timestamp: July 24, 2020, 01:00:44 pm
content: Consider how much a first order RC low pass filter with a 1.7Hz corner frequency attenuates 100Hz ripple. Seehttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/filcap2.htmlNext, consider the loop Q512, D1, Q511. Negative feedback from the output (B+) goes via D1 to Q511 emitter (e). Q511 is effectively in common base, and Q512 is common emitter. Treating Q511 base (b) as a fixed voltage, consider how a small variation in its emitter voltage (e) can take it from turned off to fully turned on, then think about how the loop will settle with Q511 getting *just* enough Vbe, to turn it on enough to give Q512 just enough base current to pass enough current to maintain Q511 emitter (e), which sits one diode Vf drop below (B+) at that minimum Vbe below its base voltage (b).