Project power supply questions. - Page 1
Forum section: electronics / beginners

author: paulca
author_rank: Super Contributor
timestamp: December 09, 2022, 05:49:33 pm
content: If the following more likely to filter a power supply or start a fire? 200mA or 1W load.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 10, 2022, 08:20:55 am
content: I would not expect it to start a fire, but chances are it can blow the pure JFET. Most JFETs are low power and would not handle more then some 10 mA thourgh the date well. During power off transients and possibly load transients this could happen.

author: paulca
author_rank: Super Contributor
timestamp: December 10, 2022, 05:15:38 pm
content: Sorry, wrong FET symbol I was going for basic MosFet.Anyway. I made it up on the breadboard without the LC filter portion. Ended up with a 1K into a 400uF LSR cap + a 100nF across it.As input I configured the signal gen to output a 1Vpp 10kHz sine wave with a +5V DC offset. I AM modulated its channel 2 of a 1Vpp 100kHz CMOS squarewave and the result was what looked awfully like a nasty power supply to me.The output of the RC LPF went to the gate of a BS170 with the original dirty signal on it's drain and as load on the source I put... a 47R resistor to ground.To convince myself it was working I put an LED + 47R resistor and the LED did in light up a bit too bright!I mean, am I doing this right? To me it looks like it is in fact working!?Scope image shows the input artificial power source from the siggen Yellow. Purple is the output of the filter. Green is the voltage across the 47R load resistor.

author: Peabody
author_rank: Super Contributor
timestamp: December 10, 2022, 05:24:14 pm
content: Could you please explain how it works? And how much voltage drop is there across the mosfet? Sorry but I just don't see the load resistor in the schematic, or the purple in the scope image, and trying to figure out what's going on.

author: Ian.M
author_rank: Super Contributor
timestamp: December 10, 2022, 05:26:00 pm
content: Now put the scope inputs to DC coupled and you'll see why MOSFETs generally aren't used in capacitance multipliers!You'll need to increase the X volts/div a lot and probably offset the X positions to get the traces both on screen. Use the same settings for both.

author: Ian.M
author_rank: Super Contributor
timestamp: December 10, 2022, 05:33:21 pm
content: Quote from: Peabody on December 10, 2022, 05:24:14 pmDave explains it better:https://www.eevblog.com/forum/blog/eevblog-1116-the-capacitance-multiplier/Note the use of a Sziklai pair in the video to minimise voltage drop, as a Darlington pair would have had excessive voltage drop.An enhancement MOSFET will drop at least its Vgs threshold voltage, rising to a drop of whatever gate voltage is required to support the load current, in most cases significantly worse than a Darlington pair.

author: paulca
author_rank: Super Contributor
timestamp: December 10, 2022, 05:34:03 pm
content: Ah!Booooo!I knew I should be DC coupled to check the voltage drop, I just got excited

author: paulca
author_rank: Super Contributor
timestamp: December 10, 2022, 05:57:24 pm
content: I only have a handful of transistors. All my Mosfets are 2-4V VGSon.So I tried a 2N3904 NPN Bipolar.The voltage drop is much less. I'll see if I can get them into a rough and ready darlington config next.

author: Terry Bites
author_rank: Super Contributor
timestamp: December 10, 2022, 05:58:38 pm
content: The component values shown won't work for low frequencies and the inductor will cause a peak in the low pass response.The multiplying effect is proportional to the tranconductance of the pass transistor. A jfet wont cut it. A mosfet will work if you can live with the voltage drop (Vgs +IRon) across it.The simplest solution is a darlington. Its very high current gain lets you use high value resistors in the low pass filter and only drops 2Vbe. So at 9V you can expect 7.5V at the load.sound-au.com/project15.htm

author: paulca
author_rank: Super Contributor
timestamp: December 10, 2022, 06:19:10 pm
content: For some reason I couldn't get the darlignton pair to work. I gave up trying because my new breadboard is one of the junk ones with no grip. WHen I had 2 or 3 connections it was maddening. The darilngton made it impossible to get it all stable at once. I don't think the price matters it's just the roll of the dice, there are factories making good ones and factories making rubbish ones. They "looked" nice, but 3 more of the junk pile.

author: Peabody
author_rank: Super Contributor
timestamp: December 10, 2022, 11:02:27 pm
content: So the capacitance multiplier is not a regulator, but it seems you could put one inside the feedback loop of a switching converter and have almost ripple-free regulated output. Why is this not done routinely? Is it the voltage drop and resulting wasted power?As for breadboards, I've have very good luck with these:https://www.amazon.com/BB830-Solderless-Plug-BreadBoard-tie-Points/dp/B0040Z4QN8/I think Breadboard Prototype Systems makes quality boards.

author: paulca
author_rank: Super Contributor
timestamp: December 11, 2022, 11:07:46 am
content: Thinking about putting it in a feedback loop. You might have trouble with the slow start up time. The base/gate drive will rise with the charge time of the RC at start up and when the voltage changes. It would effectively try and make the feedback circuit unresponsive by the RC charge time... ie. filtering the feedback back portion of the feedback so the buck/boost/chargepump whatever will not be stable?On practical uses. I think we have to accept in a linear power supply there will be voltage drop. No free lunch. How much, where and at what current become the question then. This answers part of my general question around "Why don't these kind of circuits just exist in a library like cookie cutter components?". The answer invariably is.... "because it depends" and... "they do exist in libraries, it's just you need a degree in electrical engineering to understand them". Which makes sense. It' sonly going to be academics who bother to write this stuff down in any consumable form and mathematics and electrical engineering is the domain.Specific use case I personally have in mind here is providing a separate power rail for a small section of PCB which will have analogue noise sensitive components aka, amplifiers, ADCs and DACs (AGND/VAA/AREF etc.).Input voltage is unfortunately going to be variable. 9-16V. The good news is the only thing concerned with the upper end of that voltage is a final amplification stage (opamps)... and they don't really need a fixed voltage, I have found they still work right down to about 6V before they just can't drive the headphones at anything useful. The wider the better and if I find 40V caps I like, I might just raise that limit up to 30V.The analogue supplies for the split domain devices will need to be regulated of course. I'll draft a schematic later.If I have 16V coming in. A capacitive multiplier circuit with a set of actual calculated and choosen components, like an approriate darlington power transistor and low ESR premium caps etc. I could live with the voltage drop through the bipolar.If I can size it to only drop that 1.2V (or less if I use a fancy tranny), at 200mA current. I would still have 14.8V MAX, 7.8V MIN. After split railing that linearly to give a VGND for the amplifier Have +/- 3.9V - 7.4V amplifier rails, which is "fine". After the opamp chops the top and bottom of my rails I can still get to +/- 5V which is fine for a headphone amp.... yes I'll do the sums and make sure. Finger in the air, 200mA current limit on the opamp @ 4V RMS is something new a watt, fine. 10 times a normal reasonable output.An idea:Given noise addition comes with gain, there is the potential to stagger the gain into stages. For example a Baxandal active preamp which is feeding high impedance audio opamps... it can run on a gnats fart of current (opamp quiescent might case issues). So I can have 2 capacitive multipliers. The first one drops a little and feeds the final current amp stage. The second one filters much more and drops more volts, provides way less current and supplies the pre-amp.Whichever one still has 4.5V can be linear regulated down to the 3.3V analogue ref rails/supplies.How many transistors have to die this time I wonder.

author: paulca
author_rank: Super Contributor
timestamp: December 11, 2022, 11:28:06 am
content: In Dave's video he mentioned something about being tempted to put the RC circuit an opamp to buffer the voltage, but the limitation of the output current becomes too great... If I am only running a preamp... would buffering it's rails with other opamps from a higher voltage makes things more complicated than being worth it? I could power the pre-amp from an opamp with a LPF in it's feedback loop? I suppose the question would be... what would an opamp give me over a single power transistor?

author: Peabody
author_rank: Super Contributor
timestamp: December 11, 2022, 03:39:39 pm
content: Quote from: paulca on December 11, 2022, 11:07:46 amI guess that's right. It would be unresponsive to sudden load changes. That's in contrast to putting in a real giant capacitor, which actually stores and releases energy in response to load changes. Well, I'm just guessing here.