Low Power isolation via Caps - Page 1
Forum section: electronics / beginners

author: Jebnor
author_rank: Regular Contributor
timestamp: August 22, 2013, 08:13:56 am
content: Hello all,I am planning to 'use up' some of myMC1446LMC1466L floating regulators. They require a floating supply of their own. I could add a separate transformer for it, or something, but I'd like to see if I can get away without doing that.My Transformer puts out 40VAC. It's from an old stereo amp. I was thinking something like the attached schematic. See any major (or well, minor too) pitfalls?

author: Rerouter
author_rank: Super Contributor
timestamp: August 22, 2013, 08:17:49 am
content: your going to pop your caps that way, you want them in series,and yes you can do it so long as the point you tie the 2 circuits together does not then exceed the capacitors rated voltage,just be aware at mains frequency 47u of capacitance equates to a 70 ohm impedance per capacitor, if you want that lower you need bigger values,

author: Codemonkey
author_rank: Regular Contributor
timestamp: August 22, 2013, 08:21:49 am
content: What you're doing is dropping the voltage, the capacitors don't provide any isolation from mains.

author: Rerouter
author_rank: Super Contributor
timestamp: August 22, 2013, 08:23:58 am
content: i figured the controller just required a level shifted control rail, but yes it is not true isolation,

author: Jebnor
author_rank: Regular Contributor
timestamp: August 22, 2013, 08:27:59 am
content: Quote from: Rerouter on August 22, 2013, 08:17:49 amIn series it is. Back to front right? +ve or -ve 'out'? does it matter.I just put in 47u because I have a bunch of them in my junkbox. It's only drawing 15mA Max, so I'm not overly concerned.

author: Jebnor
author_rank: Regular Contributor
timestamp: August 22, 2013, 08:30:31 am
content: Quote from: Rerouter on August 22, 2013, 08:23:58 amYes, the chip requires a 21-30V floating supply.

author: Andy Watson
author_rank: Super Contributor
timestamp: August 22, 2013, 08:45:50 am
content: Yes, with the series capacitor connection you will probably get away with it. But what is your overall aim? If the equipment at the at the output is floating then I don't see the point in floating the inputs. On the other hand if you are trying to float the inputs so that you can stack the outputs on top each other, there are better ways of doing it. Beware that any difference in capacitance (and large caps are not known for their good tolerance), will show up as ripple between the "isolated" outputs.

author: Jebnor
author_rank: Regular Contributor
timestamp: August 22, 2013, 03:42:27 pm
content: To clear up any confusion,I dismantled an old stereo amplifier and got some beautiful heat sinks, power resistors, monster capacitors and other goodness. I am thinking on converting said components, along with parts from my junk bin into a +/- 0-50V 2.25A tracking power supply. Before you ask, yes, I have ML1466L's in my Junkbin. The set voltage is going to come from a filtered PWM from a microcontroller.The point of my question was the isolation of the stepped down (40VAC) voltage via capacitors(C2-C5) to provide a little floating 21-30Vdc <50mA supply rail to power the ML1466L. There are labeled block on the schematic.This is only a quick slapped together schematic, hence no values on any of the components. The circuit is a mashup of a couple of the examples from the ML1466L Datasheet.

author: SeanB
author_rank: Super Contributor
timestamp: August 22, 2013, 06:11:35 pm
content: You will have ripple current injected into the output of your power supply though. Easiest to add in a small 1VA 24VAC mains transformer to the original circuit and use it to supply the rail, that will have the least current injection into the output. The problem with the capacitor supply is the current through the capacitors will appear on the IC between the supply and the feedback, control and output as a varying voltage.

author: Jebnor
author_rank: Regular Contributor
timestamp: August 22, 2013, 08:31:12 pm
content: Quote from: SeanB on August 22, 2013, 06:11:35 pmI hadn't fully considered that. Now I need to scrounge around for a 1VA transformer or something. Hmmm...

author: megajocke
author_rank: Regular Contributor
timestamp: August 23, 2013, 05:22:49 am
content: The most immediate problem I see with that last schematic is that the anode of D1 only connects to capacitors. The diode will only pass current in one direction while the capacitors will not pass DC. So the only thing that will happen is that the capacitors will charge and no output current will be available after the initial charge-up. So it will definitely not work.Going for a small auxilliary transformer sounds like a good idea. Or adding a winding to the existing transformer, if possible.

author: Jebnor
author_rank: Regular Contributor
timestamp: August 23, 2013, 08:06:50 am
content: Quote from: megajocke on August 23, 2013, 05:22:49 amI had considered this in the first schematic, hence the Fullwave BR.Quote from: megajocke on August 23, 2013, 05:22:49 amProbably what I'm going to do. Adding a winding is not an option.

author: megajocke
author_rank: Regular Contributor
timestamp: August 24, 2013, 06:45:55 am
content: Right. The more subtle problem with the capacitor solution then would be that there is nothing guaranteeing that the current going through one of the capacitor branches would return through the other, so the charging pulses could go through the IC/to the output in unintended ways as was mentioned before.And if are using the center tap connected to one of the capacitor branches, you will be guaranteed no current in that branch as both ends are at DC potentials. So you will be guaranteed that the charging pulses will return through unintended paths, sadly.(also these capacitors would discharge through the MC1466 if the output voltage every abruptly changes, such as when a short or a battery is connected to the output)One option could be to generate an extra +110V supply using a doubler, and using it in conjunction with the -55V supply to feed two current sources, respectively sourcing current to the cathode of your shunt regulator and sinking current from the anode. The balance of the currents needs to be chosen so nothing bad happens inside MC1466. By selecting the sink current a little higher than the source current the internal zener between pins 7 and 9 would establish the correct working condition, generating the required floating supplies of roughly +11V and -7V.But then startup/shutdown might provide surprises, with incorrect operation if the source current ever goes higher than the sink current, so maybe a separate transformer for the control circuitry is the easiest way after all.

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2013, 09:20:32 am
content: I know this isn't helpful here but just for educational purposes.It is possible to perform mains isolation using only capacitors but it only works at high frequencies and low currents. This is how RF suppression Y capacitors work. They allow through the high frequency RF noise whilst acting as a high impedance to mains frequencies.You can make a solid state relay using Y1-rated capacitors for isolation.