Bridge rectifier + voltage doubler - Page 1
Forum section: electronics / projects

author: fabiodl
author_rank: Frequent Contributor
timestamp: March 26, 2018, 09:31:32 am
content: Hello, I tried to combine a bridge rectifier with a voltage doubler, to obtain a negative voltage.I attach the circuit I came up with (cirtuitLinked.png)If I disconnect the voltage double (circuitOpen.png) and simulate all works as expected, I get a vp which is close to the peak value of V1 (open.png).However, when I link the voltage doubler, vp increases (linked.png).The bridge rectifier should provide a "stiff" power to C1. Why does the voltage double part influence vp?

author: Ian.M
author_rank: Super Contributor
timestamp: March 26, 2018, 11:03:40 am
content: Consider the current flowing into and out of C2 over the course of a whole cycle. For the sake of conciseness, lets call the the top terminal of source V1, node AC1 and its bottom terminal node AC2, and define the positive half cycle as when AC1 is positive with respect to AC2.It is obvious that on the rising slope of AC1 (from peak negative half cycle to peak positive half cycle, the current into C2 flows through D4, and V1. However on the falling slope of AC1, the current out of C2 must flow through V1, D4 and the load on the bridge rectifier.During the rising slope of the first half cycle, C1 charges to 10V-2*Vf (2 diode drops), and C2 charges to the same voltage across it. During the following falling slope, C2 partially discharges, charging C3 to 1/3 of C2's peak voltage (minus Vf), however C1, C2 and C3 are effectively in series (as the diodes block all other current paths apart from the 100K load resistors which have negligible effect) so the change of voltage due to the current through them is shared equally between them, so C1 gains 1/3 extra voltage as well.On subsequent cycles, the charge transfer through C2 is much less and the excess voltage on C1 rises more slowly until C3 is fully charged, at which point the excess voltage settles down to 1/3 the ripple across C2 due to its charge pump action supplying the load current in R2.It should be noted that for much lower load resistance R1 than R2, the duration of the excess voltage is much shorter, but if you were building a PSU and needed to stay under the regulator's maximum input voltage, but were already close to it, you'd need to also make C2 an order of magnitude smaller than C1 to limit the effect.LTspice sim attached

author: fabiodl
author_rank: Frequent Contributor
timestamp: March 27, 2018, 10:30:46 pm
content: Thank you so much!I just reasoned in terms of higher-lower voltage and diodes as opened or closed, and that was insufficient, but reasoning in terms of currents completely explains what is happening, thank you!

author: Ian.M
author_rank: Super Contributor
timestamp: March 27, 2018, 11:11:29 pm
content: You are welcome. I must admit, the magnitude of the excess voltage surprised me at first - I was already well aware that such a charge pump needs a decent load on the main bridge to work properly - but I had to consider exactly where the current could go to understand the simulation results.It should be noted that a balanced charge pump (aka: capacitively coupled bridge) doesn't cause this objectionable overshoot behaviour on the main unregulated rail, and since for the cost of two more diodes and an extra capacitor, you also get double the current or half the ripple, its well worth doing anyway if the pump is to be driven at mains frequency.

author: amyk
author_rank: Super Contributor
timestamp: March 28, 2018, 12:32:14 am
content: Why do you want both a bridge rectifier and doubler? Something more similar to the input stage of the typical PC PSU might be more suitable if you need doubling and rectification.

author: fabiodl
author_rank: Frequent Contributor
timestamp: March 28, 2018, 01:47:02 am
content: @Ian.MThis balanced pump is really interesting, thank you, and thank you for the LTspice files!@amykI want to control a magic eye tube. I need DC 250V and -15V.I am looking for several solutions (maybe lnk304?) but for now I wanted to try something with the parts I have at home. I have a 100 to 200v transformer, so the simplest approach for now is rectifying the 200v, get a negative potential (-200 in this case) and then use a resistor divider to get the -15.Do you have any link to the input stage you are referring to?

author: Ian.M
author_rank: Super Contributor
timestamp: March 28, 2018, 05:38:56 am
content: If you've only got one magic eye tube (or other tube) to run, you'd probably do better to bias it with the cathode at +15V to get the required control grid negative voltage swing with respect to the cathode.I'd start with a 15V PSU to power the rest of the circuit, a buck converter for the filament supply, (assuming an indirectly heated cathode - for directly heated you need a floating AC filament supply, preferably center tapped) and one of the many HV boost circuits popular in the Nixie scene to get +215V. Which type of boost circuit you go for depends on the local availability of the required magnetics and HV chopper device. Possibilities include a miniturel 220V:15V-0-15V mains transformer used backwards as a step-up transformer with push-pull drive, a 220V:15V transformer with H-bridge drive , various flyback trabsformer circuits or a simple boost converter with a voltage doubler on its output (as single inductor boost circuits that directly step up more than 1:10 tend to be problematic).

author: fabiodl
author_rank: Frequent Contributor
timestamp: March 30, 2018, 09:17:22 am
content: Dear Ian.M, thank you for all your suggestions, I will look up for them during the weekend. (I do not know many terms yet, like push-pull drive, etc.)The tube I want to drive is thishttp://www.r-type.org/pdfs/em87.pdfIndeed, I though to connect the cathode to +15v, and then move the grid in the range 0-15.I was thinking of using this IC, which is dirty cheaphttps://www.power.com/sites/default/files/product-docs/lnk302_304-306.pdfto bring up directly the mains voltage (100V here) to 250V and then use a voltage divider to get the 15V.Not much current should be drawn here, so I am not particularly worried.The heater instead requires 6.3v and 300ma. I'd like to again obtain these directly from the mains, but even the end of the line IC (lnk306) may not be ideal for such currents.What do you think?

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2018, 12:20:27 pm
content: It is extremely unusual to use a direct from mains PSU that doesn't provide isolation (e.g. capacitive dropper type, or the LinkSwitch-TN IC you linked to). You only normally see it in applications like HVAC, home automation and some heating and cooking appliances where the powered circuit is going to control a TRIAC or similar for direct mains switching or control, and there are no user accessible connections. From a safety viewpoint, the PSU and the whole circuit it powers must be treated as if it is at mains live potential, so either reinforced insulation or basic + supplementary insulation (aka 'Double Insulated) must be used between the circuit and any surface or control the user can touch, or if that surface or control is conductive, it may be securely grounded, and basic insulation used. (Seehttp://www.reo.co.uk/files/safety_7_-_low_voltage_directive.pdffor a quick overview of classifications of insulation - similar regulations to the EU LVD are in place in nearly all countries)For experimental use and general bench work, I would strongly discourage you from using any sort of non-isolated PSU circuit.The LinkSwitch-TN IC wont help you get a 250V DC rail from a 100V RMS mains supply and is unsuitable (even the LNK306P version) for supplying a 300mA 6.3V valve heater as its initial (cold) current draw may be several amps.You only need about 3mA at 250V for an EM87 magic eye tube. That's easily within the range that simple flyback or boost (+ multiplier) circuits can achieve from 12V to 18V DC in, without needing a custom transformer or inductor. For the filament supply, as the cathode is indirectly heated, any of the many 3A DC-DC buck converter modules you can find on EBAY, with the output set to 6.3V, would do nicely.