Junkbox DC-DC Converter

Oct 13, 2008 6 Replies

I worked up the circuit below to create an isolated DC supply, and it works (sort of). It's a 555 astable running at ~500KHz into a voltage doubler.



I get an unloaded voltage of 7v, dropping to 1.5v under load. My problem is that I'd like to get >5v at 20ma continuous out of it. Any suggestions on how to improve the current output (assuming it's possible)?



Failing that, can anyone point me to a simple 5v-5v DC converter circuit w/isolation?



+5v |-------+-+ .-. | | | | .----------. 1k | | | 4 8 | 100nf '-' | | || +----|7 3|-||-+--->|--+------+ X---| .-. | | || | | | | | | | LM555 | | |10uf | | 10k | | | | | --- | | '-' | | - --- | | |--+-|2 | ^ | |10uf |
100 --- +-|6 5 1 | | | --- .-. pf --- '----------' || | | --- | | | | +---||-|-------+ | | |Rload GND --- | || | |10uf | '-' --- | 100nf| --- | | | GND | --- | | GND | | | | |-------+------+ X---|

(created by AACircuit v1.28.6 beta 04/19/05 www.tech- chat.de)


Do you have an O-scope? If so, check the voltage on pin 3 under load -- I wouldn't be surprised if you found not able to drive the thing.

Also, what diodes are you using? Something like a 1N4148 should be fast enough (although I don't know if it's rated to 20mA); something like a

1N4001 would be _way_ too slow.
Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html

With that circuit configuration, there is NO WAY to get more than the output swing of the 555 minus two diode drops at any real load current (the 7v observed at no load is almost certainly due to ringing on the

555 O/P).

With a 5 V supply, I'd be amazed if it could even manage 4V into a 10K load.

Also the two series 10uF caps are not needed at all. The 100nF to ground might as well go to the negative side of the load instead of their junction and if you need more reservoir capacitance, increase the value of the remaining 10uF cap. It would also be pretty much essential to have a decoupling capacitor across pins 1 & 8 of the 555. The 555 is operating at somewhere close to a 45% duty cycle so to get 20mA out it must sink & source over 45mA. At that level of output current, from the curves in the data sheet, your output swing is from about Vdd -1.5V to Gnd +1V (curves only valid for Vss>5v.) Thus you are loosing another 2.5V. This is consistent with your 1.5V measured under load.

Converting it to a voltage doubler will loose you two more diode drops and halve the load current so its fairly obvious that you cant make a

555 based charge pump converter do what you want.

If you can supply the 555 with a minimum of 9V you might be fairly happy with the output.

Does the 5V out need to be regulated? Why do you need the isolation? What other constraints are there?

Might be easiest to tell us what its for.

Ive made some changes to your schematic that might help. The two diodes should be 1N914, 1N4148, 1N4448, etc.

Regards, John Popelish
[snip]

Not an option, I'm afraid...

Yes.

There's a possibility (low, but nonzero) of a ground fault or other electrical malfunction putting line AC onto the load side ground line. This would pass back through cables and equipment that really aren't designed for it.

A USB-powered device. A usb-capable PIC communicating via opto-isolators with the load side. This was my first attempt at getting power across the divide.

I was hoping to avoid a bulky isolation xfmr, but I can bite the bullet if I have to.

I dont see any way you can win with your 555 based circuit then. The capacitive coupling would also not have been a great idea if it is important to keep mains spikes and other transients out of the USB side

Is there any way you could use an AC line derived supply on the load side? You dont actually need a transformer then but could use a capacitive dropper feeding a rectifier and shunt regulator. If you are doing any power control on the load side, a line derived supply is pretty much essential so that you can lock out the drive to the output stage if the USB side looses power. You'll probably also want to optoisolate a 'power good' signal to the PIC to let you manage startup and shutdown cleanly.

OTOH if its just sensors on the load side, a DC-DC convertor solution off the USB power makes sense. I doubt I'd roll my own though. Unless you are producing a large number of these devices, buying in an isolated switched mode 5V converter module such as:

looks rather attractive. Reasonable price and you aren't going to get much more compact.

Beware of limited current availablility on the USB bus. If the device hasn't enumerated yet or someone has plugged it into a bus powered hub, you may have to keep most of your device powered down, indicating an error code if you have to.

For prototyping purposes, a small 'wall wart' and a 5V regulator should be findable in your junk box while you are waiting for yor converter module order to come in.

This might work for you, depending on what you have in the junkbox:

To bridge rectifier | | [Xformer] | | | +-------------+---+ + +---+-----------------+ | | | | | | +----- | ---- | ---+ | | | | | | | | [.1uf] | | | | | | | | | | | +----- | -----+ | | \\c | [R] | c/ |---[10K]---+ | +---[10K]---| /e +5 Vcc e\\ | | | | +--------------------+------------------------+ | Gnd

Limit the collector current with R.

Ed

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