# Alternative to V divider for high currents

I am looking for suggestions as to how to supply power to a gate drive optocoupler i am using.

The supply being used is a 25V (rms) ac supply and intend to supply 10V DC at around 500mA+ from my supply. When i have reduced the voltage, i will maintain it with some caps.

I was looking for an alternative to a high power resistor divider. I thought about a zener diode and i am aware of high power zeners, though this would still need a high power resistor to limit the current.

Any input is greatly appreciated.

• i am trying to use small components with a single supply (500VA toroidal). – engineeroverhere Jan 29 '15 at 16:29

I think that a full-wave bridge rectifier feeding a bulk storage capacitor which then feeds a DC-DC buck converter will work nicely for you. You can purchase a ready-made DC-DC buck converter from eBay for less than $10 and the bridge rectifier and storage capacitor will set you back another$10 or so.

• i have a rectifier and some high power diodes, so i am missing the dc-dc convertor. – engineeroverhere Jan 29 '15 at 16:28
• This could work looks like i could use some smoothing caps to input a relatively stabe DC voltage around the 34v peak of my supply. Seems like it will also handle my current requirements too. WHats the verdict? – engineeroverhere Jan 29 '15 at 16:35
• One thing to be aware of is that the loading characteristics of rectifier / capacitor circuits sometimes create problems. All of the current is conducted at the peaks when the rectifier diodes are forward biased. This is not necessarily a reason to avoid this topology. I am just pointing it out. – mkeith Jan 29 '15 at 17:45
• could you elaborate a little? Both as to what you mean and its ramifications in my application. Thanks! – engineeroverhere Jan 29 '15 at 17:50
• Suppose you have an AC source, and the only thing loading it is a full bridge rectifier followed by large capacitors followed by some kind of DC or DC-ish load. During the majority of the AC cycle, the current is zero because the rectifiers are reverse biased. When the rectified voltage exceeds the DC rail voltage, then the rectifier starts to conduct for a brief period when the AC voltage is near a maximum or minimum. So if there is 1 Watt of DC power getting eaten up, ignoring diodes and such, there is 1W of AC power going in... – mkeith Jan 29 '15 at 21:57

You could use a capacitive dropper, rectifier and a zener diode or, if the load is quite stable you should be able to get away without the zener. Here are a couple of examples from the internet: -

As you are using this for only 25VAC there's no real need to warn you about how dangerous these are when used on AC power directly from the socket.

Same principal - the capacitor reduces the voltage without incurring heavy power losses and producing heat.

• i just read that electrolytics arent used for this purpose. That had me confused until i clarified that haha. Thanks. ill look into it! – engineeroverhere Jan 29 '15 at 18:51