0
\$\begingroup\$

I am trying to design a small control circuit for controlling large currents (not too large, around 200W, AC 50Hz). I have kept in the AC circuit isolated from the DC control circuit. But the problem arises when it comes powering the control circuit. My control circuit will take maximum of 1W of power at any time during its operation. Also I am bound to not use any battery for this purpose and the size of this circuit has to very small (smaller than a typical mobile phone charger).

Since the load is very small I was wondering if can use AC supply from mains to power it by converting the supply voltage to DC using a rectifier and then using a potential divider to lower the voltage to usable form (Note: Since the space available to me is very small, I cant use a transformer to step down the voltage).

So my question is that can I use this configuration to power the small load (1W). The schematic is attached. I know that voltage regulation will be affected but I have no idea "How much".

enter image description here

\$\endgroup\$
  • 3
    \$\begingroup\$ If you do this you must consider that your whole circuit is live. \$\endgroup\$ – Transistor Feb 25 at 14:20
  • 2
    \$\begingroup\$ "I have kept in the AC circuit isolated from the DC control circuit." it makes no sense to use this kind of PSU, then. \$\endgroup\$ – Marko Buršič Feb 25 at 15:33
  • \$\begingroup\$ @MarkoBuršič "I have kept in the AC circuit isolated from the DC control circuit" by this statement I mean that the circuit that I wanna control is isolated from this circuit, not the AC supply powering this circuit. ;) \$\endgroup\$ – Aman Singh Feb 25 at 16:07
  • \$\begingroup\$ This power supply, if you can even call it that, is not capable of providing the amount of current you want out of it, and that's not even getting into the safety concerns. \$\endgroup\$ – Hearth Feb 25 at 16:09
2
\$\begingroup\$

Let's do a quick analysis:

Note that the current into a linear regulator (like the LM7805) will be the same as the load current.

You need 1 W at 5 V, that means a current of 1 W / 5 V = 0.2 A

That current needs to flow through the 100 kΩ resistor, that means a voltage drop of 100 kΩ × 0.2 A = 20 kV across the resistor! That's not going to work.

The idea of that 100 kΩ resistor is to drop around 300 V (220V AC rectified gives around 311 V, 11 V for the LM7805 that leaves 300 V). 300 V / 0.2 A = 1500 Ω, OK.

But let's calculate how much power is lost in that resistor: 300 V × 0.2 A = 60 Watt !!!

That's a lot.

It is much more efficient to use a regular AC adapter (like the ones used to charge a phone) or a capacitive dropper circuit but do note that the capacitive dropper does not provide mains isolation; your complete circuit would be mains live and cannot be touched by humans while in operation and also cannot connect to phones, PCs and laptops etc.

enter image description here

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

No, it won't work. Let's do the back-of-the-envelope math.

Suppose the entire dc voltage was across the 100k resistor. The peak voltage from a rectified 220V ac is about 308V. The 100k resistor will allow no more than 3.08mA through to your voltage regulator. If your voltage regulator consumed no current itself, then you would have a maximum power delivered to the load of 15.4mW.

Suppose you change the resistor values so you get 200mA to the voltage regulator. That means you have 200mA through your resistor divider. I don't need to do the exact math to know that a great deal of power will be dissipated in the resistor, so the resistor will need to be very large.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

This is called transformerless power supply. You have to change the circuit a little. Look at this circuit for example: https://electronicsarea.com/transformerless-power-supply-circuit/

I have made similar power supplies. Of course, the 1 W circuit has to be recalculated. Using zener for 9 or 12v is easy to achieve 1w power at 100-120mA. But more important here is the idea of limiting current by a capacitor before rectifier that doesn't dissipate power.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.