# Get the max current through relay without activating it

I have a circuit where the only current I can get to supply my battery is the one that I can get from a relay connected to Vcc.

When it is not pulled to GND I can get current without activating it up to around 12mA.

Is there some way to get more current without activate the relay?

I've done some tests connecting this to LC filter and a buck converter and I think I'm getting more current, but this circuit seems too sensible to changes (If I want to prevent the relay activation).

Edit:

simulate this circuit – Schematic created using CircuitLab

I need to power this circuit in a place where there is no power outlet or other power supply, so the only place where I can get power is from the relay (But I can't access to the wire connected to VCC, only the wire that goes to GND) . This is the relay from the intercom of a entry door building (when you press the switch you power this relay so the door opens. I tried this circuit in several places and there are buildings with AC powering this relay or more voltage so the current I can draw is enough, but other places have to low current capability (only 15mA before relay activates). The max current through the relay is 70mA when SW1 is pressed.

Edit2 (reply to Pipe and more info): Yes, it is Energy Harvesting situation, but I could not found the best way to solve it to works in any different scenario. I've tried supercapacitor approach, charger connected directly to the harvest source, and now I get the best results with buck+charger

Edit3: My circuit draws an average 48mA (connected to Vbat) and the current that I get now from the relay is 10mA at 9V-7V, but the efficiency of the buck and charger is not too high at low currents so I'm charging to 10mA at Vbat (whith 30mA or more the total efficiency is more than 60%), the battery acts like UPS and helps to handle current peaks

Edit4: I've also checked some MPPT (Maximum Power Point Tracker) way to approach this, but I did not see something valid

• A schematic is better than words. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar and "Save and Insert" on the editor an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. Commented May 9, 2020 at 20:45
• Essentially you're looking at an energy harvesting situation here, which may help figuring out what you need to do.
– pipe
Commented May 9, 2020 at 22:07
• The schematic is a huge help. You haven't told us what your power requirements are - average and peak and peak duration, if they're relevant. Commented May 9, 2020 at 22:11
• The inductance of the relay will change when the armature is closed. See my answer to electronics.stackexchange.com/a/487353/73158. Could you sense that? It might be easier to stop drawing current for a second every 20 s or so. Commented May 9, 2020 at 22:55
• hm, how are chances you can charge a supercapacitor really quickly when the button is pressed and live off that? Commented May 9, 2020 at 23:02