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I have a camera for which I bought a dummy battery to connect it with power bank or other power source. Original battery is 3.6 V Li-ion so I need around 4.3 V but the USB (from powerbanks) is 5.1 V usually.

What I can do to step down to 4.3 V without loosing much energy in the middle to not discharge battery quickly? I found the buck module MP1584 but in the datasheet there is Vin-Vout > 1.8 V so it won't fit right? Is there anything else I can do, or can I use a 5 V without destroying my camera?

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    \$\begingroup\$ Battery charger ic would do great. \$\endgroup\$ – Gregory Kornblum Mar 22 '18 at 20:31
  • \$\begingroup\$ what battery charger IC? I want to connect it in the battery tray with dummy battery, not to the charging port. AFAIK the battery connects with different pins to the charger and to the camera so this voltage I apply will go directly to the camera not through any charger ic. \$\endgroup\$ – Mordimer Madderdin Mar 22 '18 at 20:48
  • \$\begingroup\$ Here, take a look: m.linear.com/product/LTC4054-4.2 \$\endgroup\$ – Gregory Kornblum Mar 22 '18 at 20:49
  • \$\begingroup\$ What current do you need? Also what is the camera? It may be 5 and a bit volt tolerant but without details it will be difficult to advise. \$\endgroup\$ – Warren Hill Mar 22 '18 at 20:50
  • \$\begingroup\$ Oh I see what do you mean, to put battery charger IC between dummy battery and powerbank. \$\endgroup\$ – Mordimer Madderdin Mar 22 '18 at 20:51
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The "1.8V Vin-Vout" spec is called your dropout voltage. There are linear regulators with low dropout voltages that are just called LDOs for short. So look for a nice LDO. A quick look shows that: MODEL 33115 surfboard (digikey: 33115CA-ND) would hold a TO-263-5 footprint part. A TL1963 adjustable LDO (digikey: 296-24759-2-ND) would fit on that board. With two resistors you'd have a quick-and-dirty 5V to 4.45V LDO. Add a cap or two for "less dirty". Cost ~$5 plus shipping. The thermals of the surfboard might not get you to 1A. But a switcher won't help much since the LDO is already 90% efficient.

Ebay is full of pre-made LDO boards but they are all 3.3V/5V fixed output and 1.2V dropout which won't help you.

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  • \$\begingroup\$ Ebay is full of many things. It is a parts broker for discontinued or EOL parts, so normally we shy away from them. Main reason is a hiked cost and lack of documentation. Buy at your own risk. \$\endgroup\$ – Sparky256 Mar 23 '18 at 3:28
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Assuming that the powerbank voltage has a regulated output at 5.1V, then a series diode would do it, probably giving you 3.9-4.1V under load. Using a bipolar transistor as a diode (join B+C) is less spongy giving you closer to 0.7V drop under load.

Just check the no-load voltage in case your load goes to really zero, and leakage reduces the diode drop.

Alternatively just open your powerbank and either change the regulator feedback to get the voltage you want, or if it is a single cell power bank, just bring the battery out directly.

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