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I am currently building a wireless charger. I have difficulty in understanding the working of the regulator (7805) to which I connect the load and measure my output power.

My power receiver circuit schematic is like this LC circuit(receives power from transmitter circuit) -> Rectifier(converts AC to DC) -> Regulator(7805)(for stable 5V power supply) [connected to resistor load]

I have two questions to understand on the regulator:

  1. When I disconnect my regulator(leave the circuit open) I see my DC output like 30-40V, when I close the circuit (connect my regulator) I see voltage between 7 - 20V . Does the resistor (load) connected to regulator have any impact on this decrease in voltage? Is there any relation relating the load resistance and input voltage? Or any relation for impedance matching between them?

Note: I am keen on the relation between input to regulator and load resistor

  1. To the resistor(load) to receive maximum power are there any design calculations to be done for regulator?

Please help! I am new to power electronics, I have difficulty in understanding these concepts

Thanks

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closed as unclear what you're asking by Olin Lathrop, PeterJ, Daniel Grillo, Dave Tweed Jan 12 '16 at 17:25

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    \$\begingroup\$ We need to know more details on your setup. The load. The transmitter and receiver schematics and setup. \$\endgroup\$ – Passerby Dec 28 '15 at 9:50
  • \$\begingroup\$ Thanks. My setup is based on standard Wireless charging circuits, Transmitter works fine. My problem is only with receiver (can you specify which block needs more explanation). When I do not use a regulator and direclty connect my load(resistor) to rectifier I see a performance and when I use the regulator I see a different performance. That is why I want to know the impact of regulator when connected as a bridge to load. \$\endgroup\$ – hacker194 Dec 28 '15 at 10:00
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    \$\begingroup\$ We don't know which "standard wireless charging circuit" you are using. \$\endgroup\$ – Passerby Dec 28 '15 at 10:04
  • \$\begingroup\$ It is based on Qi standard. The transmitter schematic is PWM cotroller (to generate signal of desired frequency) -> H-bridge (to generate alternating current in the tx.coil) -> LC circuit ( to generate power tx. signals). If further design details are needed please check this link, I tried to follow the same sites.google.com/site/ddmcintosh2projects/inductive-charger \$\endgroup\$ – hacker194 Dec 28 '15 at 10:10
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    \$\begingroup\$ Dave Jones has blogged on a teardown of a Braun toothbrush. He raised interesting points such as what you're asking for. \$\endgroup\$ – user59864 Dec 28 '15 at 10:27
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With a wireless charger circuit (ignoring the 7805 voltage regulator part) your receiver coil is trying to pick up the alternating magnetic field from the transmitter. It can have a good coupling or a less-good coupling as per the picture in one of the links: -

enter image description here

On the right with the receive coil further away it receives less magnetic flux. So if you looked at the voltage on your receive coil you would see an AC voltage that got bigger as you approached the transmit coil.

You could do a different test that has the transmit and receive coil fixed at a certain distance. That test would be to load the receive coil with a resistor and you would find that as you try and draw power from the receive coil, the receive coil voltage would fall.

Hanging a diode bridge and smoothing capacitor on the output makes very little difference except you'd be looking at a dc voltage instead of an ac voltage.

So that's the backdrop and of course the 7805 regulator and its output load want to consume a certain amount of power - that power is determined by the load connected to the output of the 7805 and the 7805 output voltage (5V).

When I disconnect my regulator(leave the circuit open) I see my DC output like 30-40V, when I close the circuit (connect my regulator) I see voltage between 7 - 20V

Hopefully that should be clear now from the explanation above.

Does the resistor (load) connected to regulator have any impact on this decrease in voltage?

On its own the 7805 will consume a few tens of milli watts but when the load resistor is connected to the 7805, that power will increase because it's using power to charge a battery (battery voltage x charging current). This might mean 5V at 1A = 5 watts but there were no details in the question about this.

Is there any relation relating the load resistance and input voltage? Or any relation for impedance matching between them?

Yes, load resistance changing will affect the dc voltage level because even when the receive and transmit coils are adjacent there will be an imperfect leaky transformer coupling and voltage will lower when more watts are required to charge your battery.

For impedance matching, you could make an argument for developing a circuit that tried to optimize the max power output from the coils. This would toss-away the 7805 voltage regulator and replace it with a buck-boost regulator and power monitoring circuit (the same as what is used in sophisticated solar panel charging circuits to optimze the power transfer based on amount of sunlight). But this is way too complex for a simple non-contact low power charging circuit.

To the resistor(load) to receive maximum power are there any design calculations to be done for regulator?

I've designed systems like this to provide magnetically coupled power to electronics on rotating machines and the calculations rapidly become too cumbersome so I simulate the leaky-transformer effect of the transmit and receive coils. If a working prototype isn't good enough (say for transferring a couple of watts at 40mm gap I rethink, rewind and retune.

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  • \$\begingroup\$ Thanks for the explanation, I am more interested in the comments you made on power loss across regulator and load. When I did a search on regulators I found that they contain zener diodes which are connected to the base of emitter followers which give stable 5V when input voltage is within the specified range of diode (reference link is attached below). Does the same apply for 78xx series regulators? If yes then is there any relation relating input voltage to regulator and load resistor. Reference link for regulator: ko4bb.com/e102/e102-4.php \$\endgroup\$ – hacker194 Dec 29 '15 at 2:32
  • \$\begingroup\$ What is the simulation software you used for simulation of coupled circuits? Basically in my circuit, I see the alternating voltage at my receiver with Peak to peak of 96V (when open circuit) and a DC voltage of 60V (after rectifier[without regulator]). Here we have a relation that DC output of rectifier is 0.637*Vpeak-peak. Now when I connect my regulator with load I observe this effect when load is 55ohms my regulator is giving 5V output but when I reduce it 30ohms it is giving only 3V so I want to know how the load is affecting the input voltage? \$\endgroup\$ – hacker194 Dec 29 '15 at 2:33
  • \$\begingroup\$ For my setup I do not connect any battery (hence not mentioned about it) I just want know what is the least resistance(load) that is supported by the system, so that maximum current is drawn by load and 5V output is obtained. \$\endgroup\$ – hacker194 Dec 29 '15 at 2:34
  • \$\begingroup\$ You can use LTSpice (free) - it can simulate coupled coils and bridge rectifiers and smooting capacitors. The 7805 is a linear regulator - it has a voltage reference (like a zener), control/measurement system and transistors that attempt to keep the output at 5V but, if the input drops below 7V this becomes unobtainable and the output inevitably drops from 5V. Max input voltage is about 35V from memory. The electronics 102 link is far too basic to gain much knowledge for a 7805 but pls don't ask me for one because I've used them for a million years. \$\endgroup\$ – Andy aka Dec 29 '15 at 10:25
  • \$\begingroup\$ 96V p-p is 48Vpeak and after rectification (assuming a sinewave) the dc level will be \$\sqrt2\$ higher at about 68VDC. But maybe there is some flattening on the sinewave. You need to research bridge rectifiers to learn more about this - frequency doesn't matter (i.e. you can look at examples that run at 50/60Hz for the same general principle. \$\endgroup\$ – Andy aka Dec 29 '15 at 10:27
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You should re-phrase your question as it mentions "wireless charging" yet your question is about how to use a standard 7805 regulator !

With your current question you get comments about the wireless charging but nothing related to that which you do not understand yet (how to use a 7805).

Before talking about anything related to wireless charging you must first understand how to use a 7805 as that is very basic knowledge. In electronics you must split everything in small problems and solve these separately, from simplest to more complex. Using a 7805 is very simple so focus on that now, deal with wireless charging later. Rome also was not build in one day ;-)

Many questions have been asked about how to use a linear regulator like the 7805 and there also plenty of good answers. I suggest you read these first.

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  • \$\begingroup\$ You sure? Cause I understood the real question to be the transformer characteristics. \$\endgroup\$ – Passerby Dec 28 '15 at 11:16
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    \$\begingroup\$ Uhm, it says: "I have two questions to understand on the regulator:" OK maybe in the comments suddenly the question takes a whole different direction because he thought his problem was with the regulator while actually it was something different. Probably a consequence of not having a clue what he's doing ... ;-( \$\endgroup\$ – Bimpelrekkie Dec 28 '15 at 12:53

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