Hello all,

I'll preface that I'm a newb when it comes to electronics, I only know what I want to achieve but don't know how to design my own circuits. I have found an overvolatge protection circuit that works, but only on low amp devices like leds.

I want to make a dummy battery for a camera.

As a main power source I intend on using a phone charger 5V/2A (this could also be a power bank or a PC via USB), connected to MT3608 and I will set the output voltage to 8.3V as that is what the camera battery outputs.

After that I want to have a protection circuit just to be doubly shure the camera doesn't get fried. So the voltage is limited with a zener diode (I'll be using a 7.5V one as the 8.2V could go as high as 8.8V before the circuit reduced the voltage, and I don't know if cameras can take that high of a voltage).

The added LED in the circuit is lit when the voltage is in the operating range that I want. And I could also add a switch so the LED and the circuit turns off when the camera is not in use but still connected to a main power source.

The problem is low amperage at the output as the camera doesn't turn on. With my limited knowledge, I think the issue is/are transistors. I didn't use BC557 as I don't have it, but I replaced it with 2n3906 which should be an equivalent. Looking at the data sheet it seem the collector current is too low for my needs? So I',m thinking I need a transistor with a higher collector current?

I don't have any high amp transistors in my basic transistor kit, so I was looking at TIP42C or something along those lines. Would I be able to just put that in for the Q1 transistor to get the result that I want (would I make a disaster?), or is there a transistor that is better suited for the job that I haven't found yet?

If not, would I have to do something like Darlington/Sziklai pair? In that case I have no clue where and what I would have to replace and add as I just found out about these two transistor configurations.


Also on a side note, I saw some mentions that the zener diode should be swapped with R4 (based on the provided image) as the circuit is otherwise faulty somehow?

 

Welcome to AAC.

A few misconceptions here.

Firstly there is no need to limit the current. The camera will take whatever current it needs, no more, no less - as long as the voltage is within the parameters its expecting. So controlling the voltage to the maximum that the existing battery could provide is all you need to do. An 8.3v battery is 2 Lithium cells in series - max 8.4v fully charged, or it could be 8 NiMH cells - no higher than 9.6v fully charged. So determining the battery chemistry will tell you what is the likely highest voltage. What battery pack is it?

Now I understand you're using a 5v 2A wall-wart and boosting that to 8.3v. Do you know what current the camera needs? The most you'll get from the MT3608 from a 5v 2A input is about 1A (5v x 2A in = 10W, at 80% conversion efficiency is 8W out, at 8v = 1A) If you set the output of the MT3608 to 8.3v when disconnected from the camera it will never go above that on load - in fact your problem might be that it drops too low.

Turning to that 'protection' circuit, the BC557/2N3906 may be unsuitable for the purpose as limited to <200mA, but as you don't know what the camera needs its hard to say.

 

Thank you for the reply Irwing

First let me say that I have tested the MT3608 on it's own and it is not a problem. In the test I have also used the 5V/2A phone charger as the main power source, then I set the voltage output in the ranges of 7.4V to 8.3V and the camera worked. Introducing the extra circuit though makes the camera not power on, or well it powers on for a second and doesn't even properly turn off which is not ok.

When it comes to the original camera battery, it is Canon NB-10L, the specs on it say 7.4V, 920mAh, 6.8Wh(Li-ion). Based on the specs, I think the Canon SX40HS camera should work with 1A? But I would like to see this circuit provide 2A if possible, as I would then be able to use this with other cameras without worry. I do have some bits and bobs so I could probably find the right parts and components to make it work.

 

Not sure what that protection circuit is meant to achieve. A quick simulation suggests it crowbars output volts to around 5v max...



Replacing the 5v zener with a 8.2v one...



Varying the load shows its got a max output of about 160mA. This is not directly related to the actual transistors used but to a combination of variables. The crowbar nature of this circuit makes it really unsuitable for your purpose unless the current drawn by the camera is very constant.

 

Thank you for the reply Irwing

First let me say that I have tested the MT3608 on it's own and it is not a problem. In the test I have also used the 5V/2A phone charger as the main power source, then I set the voltage output in the ranges of 7.4V to 8.3V and the camera worked. Introducing the extra circuit though makes the camera not power on, or well it powers on for a second and doesn't even properly turn off which is not ok.

When it comes to the original camera battery, it is Canon NB-10L, the specs on it say 7.4V, 920mAh, 6.8Wh(Li-ion). Based on the specs, I think the Canon SX40HS camera should work with 1A? But I would like to see this circuit provide 2A if possible, as I would then be able to use this with other cameras without worry. I do have some bits and bobs so I could probably find the right parts and components to make it work.

7,4v is 2 Lithium cells. 8.3 - 8.4v fully charged, they'll drop to 7.4v (2 x 3.7v) almost immediately until 85-90% discharged then they'll drop rapidly to <3v.

The battery is spec'd at 920mAh. This is usually stated at the 20h rate, so 46mA for 20h. At 1A it'll last maybe 50min. At 2A maybe 20min. How long does it last in the camera?

You can't get 2A out of the MT3608 at 5v in without it taking 3.5-4A from the supply. Its not really up to that; well the chip is capable of 4A but not on those tiny boards featured in the cheap modules, it needs a more sophisticated setup or a different module.

 

When it comes to the original camera battery, it is Canon NB-10L, the specs on it say 7.4V, 920mAh, 6.8Wh(Li-ion). Based on the specs, I think the Canon SX40HS camera should work with 1A?

Does the battery last less than an hour? Because if the camera is drawing 1A, that is the longest it could last.

 

Does the battery last less than an hour? Because if the camera is drawing 1A, that is the longest it could last.

The camera might run for 1.5h when recording, it's been a while since I used it.


Thank you for some more insight Irwing.

 

The camera might run for 1.5h when recording, it's been a while since I used it.

so its probably taking no more than 550-600mA

 

Thank you for all the answers up to this point. Since the original circuit is not up to the task I did some more research and came upon a crowbar circuit and resettable fuses - did not knew those exist. I made some changes to the circuitry but since I'm missing a 2 major components I can't make it, yet. And I don't know what (free) software to use to easily build and simulate the circuit.

The circuit in the image was made with flux ai, but most components are not simulatable, so this is where I would like some input, whether the circuit I made whould actually work.

The only thing I'm not clear on is what value capacitors I'd need (I saw 0.47uF for C2 in some examples), the value of the resettable fuse - let's say the circuit will go max up to 2A since that is what the MT3608 is theoretically capable of - and whether D7 is actually 1N4007.

On that note, if the input voltage is 8-8.3 volts, do I still use the 8.2V zener diode, or would I need a higher/lower voltage zener since the thyristor trigger voltage also affects it in some way and that combined voltage then triggers the fuse.

 

A 7.5v zener is preferable. Use LTSpice for simulation, its free, well supported here and elsewhere, and there's not much that you're likely to need that's not got a model!

 

The circuit in the image was made with flux ai

The LED is backwards.
In normal operation the current goes through the LED from anode to cathode.