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I made a simple 9v battery nimh and nicd charger according to this Circuit

I modified it after it and i use something like the image Simple circuit with Led

I need 16mAh per hour to charge a 160mAh 9v Battery and i manage to have an led as the image shows in it, but as i used my multimeter to measure the amps as the A sign indicates it starts charging at 16mAh and it drops down by 0.1 every some seconds in the beggining and as the time goes by it slowly again drops by minutes, after couple of minutes the 16mAh went down to 15mAh. I can only imagine how the mAh will drop after an hour cause in order for my 9v battery to be fully charged it needs about 10 hours.

Is this normal or am i doing anything wrong here in the image circuit ?

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  • \$\begingroup\$ There is a difference between mAh and mA. Your multimeter is very unlikely to measure mAh. They typically measure mA. The differencebis the time factor. One measuress current (mA= milli Amperes) while the other measures current multiplied by time (mAh= milliAmpere hours.) Multiply mA by Volts and you get power (VA=W VoltsAmperes gives Watts.) Multiply mAh by Volts and you get energy (VAT=J VoltsAmperesTime gives Joules.). Very different things. \$\endgroup\$ – JRE Jun 7 '17 at 4:48
  • \$\begingroup\$ @JRE i have a small fog around this measure cause its below 200mAh i use my multimeter at the classic possition of measuring volts and amperes (not the undfused) and i set the position to 200mAh to read it, it reads as i want 16mAh but when i switch it to the position of 20mAh it reads about 4.43 and it is not so stable. Is the 200mAh position correct in multimeter to use ? \$\endgroup\$ – Nocs Jun 7 '17 at 11:28
  • \$\begingroup\$ Your meter doesn't have a mAh setting. It has a mA setting or an A setting. Notice the ever lovin' h at the end of your units, and note I don't include it. As to which scale to use, you must use a scale with a maximum value higher than what you expect to measure. Either the true value of the current is out of range for the 20mA scale, or that scale is broken - or something else entirely. \$\endgroup\$ – JRE Jun 7 '17 at 12:37
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First, D2 and R2 are completely unnecessary, and are not mentioned in your link, so why you have them is a puzzle. However, since they have almost no effect on your circuit, they aren't actually hurting anything.

Second, your power supply voltage is too low. I recommend a supply at least 10 volts over your battery voltage, and more is better. The reason will become apparent.

Now, as to your circuit performance. I suggest that you get a second meter, and monitor the battery voltage as you charge. You will see that, during charge, the battery voltage will slowly increase. Since the charge current depends on the difference in voltage between the supply and the battery, charge current will drop as the battery charges. This is made worse by the inclusion of the LED, which will typically drop about 1.5 volts at your current. So the voltage across the charge resistor (R1) will start out at about 1.5 volts for a battery voltage of 9 volts. An increase of battery voltage of only 0.1 volt will cause the resistor voltage to drop to about 1.4 volts, which will cause a change in current from about 15 mA (for a 1k resistor) to 14 mA, and the trend will continue as long as the battery voltage rises with SOC.

So the effect is perfectly normal, given the simple circuit you are using.

Unless you can come up with a good reason for them, get rid of D2 and R2 and recalculate R1. Increasing the supply voltage will decrease the change in charge current. For instance, if you increase the supply voltage to a nominal 20.5 volts, a 9 volt battery and 1.5 volts across the LED will give you 10 volts across R1. If the battery voltage rises 0.1 volts, the resistor voltage will again drop by 0.1 volts, but this is only 1% of the original, and charge current will only drop by 1%.

But also keep in mind that your charging technique is not precise. Using exactly C/10 for a charge current, for exactly 10 hours, will not produce exactly 100% charge, and using a little more or a little less current will not have much effect on the final result.

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  • \$\begingroup\$ Thank you for your good explanation and detail. I use D1 and R2 cause i need to use a led when the battery is connecting to the circuit, its the only circuit i could find for charging and having a led indicator i use it in other as well, here is a link electronics.stackexchange.com/questions/293231/… \$\endgroup\$ – Nocs Jun 6 '17 at 22:53
  • \$\begingroup\$ @Nocs - The linked circuit is a very, very bad circuit, and there is nothing in the link which explains the resistor/diode combinations, and the linked page within the link doesn't even use such a circuit. So where you got this from remains a mystery, and there is no reason to use it. That diode/resistor will have almost no effect on the LED, so your explanation makes no sense. \$\endgroup\$ – WhatRoughBeast Jun 6 '17 at 23:00
  • \$\begingroup\$ then how could i use a led indicating that the battery is connected to the circuit ? do you know any link i could learn better with practical circuits ? i have this circuit d1-r2-led from a usb charger for 2 aa 1.2v batteries \$\endgroup\$ – Nocs Jun 6 '17 at 23:02
  • \$\begingroup\$ @Nocs - For the third time - just get rid of them. The LED will work fine. The circuit you linked to in your first comment is crap. It is also not a USB charger, and it handles 4 batteries rather than 2, so it can't be the circuit you are talking about. I am not going to advise you on circuits which only you can see. \$\endgroup\$ – WhatRoughBeast Jun 7 '17 at 0:32
  • \$\begingroup\$ @RoughBeast yeap you are correct, the led works fine without the D1 and R2, i really dont know what they do and its better without them as it seems since i have to calculate only R1 instead of combining 2 resistors. Thanks Beast. \$\endgroup\$ – Nocs Jun 7 '17 at 11:26

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