Zener diode, is this right? Pictures attatched

[SammyG14]

Hey guys,

I have been experimenting with a zener diode circuit all day and I still can't make much sense of it. Mathematically it does not add up if If I am doing my maths right based on zener diode voltage regulator google results. The circuit is from a USB supply, dropping the voltage down to 3.6V (3.7V which is fine) and allowing 200ma to go through the circuit to wear some batteries will be connected for reverse charging at 1/10th of there capacity.

I have used the followingg configuration for the pictures. The zener diode is 3.6V rated at 1.3W. The resistor I have used is 25ohms.

+'ve (5V) --------------------[25ohm resistor]------------------------------------->
|
[zener diode: 3.7V @ 50ma] (Voltmeter reads 3.69V at 190ma from these points)
|
-'ve ----------------------------------------------------------------------------------->


I don't understand why a 25ohm resistor seems to be giving me what I want. According to zener calculators and my own calculations it should be a 5ohm resistor, but when I use this the resistor just gets really really hot and the zener does not operate properly. Here are some pictures, btw the black wire from the USB is the positive pin:

CURRENT ACORSS WHERE BATTERY WOULD BE CONNECTED

USB VOLTAGE

VOLTAGE WHERE BATTERY WOULD CONNECT (ZENER VOLTAGE)

THE CIRCUIT


Thanks everyone, If someone could explain to me how this is all working I would appreciate it, really keen to understand this properly.

 

[Grossel]

Too much current through the zener will kill it. Also a zener will have higher voltage drop when conducting higher current.

 

[geko]

In your first photo you are bypassing the zener diode since you have the meter connected between the resistor and ground

You measure voltage across devices, current through them.

Disconnect the ground end of the zener diode and then connect the meter between ground and zener diode so you are measuring the current 'through the zener diode and resistor.

 

[SammyG14]

In your first photo you are bypassing the zener diode since you have the meter connected between the resistor and ground

You measure voltage across devices, current through them.

Disconnect the ground end of the zener diode and then connect the meter between ground and zener diode so you are measuring the current 'through the zener diode and resistor.

Thanks for the reply. Measuring that in that way, the zener is drawing 50ma. However, what I am trying to determine is the current I can gain from the two points which I measured in the pictures which can then by connected to a battery. **broken link removed**

I have found this website the most useful throughout the day. From what I understand form that the resistor I have in series sets the maximum current the diode will take. so (5-3.6)/25ohms = 56ma.

That website also leads me to believe that this has an impact on the load current? so the voltage now running across the battery is 3.6V because of the zener. So 3.6/25ohms = 144ma. Hmm this might be starting to make sense now, should I think of it like this. The resistor sets the current for the diode, settings its voltage which should be around 3.6V unless its max current is exceeded. The same resistor then impacts the load by simply dividing the load voltage, 3.6V (because of the zener) by the resistor, giving 144ma to the battery. Is this right? Sorry for rambling on, just trying to understand! I think the thing which is confusing me is that website talks about a load current/resistance which effects the current running through the zener diode.

So if I want exactly 200ma to get to the batteries I need ab 18ohm resistor, giving the zener 70ma??

thanks so much.

 

[audioguru]

In your first photo, the meter is set for measuring mA and it is shorting the zener diode. If the voltage source is 5.0V and if the meter is 1 ohm then the current is 5V/(25 ohms + 1 ohm)= 192.3mA. Yours shows 192.4mA. The resistor is being fried with 192.4mA squared x 25 ohms= 0.925W but it is too small to dissipate such a high amount of heat. Its maximum allowed heating is probably only 0.25W.

Why are you overcharging the battery cell with a voltage way too high for it? The voltage should be limited to about 1.4V to 1.5V (not 3.69V) and the current should be limited to about 245mA. The charging must be monitored by a circuit that detects a full charge then disconnects the charger.

Basically, your zener diode does nothing when the battery cell is charging and is overcharging. The battery voltage might rise to 1.4V then its charging or overcharging current will be (5V - 1.4V)/25 ohms= 144mA. You measured 143.5mA. A Ni-MH cell must never be overcharged, a fully charged cell must have the charger disconnected.

 

[SammyG14]

In your first photo, the meter is set for measuring mA and it is shorting the zener diode. If the voltage source is 5.0V and if the meter is 1 ohm then the current is 5V/(25 ohms + 1 ohm)= 192.3mA. Yours shows 192.4mA. The resistor is being fried with 192.4mA squared x 25 ohms= 0.925W but it is too small to dissipate such a high amount of heat. Its maximum allowed heating is probably only 0.25W.

Why are you overcharging the battery cell with a voltage way too high for it? The voltage should be limited to about 1.4V to 1.5V (not 3.69V) and the current should be limited to about 245mA. The charging must be monitored by a circuit that detects a full charge then disconnects the charger.

Basically, your zener diode does nothing when the battery cell is charging and is overcharging. The battery voltage might rise to 1.4V then its current will be (5V - 1.4V)/25 ohms= 144mA. You measured 143.5mA. A Ni-MH cell must never be overcharged, a fully charged cell must have the charger disconnected.

To clear a few things up which I should have said, I just used 1 battery as an example. It will in fact be charging 3 x AA batteries wired in series which = 3.6V. The current should be limited to 1/10th of the battery capacity which in this case is 200ma. The batteries are 1.2V rechargeable batteries with a capacity of 2000mah, Nimh. Therefore the charging does not need to be monitored as it is below 1/10th and any over charging will be dealt with the battery via gas recycling which is built into most Nimh batteries these days. this is only effective as long as you don't exceed 1/10th of the capacity. Therefore a circuit is not required for the charging at the moment, although I do eventually intend to add one of these to the project when I start looking at charging the batteries higher than 200ma, reducing the charge time from 10 hours to just 2 or 3 hours.

So with this in mind, 3 Nimh batteries in series = 3.6V, the zener has an important job.... I think???

 

[JimB]

the zener has an important job.... I think???

The zener is irrelevant and not necessary. Think again.

JimB

 

[SammyG14]

The zener is irrelevant and not necessary. Think again.

JimB

This has just made it click for me! The zener was previously used to directly power the device as well from USB.

So for just charging you would not need the zener, you simply step the current down to what ever you need your charge current to be using a resistor.

If I was still powering the device directly from USB at the same time, the zener stepped the voltage down to 3.6V for the other components on the PCB (not in photograph).

Thanks all!

 

[audioguru]

A Ni-MH battery cell charges to 1.4V. Each cell is about 1.2V when it is about half charged.Then three need 4.2V. The 3.6V zener diode will prevent a full charge.

Energizer and a Japanese battery manufacturer say that the trickle charge current for a Ni-MH cell should be its mAh rating divided by 40 or more so for your 2450mAh cell it is 61.3mA or less. The 25 ohm resistor will have about 6V - 4.2V)/25 ohms= 72mA which is too high. Use older Ni-Cad cells if you want a trickle charge at 10% of its mAh current but a Ni-Cad cell has less capacity so the trickle charge current will be the same.

You will not save much power by using an LM317 regulator to reduce the voltage because it wastes power getting hot.

 

[SammyG14]

Wow I did not know this ^^^ thanks for the information. Never got taught that at University, I think they just kept it simple and used the nominal voltage of 1.2V by the looks of it.

If this is the case then I might as well start looking at smart charger IC's now. For now, I might as well use one normal diode which will cut the voltage to 4.3V and stop any reverse leakage going back down the USB. This will mean the device itself will also self power off 4.3V but this should not effect anything in combination with the capped 500ma current from the USB. The main IC on the board can handle anything up to 5V so this should be fine. Going to start looking at charging IC's, I think I need to now with this extra information.

Thanks again!