The BAT41 is a Schottky diode which requires a forward voltage to switch on, i.e. 200mV at 25deg C.

It is only rated at 100mA so you might need something that will carry a bit more current. but before buying diodes I would check the data sheets for the solar panels as a lot of them have a built in diode.

The circuit's OK other than BAT41 being a small signal diode and a bit delicate. If the battery is completely flattened by a few dull days, followed by a bright hot dawn, I suppose the solar panels might deliver enough current to pop one. How much current can they produce? I guess they're tiny if the load is a single LED, perhaps a few square centimetres as found in garden ornaments, so might not manage 100mA into a short circuit, in which case a BAT41 would do.

I suspect either a mistake measuring 270k or the diode is broke.

Paralleling diodes isn't a particularly good idea because they're not balanced. One of them might take current before the others wake up, so there's a risk of them popping one after the other rather than sharing the load safely. Easier to upgrade to a bigger device, possibly an SR150, like this example from Farnell. Depends on the current output of the panel, but an SR150 is good for 1A. Plenty of larger schottky rectifiers available if the panels are monsters.

Dave

Hi Glyn,

You can't just measure the resistance of a diode using a meter as diodes are not simple resistances but are non-linear devices.

No component is perfect, it always has resistance, capacitance and inductance. There is the concept of equivalent circuits, where a circuit of perfect components can represent a component.

A diode can be thought of as an ideal diode in series with an ideal resistor. The "low resistance" in the specifications refers to the value of the equivalent perfect resistor.

Diodes need a minimum forward voltage to start conducting, as said above for yours, this is 0.4V. This voltage is nearly constant and is always dropped across the diode.

So as the forward voltage(Vf) of a diode increases the resistance changes. Below Vf the resistance is very high, but as Vf is reached the resistance rapidly drops. But there will always be the diode voltage drop across it plus the voltage drop of the equivalent resistor.

From the specs sheet at 0.1A the Vf across the BAT41 is 0.9V @25C So the equivalent resistor is (0.9-0.4)/0.1 = 5 Ohms

I don't know the power output of your solar cells, but if they can supply more than 0.1A you will likely burn out the diodes. Also if the voltage of the solar cells is more than about 4.8V you will have the chance to overcharge the batteries. The max voltage across the batteries should not be more than about 4.5V ideally about 4V.

Although with the LED load constantly across the batteries will limit the max voltage so you may be OK.

I wouldn't worry about the resistance of the diodes; for a circuit like this it's usually enough to confirm the resistance is high on one direction and low in the other, showing the device is working. Measuring the actual resistance requires an experimental set-up measuring amps whilst the volts are accurately varied over a range. As semiconductors don't obey Ohm's Law, the result is a curve on a graph, not needed in this case.

How well or badly Glyn's circuit behaves depends on the properties of the components:

• The Solar Cells deliver 2W at 5V, that is a maximum of 0.4A each, 0.8A total.
• BAT41 diodes support a maximum of 100mA. Therefore diodes are at least eight times too small to cope with the maximum output of the solar cells, which is a risk
• An AA-size NiMh cell has a capacity of about 1900mAh. Three of them in series (4.6V) will charge reasonably well from a 5V solar cell so that's OK.
• The power requirement of Glyn's flickering yellow LED is unknown, but the small tea-light type run for a claimed 100hours on a CR2032 cell. As the capacity of a CR2032 is about 230mAh, the current drawn by the LED might only be 2.5mA. It would run for about 760 hours on a NiMh AA.

So, if Glyn is using a small LED, and the AA batteries are fully charged at the outset, there's a good chance in normal service that the batteries will never discharge to the point that the solar cells will exceed 100mA output.

All good, unless 'normal service' doesn't happen. If the NiMh batteries go flat in storage, or deteriorate with age, then their resistance will drop massively, causing the Solar Cells to deliver a heavy current, with a high risk of popping a BAT41. (Cells don't obey Ohm's Law either!)

The component values really depend on the current drawn by the flickering LED. In the good old days, LEDs rarely consumed more than a few tens of milliamps, but modern devices can be much more powerful. It would help to know how much current in drawn by Glyn's LED.

From a 'common-sense' perspective the circuit is reasonable, but the operating conditions need to be defined and the sums done. Here common-sense has delivered a solution likely to work, perhaps for years, but – assuming a small LED – there's risk of failure depending on how it's operated. For example, the diodes are likely to fail if the batteries go flat in storage and the unit is suddenly exposed to bright sunlight. Or they might survive if the solar cell output is limited by weak sunshine and the batteries recharge slowly. As always common-sense relies too much for comfort on luck – real engineers work to requirements, do the maths, and set specifications!

Doesn't matter for Glyn's purposes, but – again assuming the LED is small – on the face of it the Solar Panels and Battery are both over-engineered; bigger and more expensive than needed to do the job.

A professional designing a commercial product like this would start from the current requirement of the Lamp, from that decide the capacity of the battery (perhaps assuming long Scottish nights and short winter days rather than Florida), and derive the size of the solar panels needed to charge the battery in the target conditions. The diodes would probably be rated to take the maximum charging current drawn by a completely flat battery from a Solar Panel at peak output.

For cheapness, diodes are fine, but if the device requirement included long-life, then a more or less clever recharger of the type recommended by Bruce Edney would be used. They extend battery life considerably by matching charging volts and amps to the batteries ideal recharge cycle. A requirement for long-life, say the light on a life-jacket, would trigger a bunch of other specifications, such as water-proofing, liable to push the price up considerably. Daft to pay safety-critical equipment prices for a garden ornament, or to expect a garden novelty to last forever!

Dave

Edited By SillyOldDuffer on 11/07/2022 10:53:40