0-2v variable power supply

[Tony27]

Hi,
I am looking for a circuit for a small variable power supply. It is required to test watches. The specs are 0-2V DC, current is very small say max of 100ma. Input would be a 9v batt. I am not sure where to start.

Anthony

[digital_silence]

I've hand-sketched the simple cct of variable linear regulator that will fit the bill - please excuse the quality of the drawing...

Note that the efficiency of such regulator is going to be less than 30%, so it will drain your 9V battery fairly quickly.

If you want the more complicated solution, like say DC/DC converter, in order to increase the efficiency, please indicate that here.

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Addition: Just realised looking at my cct that, if the current drawn by the load is ~100mA, then the transistor power dissipation is going to be just under 1W, so BC547 may not be up to the task... in which case, use any NPN bipolar transistor rated appropriately (just ask the sales guys at Dick Smith or Jaycar, or wherever you will shop), and MOUNT it on the heatsink.

BC547 should be OK if the current consumption is limited to about 40 mA.

[Tony27]

Thanks very much for that.

The actual current drawn would typically be ~less that 20micro amps so the efficiency isn't a problem.

What about short circuit protection, I assume that the transistor would zap. Is there a simple way to protect it.

[digital_silence]

First, I think there is a confusion about the units of current used in this topic.

You mentioned that "current is very small say max of 100ma"

Did you mean 100 microamps or 100 milliamps?

Normally, mA or ma refers to "milliamps", and uA or ua refers to "microamps".

If all you are using is 100 microamps max, then you will be fine with BC547.

Regarding the short-cct protection, the easiest way to do it for this circuit is to insert a 150R (one hundred and fifty Ohms) resistor to the upstream side of the transistor, that is between its Collector and the incoming 9V rail.
It will limit the short-cct current to approx. 60 milliamps and prevent the transistor from damage.
The resistor should be rated at least 2W to be able to dissipate the short-cct power.

If you want to be a bit more sophisticated, you can also add the red LED in series with that current-limiting resistor (LED's Anode (+) is on the left, Cathode (-) is on the right), and it will light up when the short-cct event occurs in the load.

[Tony27]

I did initially wrongly say 100mA but in reality it only needs to provide ~20uA.

I think what you have proposed will be good. I will add a 150R for the short cct protection. (why 2w, isn't the power v2/r = 81/150 =.54w?)

Off to try and find a 10k linear pot with a switch, Jaycar don't sell switched pots anymore.

Thank you.

[gordonwh40]

Hi Tony
Why are you going to the expense of a switched pot? You do not need a switch in the circuit, all you need to do is remove the battery when you have finished working. If you want the convenience of a switch a much cheaper option would be a spst swich if you want to switch the supply, and much easier to get.
gordonwh40

[digital_silence]

why 2w, isn't the power v2/r = 81/150 =.54w?)

...just a "silly" habit to always add a big headroom to the resistor power rating, because if the 1W rated resistor dissipates 1W during operation, it will get hot.

In your case, I agree, you should be OK with using just a 1-watter, or even 1/2-watter, because short-cct is not a normal operation mode... but I prefer to be safe than...
If sourcing power resistors is an issue, you can always combine the lower rated ones - use two 75R resistors with only half of the required power rating in series, or 2 x 300R in parallel, or 4 x 620R in parallel, and so on... you got the idea I hope...

[Tony27]

I just finished and it works like a beauty - perfect. Replaced the 33k with a 50k trim pot set to ~20k to get the max 2v out. 33k was a bit high. Thanks again for the help.


This is a great forum - I was a member in 2002 under different ownership. Glad I joined up again.

[digital_silence]

With no load there is a 1mA draw on the 9v battery

Hmm... it shouldn't be that much without the load...
should only be 9V/(50K+20K)= ~120uA

You don't have a polarity reversed for either (or both) of the electrolytic capacitors, do you? Please check that the polarity is exactly as shown on the schematics.
Even if the polarity is correct, the electrolytic caps can leak a bit - if you really want to get to the bottom of this, disconnect both caps and measure the standby current again.

What is the voltage on the collector of the transistor, referenced to the ground (negative battery terminal) ?

Also, check the accuracy of your current meter... Cheap multimeters give you atrociously bad accuracy in submilliamp ranges... The easiest way to calibrate your meter is to connect the battery across 50K resistor via your current meter - the reading should be 9V/50K = 180 uA.

do I even need a switch compared to shelf life of the battery?

It's entirely up to you - the typical capacity of alkaline 9V battery is approx 500mAH, so if you can afford changing the battery every 500 hours of operation... :-)
The good news is that you will use the full capacity of the battery, as the circuit will operate even with the battery voltage down to 5-6V...

Good luck.

[Tony27]

I have a 0-2v panel meter across the op, without it the no-load current is ~280uA which is in the region of 8.1v/20k+10k (20k of 50k trimpot and 10k pot)

Thanks again for your help on this.