I love light up gizmos, and what better, and potentially more useful, than a voltage readout on the panel?
Component Shop sell something suitable, but I went down the eBay route as I wanted one with bezel and ordering from weird people was part of the fun of this project. How you can have a business that sells components so cheaply, including minimal or no postage, and make a profit is a mystery.
Anyway, the meter needs a feed from the track output that is the correct polarity, so it's fed through a tiny bridge rectifier.
You're supposed to solder these to a circuit board, or some veroboard. Mine floats in mid air, although I have used some heat shrink on many of the connections to both support them, and avoid short circuits if the wires move around.
Track power goes in through the inputs that would normally take the AC power, and it ensures the outputs are always the correct polarity.
The good news is, the panel lights up and changes value as you twiddle the controller knob.
The bad news is that according to my old dial multimeter, it's reading 2V too low. Are those volts being lost in the rectifier? I don't know, but I'm not losing much sleep over this, the display is mostly for show anyway.
You'll notice that I've added a couple of LEDs which hadily tell you when you've reversed the polarity to the track, in case you didn't notice you'd flipped the switch. They are petty though.
Talking of switches, I've added another one...
I seem to recall that there is a power drop across a diode of 1.4V. A bridge rectifier will be in the same ballpark I would imagine.
ReplyDeleteA silicon diode (as used in a bridge rectifier) has a forwards voltage drop of about 0.7V, the exact figure will vary depending on load current and temperature. A bridge rectifier will have two of those conducting at any one time. You'd need to reference the datasheet for the one you've selected for precise information, but an example figure for a similar part gives the maximum voltage drop per diode at 1.5A to be 1.1V (thus 2.2V total). You're only powering a low power meter, so I'd expect it to be closer 0.7V per diode (thus about 1.5V overall) - but specs vary.
ReplyDeleteTo improve the situation you could make up a bridge of your own using four Schottky diodes, these have a lower voltage drop of nearer 0.4V per diode (but do have greater leakage current, which wouldn't really bother you here). The alternative would be to wire up four field effect transistors to act as 'ideal diodes', this would change a fairly high voltage drop into a fairly low resistance. Texas Instruments have a datasheet covering the principles. You can buy such transistors already configured as ideal diodes for a premium, but I'm guessing you'd prefer through-hole components rather than surface mount which radically reduces the options. A series of four FETs and some resistors wired manually would still be in the sub-two pound region (given you can pick up N-channel FETs for around 30p each).
A good starting point would be to check the voltage at the terminals of your panel meter with a separate voltmeter (ie. multimeter), this will show what the panel meter is measuring after the bridge has dropped a volt or so.
Phil:
ReplyDeleteWill the voltmeter give a negative reading if powered backwards? You could then dispense with the diodes.
Some people used the voltage drop over the diodes to power LEDs which needed 1.5V and would then give directional lighting.
Sadly, the voltmeter just goes blank with a reversed voltage.
ReplyDelete