What value of resistors do I need and a sanity check

This schematic has given me a splitting headache. It has obvious numerous errors. Please supply the link to the original pdf schematic.

 

I only read halfway through your post to preserve my sanity. It would be helpful to just tell us what you're hoping to achieve.

 

This seems to be a constant current power supply with switched output range.

The leds will need a voltage to work which will completely ruin the supply.
If you want an indication of what you are doing, use a switch with another pole to select a led independent of the control circuit.

Are your leds the wrong way round?

 

Well, I consumed quite a bit of suds. Then went and had some great Greek 'Tuck',.. I hope I got that right????.... Anyway, washed that down with Greek wine, came home and looked over the newly posted schematic.... Anyone have an Excedrin? I know this is going to sound trivial but besides the reversed LED my old eyes can't focus on schematics with negative on top. Yes, I know it's lame but I can't help it!

Has anyone asked what this current source is going to be used for? If not, I'm asking.

Chris

 

To summarise the comments so far and add some of my own.

1. Your schematic, and the original schematic, are drawn in a confusing way. It is normal to put the main negative rail at the bottom of the schematic (unless it uses all PNP transistors, but that's another story). That means the negative terminal of the battery should be at the bottom, and the battery should be at the left side, with the output at the right, and everything else in between.

2. The LED connected in series with R5 is around the wrong way. Also, it's marked as high-efficiency with a 4K limiting resistor. That's OK, but high-efficiency LEDs are more expensive than standard LEDs and using a high-efficiency LED at low current is only important if you're trying to minimise the load current on the battery. In any case the anode should be towards the positive side of the battery.

3. You can't put LEDs in series with the current setting resistors. You need an exact resistance between the -V and Rset pins of the LM334Z to set the current. Putting LEDs in series will totally screw up the current regulation.

The only option I can see is to use a double-pole rotary switch, as duke37 has already suggested. You can replace the power indicator LED with five LEDs, one for each operating mode, and switch the supply to one of the five LEDs using a second pole on the rotary switch.

4. I also suggest using a 5-pole rotary switch instead of a 4-pole rotary switch and an SPDT switch. It's just tidier.

5. For your adjustable current option you need your resistance between -V and Rset to be adjustable down to about 25.6 ohms (for 2.5 mA). That means you should use a potentiometer in series with a 25.6 ohm resistor (approx) so that even at the minimum resistance end of the potentiometer, there is 25.6 ohms between V- and Rset.

You can't get down to 0 mA out of the LM334. If you aim for a minimum output current of 0.2 mA, for example, you need the resistance to be adjustable up to 320 ohms, so you would use a 300 ohm potentiometer. The closest available value would probably be 500 ohms.

All of these values are calculated from Ohm's law, using V = 0.064 (the nominal reference voltage of the LM334 at 25 degrees Celsius). For example to get 2.5 mA, use R = V / I where V=0.064 and I=0.0025. The answer is 25.6 ohms.

Ideally you want a smooth even adjustment of output current as you rotate the potentiometer, but no type of potentiometer that's available will give you that - no matter which "law" you choose, logarithmic or linear or something else, the current markings will not be evenly spaced. A linear pot might be your best bet, but be aware that the markings won't be spaced evenly.

6. Monitoring the output current is a good idea but you need to connect your current meter in series with the output. That is, between the V- terminal of the LM334 and the left side of the output fuse. Also the extra diode you've added from LED5 to the fuse is not needed and not appropriate.

7. The LM334 is not just a current regulator - it is a temperature sensor. The reference voltage, which controls the output current in conjunction with the resistance between V- and Rset, is nominally 64 mV at 25 degrees Celsius, and has a temperature coefficient of nominally 0.214 mV per degree Celsius. That means that at 15 degrees Celsius, for example, the reference voltage will be 2.14 mV lower, which works out to a drop of about 3.3%. So your current source will be temperature-dependent. Just letting you know!

8. The 470 uF capacitor across the output, which according to the original schematic "creates a soft start and stop for the device", will have the opposite effect if the current source is turned on before the load is connected. It is a bad idea generally. That depends on exactly how the current source will be used, and what you will be supplying from it.

9. Please tell us what you intend to use this current source for. The more information you can give, the more useful responses we can give. Without knowing what you intend to use this circuit for, we are mostly stabbing in the dark.

Edit: corrected "Radj" to "Rset" in point 7.

 

jjanes, I don't think you need to worry about being "banned from the forums" for posting a thoughtful, rational and measured comment like that. But I can tell you for sure that TELLING US ABOUT YOUR PROJECT will be a big help towards getting people "on your side".

Edit: actually, your original post does contain a lot of good information. But it's always important to tell us what you're actually trying to achieve! What do you want to use it for?

We regularly see requests for help with absolutely minimal background information, and it is often literally a complete waste of time to respond with any suggestions in these cases.

It's good that you came forward with the original schematic, but we shouldn't have had to ask. Don't be afraid that extra information will overwhelm us, or that your idea will be stolen and patented by someone else!

You're obviously smart enough to learn - you just need some pointers. Read up about voltage and current, and Ohm's law - it's about the most basic and important formula you can know in electronics. Try Wikipedia. Look at the "related" pages. Have a look at the LM334 data sheet. It may not make complete sense but try to understand what you can. Ask questions here; if you show you've made an effort to learn stuff for yourself, people will be much more willing to help you understand something you're stuck on. I hope you continue to learn about electronics - it's a very interesting field!

 

Nice work davenn

I have a few corrections for you to make if you want to.

The centre pin on the LM334Z is called "Rset" not "RESET".

The potentiometer needs a resistor in series with it to limit the maximum current, and needs to be returned to the same place as the other resistors, without the diode. I calculated the series resistor at 25.6 ohms for a maximum current of 2.5 mA at 25 degrees Celsius.

The voltmeter should be replaced with a current meter connected in series with the output. It's not practical to make the meter measure the potentiometer setting; the only reasonable way to use it is to measure the actual output current.

The 470 uF capacitor should be removed. It won't help performance and could be unsafe - the load is a person's head; if an electrode becomes disconnected, the capacitor will charge up, and when the electrode is reconnected, a high current will flow until the capacitor discharges.

The "power on" LED could be removed, since one of the other five LEDs will always be ON when the circuit is powered up.

The five individual current limiting resistors for the LEDs could be replaced with a single one.

 

jjanes, I want to clarify something.

Connecting ANY circuit to electrodes attached to your head has certain risks. I and others on this thread have made suggestions in good faith and on the basis of the information you have given us so far. You must accept the final responsibility if you connect this circuit to your own head, or anyone else's. I, everyone else who has advised you, the manufacturers of the components you use, and anyone else who may be involved in any way CANNOT be held responsible for anything that happens as a result of you connecting an electronic device to your head or anyone else's head. You should understand the risks, and you must take responsibility for your decision to do that.

 

The right side of the potentiometer needs to be connected to the same place as the other four resistors, i.e. to the left side of the fuse, and you also need a resistor in series with the potentiometer.

Thanks Dave.