Learn me something electronical...(Flexmod)

[Bradfo69]

Hopefully last question...

Instructions say:

5) Short the modulation input wires together and power on the driver, you should get a reading on your multi-meter of 50-
150mA. Adjust 0Bias until you reach the desired standby current. Clockwise turns the current up, counterclockwise turns
it down.
6) Un-short the modulation input wires and while maintaining power to the driver apply 5V to the Mod+ input and adjust
gain to the maximum desired current. A clockwise adjustment increases current.

So... taking step 5, does that mean I really should have two leads soldered to "G"? One to run to the (in my case) 12V supply, AND one for the 5V modulation source? As I read it, if I am supposed to short the modulation line"s" (meaning two). That means I should have a lead from the + on the flexmod to the V+ on the 12 volt supply and one from the G on the flexmod to the FG on the 12 v supply. And then... a lead from the M+ on the flexmod and another lead from the G that need to be twisted together on this step to short them out. (Or can I just insert the lead from M+ along with the lead from G into the FG screw terminal on the 12v supply and effectively do the same thing?)

Sorry to sound so dense to people who, for them, this is probably second nature but, I've watched someone who KNOWS electronics inside and out make the flexmod put out a $40 puff of smoke doing this. They're rare as hens teeth unless you corner Andrew in a small space when he has them in his pocket so, I'm just being overly cautious.


Also, for swami... that makes a lot of sense so, to be clear, the "hot side" of the resistor is the one physically closest to the V+ input?

[swamidog]

here's the cheat sheet i use. i think this was courtesy of mixedgas.

Hopefully last question...

Instructions say:

5) Short the modulation input wires together and power on the driver, you should get a reading on your multi-meter of 50-
150mA. Adjust 0Bias until you reach the desired standby current. Clockwise turns the current up, counterclockwise turns
it down.
6) Un-short the modulation input wires and while maintaining power to the driver apply 5V to the Mod+ input and adjust
gain to the maximum desired current. A clockwise adjustment increases current.

So... taking step 5, does that mean I really should have two leads soldered to "G"? One to run to the (in my case) 12V supply, AND one for the 5V modulation source? As I read it, if I am supposed to short the modulation line"s" (meaning two). That means I should have a lead from the + on the flexmod to the V+ on the 12 volt supply and one from the G on the flexmod to the FG on the 12 v supply. And then... a lead from the M+ on the flexmod and another lead from the G that need to be twisted together on this step to short them out. (Or can I just insert the lead from M+ along with the lead from G into the FG screw terminal on the 12v supply and effectively do the same thing?)

Sorry to sound so dense to people who, for them, this is probably second nature but, I've watched someone who KNOWS electronics inside and out make the flexmod put out a $40 puff of smoke doing this. They're rare as hens teeth unless you corner Andrew in a small space when he has them in his pocket so, I'm just being overly cautious.


Also, for swami... that makes a lot of sense so, to be clear, the "hot side" of the resistor is the one physically closest to the V+ input?

[Bradfo69]

Sigh.... that assumes a test resistor dummy load, which is what I thought we were avoiding.

[swamidog]

you can ignore the dummy load.

Sigh.... that assumes a test resistor dummy load, which is what I thought we were avoiding.

[Bradfo69]

See... no second wire available to "short" the modulation wires for step 5. That's what I was referring to.

[swamidog]

short the modulation line (M+) to the shared ground (G).

See... no second wire available to "short" the modulation wires for step 5. That's what I was referring to.

[Bradfo69]

Ok so... I got nothing on the meter. And no needle movement when turning the 0 bias pot. (and yes, it WAS set to the DC 250mA scale, lest you think I'm a total moron.)

The lead for G (and for the first setting, the M+) should be going to the FG (field ground) on the power supply correct? Or... should it be the V-? I had them going to the FG.


I also then disconnected everything and double checked to make sure the 12v power supply worked and was outputting 12v. It is.

[DZ]

Hey Brad, I've never really liked measuring current in line with a meter. Your best bet is what kecked already mentioned, to measure the voltage drop across the big power resistor. It's been a while since I messed with a P3 so I don't remember the value of the resistor, but it's just the voltage you measure across the resistor divided by the resistor value to get the current. Example, if you measure 0.25V and the resistor value is 1.0 ohm, you have 250mA running thru the diode.

[Bradfo69]

What's written on it is .200 ohm. So, if it were all 12 volts divided by .2 that would only be 60mA, which doesn't make sense to me. Buffo spoke about running these blue modules w/ 2 in-series M140's at 1.8 amps. Am I missing something about the resister value?

[DZ]

What's written on it is .200 ohm. So, if it were all 12 volts divided by .2 that would only be 60mA, which doesn't make sense to me. Buffo spoke about running these blue modules w/ 2 in-series M140's at 1.8 amps. Am I missing something about the resister value?

You're seeing 12V on the 0.2 ohm resistor?!?! That would be about 60A, no 60mA!