analog modulation driver. Problem (Graphic heavy)

Discussion in 'The Projects Forum' started by kiyoukan, Oct 13, 2010.

  1. kiyoukan

    Thread Starter Member

    Nov 17, 2009
    I have been building a few drivers and recently tried my luck at an analog driver.
    Part Value
    C1 100nf
    C2 100uf
    D1 1N4148
    D2 1N4148
    D3 1N4148
    IC1 LM358N
    IDLE 10K RTRIM64W resistor
    LED1 power LED3MM led
    MAIN 10K RTRIM64W resistor
    R3 10K
    R4 10K
    R5 10K
    R6 10K
    R7 1K
    R8 10K
    R9 1R
    R10 10K
    R11 10K
    R16 calculate for diode type
    T1 BD437
    Thats all the schematic info i have for it.
    This is what i built. I did make a change my transistors anre now turned 180 due to the pin outs are different.
    This is what my board looks like.

    Okay and it called for 1ohm resistors for one and a .33 ohm resistor on the other. all i had was smd resistors so i used 3 1ohm on the one board and 1ohm on the other.
    I have one running at 12v and its outputting 1.3A and another running off 5.5v and it outputting 400ma.
    But i cant seem to get the signal wire to work. i put it to 5v high and nothing happens.
    Also i did not have the same npn transistor so i used,
    MJE3055Tjp instead of the BD437. (i suspect this could be my problem.)
    Also i had to mount my pots vertically as i did not have the same kind.
    i am using a pc power supply.
    Sorry for the long post but i just wanted to make sure that everything was listed and shown for the best possible answer.
    Last edited: Oct 13, 2010
  2. kiyoukan

    Thread Starter Member

    Nov 17, 2009
    I need some help. if any one is good with transistors. my knowledge is very very limited my main thing is the transistor i used will work right?
  3. wayneh


    Sep 9, 2010
    Without really understanding your circuit (what are you driving?) I think your replacement should be fine. Are you aware, though, that the two transistors have different pinouts when viewed from the "front"?
  4. kiyoukan

    Thread Starter Member

    Nov 17, 2009
    yes thats why mine is turned around 180 from its position in the picture.
    It seems to work but i cant get the modulation right.
    I am driving lasers. they are kinda like leds but alot more picky on there current.
  5. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    Bypass the emitter resistor with a ceramic or polyester .1, or .01, or something around there. ;). Will depend on the freq.
    Also try some bypassing right on top/bottom of that IC too.

    Just something to try before major reworking is attempted.
  6. gootee

    Senior Member

    Apr 24, 2007
    Yeah, if you don't have a good reason not to, always put a 0.01 to 0.1 uF ceramic and a 10 uF to 100 uF electrolytic across the power pins of each IC, in a single-supply circuit (or from each power pin to ground if dual supply). But they must be connected VERY CLOSE to the pins!

    You should be able to solder them onto the bottom of your existing board.
  7. marshallf3

    Well-Known Member

    Jul 26, 2010
    I keep a supply of 0.1 uF AXIAL caps around for those times I may forget one, underestimate or am working on something where that could be suspect.

    They're only ~3 mm long, 1.5 mm thick and tack nicely to the pins on the back of the board.
  8. retched

    AAC Fanatic!

    Dec 5, 2009
    Speaking of bypass caps:

    The SMD varieties can solder right to the pins on the bottom of the board. Great for placement or after-the-fact work.

    You can just put them in a salt shaker and solder them where they land ;)

    not really, but do get yourself some bypassing. Your lasers and ICs will thank you.
  9. gootee

    Senior Member

    Apr 24, 2007
    Listen to these guys about using bypassing capacitors, Kiyoukan. Bypass caps can make poorly-operating or non-operating circuits into nicely-operating circuits.

    Some people think of them as low-pass filters, or noise/spike shunts. But I like to think of them as small local "transient" power supplies. They tend to compensate for the inability of the main power supply lines to provide fast-changing currents that might be demanded by the ICs, due to the unavoidable inductance inherent in the supply conductors. They supply the fast-changing current demands more-quickly than would be possible with the supply lines' inductances, which also tends to prevent voltage spikes from developing on the supply lines, which might occur if the fast-changing currents that were demanded had to be moved through the supply lines' inductances.

    Side note: It's often best to use lossy caps, or a small series resistance, to try to avoid setting up a resonant network with the bypass caps and the supply lines' inductances. So for the small-valued bypass caps I usually avoid film caps as well as NPO/C0G ceramics, using cheaper ceramics instead. And for the larger parallel cap, I usually use a cheap electrolytic, since they usually have a fairly-large ESR. Also, 0.1 uF and 10 uF in parallel is typical but other values might be more ideal for certain circuits or devices.
  10. eblc1388

    AAC Fanatic!

    Nov 28, 2008
    This is how you can setup your current controller for the laser diode. Do not use a real laser diode until the function of the circuit has been confirmed.

    The circuit consists of two parts and needs to be checkout separately, as follow:

    Part A: Current drive


    0. Power off
    1. unsolder and lift one end of R6(10K) from the circuit board
    2. adjust IDLE VR to 50% midway, MAIN 10K VR all the way to 0%(zero ohm between 1&2)
    3. solder a short wire from wiper of MAIN point A to end of R6
    4. connect a load of 4.7Ω¸(3.3Ω¸ or 2.2Ω¸ will also works) to the two load terminals
    5. connect a voltmeter(DVM) to the monitor points

    6. power ON, voltmeter should read near zero, if not power off quickly and check circuit connection

    7. slowly adjust MAIN 10K VR and watch voltmeter reading increases. With one ohm as the monitoring resistor, you will get 0.1V form current of 100mA and 0.4V for 400mA etc...For 0.33Ω, you will get 0.33V for 1A.

    8. stop at the desired current you want to drive the laser diode with.(say 450mA)

    9. carefully check point A voltage. Let's say 0.45V or whatever. Record this measurement on paper for use in Part B test.

    10. power OFF and Part A check completed.

    Part B: Current setter and modulation drive


    1. move the jumper wire on R6 from point A to ground 0V, as shown
    2. remove voltmeter from monitor point and connect it to measure point B instead(ignor step if you have two voltmeters)
    3. power up, it is normal to get a higher voltage reading(e.g. 0.8V) on the meter
    4. slowly adjust MAIN 10K VR to get back the same voltage as measured in "Part A test" step#9, i.e. 0.45V say.
    5. apply a DC voltage to the modulation signal input and confirm it will change the voltage on the voltmeter
    6. if the voltage change is too (small/large), adjust the IDLE VR towards(100%/0%) accordingly
    7. re-adjust the MAIN VR to get back the same voltage as in last test step#9
    8. power off and Part B test completed

    Part C: Overall function test

    1. restore the circuit to that before any test, except leaving the 4.7ohm resistor connected as load
    2. power up the circuit and confirm same current, i.e. 0.45V (say 450mA)
    3. power off and remove load resistor, connect laser diode in place
    4. power ON
    5. apply modulation signal to see effect
    6. DONE
    Last edited: Oct 15, 2010