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advice on replacement power transformer for vintage dbx 163x

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  • advice on replacement power transformer for vintage dbx 163x

    Hi all- I have a dead dbx 163x rackmount compressor. The power transformer is putting out 0 volts when unplugged from the rest of the circuit, so no drain from somewhere else upstream. It has a switching 110/220 voltage selector fed from the wall plug, that connects to the transformer primaries, and the transformer secondaries are three wires to a molex connector. I have verified that continuity from the wall plug through the functions of the switch are all solid. At the molex plug I get 0-0-0V So, the transformer has died.

    I have another working unit, I pulled it's power section and tested it unplugged from the circuit, and I get 24.1-0-24.1 VAC.

    So here's where I need help. I think I have done the math and found the solution, I just want someone to check my math and my logic before I pull the trigger on purchasing a replacement.

    The rear of the unit lists the power requirements as AC 110V, 7 watts. So by Ohm's law (I=P/V) that means it draws 0.058A of current.

    If I am reading everything correctly, I am now looking for a dual primary 48V center tapped transformer, with a current rating of upward of 60-70mA, yes?

    Looking for those specs, I find a Hammond 186B48 that seems to fit the bill. Secondary current rating of 120mA, and it's almost a perfect match size-wise. The mounting holes are almost exact.

    So, would someone please check my numbers and let me know if I am way off the mark, and if so, where did I screw up? Or, if my numbers are correct, any other options for a good transformer? At ten bucks, I would be very comfortable buying the Hammond. But if there's any other reliable options out there, I'd be interested to hear about them.

    Much thanks in advance for your time and advice.

  • #2
    I think you've found a suitable replacement and the numbers make sense. Hopefully there's no major difference with radiated stray field on the replacement xfmr. I didn't look up the Hammond xfmr, but theirs usually have universal primary, so should be a good replacement. Best of luck on the surgery.
    Logic is an organized way of going wrong with confidence

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    • #3
      Firstly, thank you so much for the response.

      Secondly, this is why I ask the question. You bring up factors I don't even know to look at.

      I must admit, you are way above my paygrade with the "radiated stray field" reference. I will go do some home work on that one.

      Also, thanks for the good luck. I am mostly self taught, so I apologize if my inquiries might be no-brainers, but I have found it's good policy to check with those who know better. I just want to make sure I am not overlooking something important I am not yet aware of.

      I have a particular fondness for these units in my studio. Fun little color units. Hopefully this works. Thanks again.

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      • #4
        Here is the power supply schematic.
        Although I do not see a 'center tap'.
        Attached Files

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        • #5
          I will point out that you can't exchange primary and secondary currents unless the voltage is the same. Use the wattage and solve for current.
          "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

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          • #6
            Jazz P Bass, I'm pretty sure that's the 163. I am referring to the 163x. Different units.
            Last edited by MichaelNuzum; 01-13-2021, 06:14 AM.

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            • #7
              g1, I was curious about primary and secondary current. I will go back and do some more research on that. Thanks.

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              • #8
                So, do I do the math like a ratio? If the unit requires 0.058A at 115V, then it needs X at 48V? So if that formula is correct I need a secondary current of 24mA or above? So this one should be fine, right?

                Or is it inverse? So if the transformer has a ratio of roughly 2.4:1, then I'd need 139-ishmA, which means this one barely cuts it.

                Again, sorry if this is all elementary, so I really appreciate the help.
                Last edited by MichaelNuzum; 01-13-2021, 06:26 AM.

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                • #9
                  The correct approach for calculation is based on the assumption input power equals output power.
                  Using 7W gives a secondary current od 7W/48V = 146mA .

                  Now all transformers have losses, meaning that input power must be somewhat larger than output power. Typical losses of a small PT are around 20%.
                  This gives a secondary power of 0.8 x 7W = 5.6W and a secondary current of 5.6W/48V = 117mA. E voila.
                  - Own Opinions Only -

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                  • #10
                    As you have a second working unit, if you're equipped with some test equipment, you could make some measurements to get closer to what the actual load current is on the secondary of the power transformer.
                    Assuming all you have is a multimeter, you could acquire a couple 1 ohm 1W resistors.....1% would make it easier, though measuring actual resistance to start would help. Install the 1 ohm resistors in series with each secondary lead ahead of the rectifiers of the power supply. You'll be measuring the AC voltage drop across the resistors. Assuming the resistance is actually 1 ohm, if you measured 50mV, for example, the current thru the resistor is 50mA. Assuming the secondary winding is 48V C/T and feeds the input of a full wave bridge rectifier circuit, then that same current is flowing from the upper winding thru the bridge and out the lower winding, so the total current flow is 50mA.

                    You could also place a 1 ohm resistor in series with the primary....though here you need to be very careful and clinical, as this IS dangerous, working on the AC mains side. Your multimeter must be floating...battery powered meter makes sure of that. Voltage drop across this resistor shows the AC mains current flow thru the primary windings. Then, measure the AC voltage across the primary....or insert the two probe tips into the adjacent AC mains outlet safely and record that reading as well. Multiply the AC voltage times the calculated AC current (ACV reading divided by the actual resistance, which will yield VA (Volt-Amps). VA is similar to Wattage, and for a low current product like this, close enough to compare to the specified Wattage rating shown on the panel. The secondary current reading measured will be sufficient for guide in selecting the replacement power transformer. You could, if possible, put the dbx 163X into operation to see if there's increase in current flow in the secondary. It probably won't vary that much, it being signal processing gear. If it was a power amplifier, for example, then you'd have to be assessing current flow under full load operation to know what current rating you'd need for the xfmr.
                    Logic is an organized way of going wrong with confidence

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                    • #11
                      Helmholtz, thanks for that! This is what I am looking for. So I guess I have more questions, and sorry if they are basic stuff. The value you provided of 146ma makes sense. I am assuming this is a theoretical description of the relationship? The next statement, I am assuming, describes the practical limits in that relationship with regards to loss in the transformer?

                      So, when the manufacturer states that the unit draws 7 watts/110V , am I to understand they are listing real world practical current draw (including the 20% loss of the transformer), not theoretical current draw (of power in equals power out)?

                      I would assume it would be the practical draw as measured with a current meter, but I have long ago learned it's bad policy to assume with electronics. I'd rather ask a silly question and look stupid but end up knowing the correct answer, than assume and blow something up.

                      Thanks again everyone for all your help with this. I promise I am doing my homework in between these posts trying to figure as much of this out as I can on my own, but it is daunting.

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                      • #12
                        Originally posted by nevetslab View Post
                        As you have a second working unit, if you're equipped with some test equipment, you could make some measurements to get closer to what the actual load current is on the secondary of the power transformer.
                        Assuming all you have is a multimeter, you could acquire a couple 1 ohm 1W resistors.....1% would make it easier, though measuring actual resistance to start would help. Install the 1 ohm resistors in series with each secondary lead ahead of the rectifiers of the power supply. You'll be measuring the AC voltage drop across the resistors. Assuming the resistance is actually 1 ohm, if you measured 50mV, for example, the current thru the resistor is 50mA. Assuming the secondary winding is 48V C/T and feeds the input of a full wave bridge rectifier circuit, then that same current is flowing from the upper winding thru the bridge and out the lower winding, so the total current flow is 50mA.
                        Problem with measuring secondary transformer current is that it's pulsed and would require a true RMS meter with sufficient crest factor rating.

                        - Own Opinions Only -

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                        • #13
                          nevetslab - wow, this is awesome, thank you so much!! I am gonna wait until I have a window later today and then I am going to try this test procedure.

                          Again, thanks to all of you. These are the bits of info I was hoping to find. Thanks s much for all of your insight and guidance. Much appreciated!

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                          • #14
                            oooh "crest factor rating" We're going above my head here. Time to hit google again.........

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                            • #15
                              Originally posted by MichaelNuzum View Post
                              So, when the manufacturer states that the unit draws 7 watts/110V , am I to understand they are listing real world practical current draw (including the 20% loss of the transformer), not theoretical current draw (of power in equals power out)?
                              I'm sure the rated power of 7W means max. mains power draw, which includes PT losses
                              Last edited by Helmholtz; 01-13-2021, 06:28 PM.
                              - Own Opinions Only -

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