Transformer Problem

I’m having problems with my SMPS. I checked the transformer (749118105) and it only outputs 1.1V. I calculated that it should output approximately 8.83V but I don’t know why it doesn’t. Pin 3 is connected to 167V DC, Pin 1 is connected to the drain pin of NCP1070P065, Pin 4 is connected to ground, and Pin 6 is the output.

Hi @thebutterminecutter ,

The stated transformer should work with the 1070 although the saturation current is lower than the peak current of 1070.
Is there any chance to share the complete circuit diagram of your setup?
Cheers, heke


Hello thebutterminecutter,

The 749118105 transformer has an 18.9:1 turns ratio. You have correctly connected the high voltage to pins 1 and 3.

\color{red}\Huge \fbox{WARNING}

This post explores line-powered circuitry. We should take a moment to recognize the serious nature of our actions. A mistake could cost money, production time, damage to equipment, fire, or even hurt a person. Be sure to follow all applicable employer, state, and federal guidelines.

Let’s look at a typical application for the NCP1070P065:

The concept of “ground” is problematic in this type of circuit. A careful exploration of the schematic suggests three “ground” connections including:

  • AC line neutral
  • floating ground associated with the NCP1070P065 e.g., pins 3, 7, and 8
  • isolated right-hand side output ground


  1. Start with a clear understanding of the circuit’s “grounds” and the isolation actions performed by the circuit.
  2. Verify VCC (pin1) relative to ground (pin 3) is within spec. Note that the chosen transformer does not match the application note as it is missing the tertiary winding. However, you may have another method of providing voltage to the driver chip.
  3. Verify the switching signal is present on the drain - see video below to assist with oscilloscope grounding.
  4. Verify feedback signals through the optocoupler
  5. Other?

Best Wishes,


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I’d politely suggest that these two concepts are in conflict, unless the application in question is generation of smoke signals and/or testing of overcurrent protection devices.

The output of a flyback supply is determined by the feedback loop. One picks the turns ratio to yield a reasonable duty cycle and limit voltage stress on the switch.

Getting such things correct is rather important. If the input and output “grounds” are connected to produce a non-isolated supply, the resulting output could only be (properly) used internal to a device with appropriate insulation to prevent user contact. Since the selected transformer will likely be saturating on every cycle, the chance of a meltdown and primary-to-secondary short in this case would appear substantial.

If left isolated as in the pictured “typical application example,” the lack of a Y-cap to the primary side means that everything connected to the output becomes an antenna that’s transmitting broadband noise coupled from the switch node through the transformer’s inter-winding capacitance. Which is the sort of thing that can get a person a fine or prison time courtesy of the FCC.

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Here is the circuit schematic :

I’m only starting out in electronics, so I would greatly appreciate guidance on how to design this circuit and follow best practices.

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Yes, I do understand, thebutterminecutter.

We all need to start somewhere. Personally, I released a considerable amount of magic smoke as I learned electronics.

Let’s take a closer look at your schematic.

  • The red squares are the dangerous grounds. They are closely associated but not equal to the neutral (return) wire in the AC line.
  • In blue we have what should be isolated.
  • We should be able to mill a cut through the PCB as indicated by the green line. The only things bridging this chasm are the transformer and the opt isolator.

With this stipulation we see a problem with VCC for the driver chip. You are attempting to power the device using a power source that crosses the divide. Do you see it? Since the two grounds are not (AND SHOULD NEVER BE) connected, it’s like powering a device using only the positive terminal of a battery. You have no power to the driver.

Referring back to the typical application example, we see a tertiary winding used to provide an isolated voltage for the driver. You will need to find a safe way to provide that functionality.

Perhaps you could bench test this using a safe 24 VDC supply. Apply the 24 VDC to the full-bridge output. Then provide 10 VDC to the driver via a current limiting resistor (see datasheet for values).

No guarantees, but that should get the circuit started. You can then safely probe the circuit using an oscilloscope.

Be safe, line powered equipment isn’t a good place to start.


P.S. Which ground did you use for the feedback diode?


How do I Isolate the two ground connections? All component grounds are connected to the ground plane of the PCB. According to the green line, would a resistor connected to the 167V and VCC of the IC (with a high enough power rating) be sufficient to power the driver?

Thank you

On an isolated power supply you DO NOT connect all grounds to a single ground plane of the PCB, instead you design the PCB with two separate electrically isolated ground planes.

The isolation can sometimes be provided by a large no copper gap between circuit sections. Other times you have to cut one or more slots in the PCB to get a high enough level of isolation.

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Hello thebutterminecutter,

Below is an example PCB from a B&K Precision 9110 power supply. The line shows a clear demarcation zone between the AC line and the isolated sections. You can also see several slots milled into the PCB to provide increased isolation.

As for powering from the primary rail, yes BUT:

  • This is not common as there will be considerable losses in the resistors.
  • The resistor(s) will be physically large when compared to the remainder of the SMD components.
  • The corresponding enclosure must be larger to dissipate the added heat.
  • Use caution when selecting the resistor. In addition to the power rating each resistor has a maximum voltage rating.

As for the PCB design. All modern PCB tools have the ability to label nodes. You can also use different symbols for the grounds as shown in the typical application example shown above. You could use labels such as “AC_side” and “DC_isolated” as names for the respective grounds.

Best Wishes,




I redesigned the schematic of the circuit. It uses a different transformer (750871110) which has an auxiliary winding which, I think, enables me to supply an isolated voltage to the driver. I just wanted to know if this redesign will work in theory. Here is the schematic :

Thank you in advance

Hello thebutterminecutter,

Close, the only obvious issue is the optocoupler comments as shown below. You should use the isolated (right hand side) ground. There is also a short on pins 1 and 2.

A few comments:

  1. Be sure to verify all resistor, capacitor, inductor values. Also be sure the transformer is compatible with the circuit and the load.

  2. Consider heat issues. For example, electrolytic capacitors have a life vs temperature specification.

  3. Mill a trench in the PCB under the transformer and the opto.

  4. Follow @rick_1976 warning about EMI.

  5. Consider Brook’s law

“Plan to throw one away; you will, anyhow.”

There is much to be learned in the design process. Often, we only see / learn as the design takes shape.

Best Wishes,


P.S. See @rick_1976 comments below - C6 is effectively shorting the output of the tertiary winding. Move D7 to make a half wave rectifier with C6 as the filter capacitor.

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Returning to the original question, the reason one would be getting ~1V out from such a circuit is because that’s what it’s programmed for, per the schematic. Flyback supplies can be thought of as filling a bucket from a faucet and dumping it into a bathtub until the feedback loop says stop.

When the opto glows, that’s the stop signal. Simply connecting the output winding to the opto’s input as shown will therefore result in an output approximately equal to the diode’s forward voltage. A common practice is to use a voltage reference IC of the TL431 flavor to drive the opto based on a divided version of the output voltage.

Also, the primary winding is connected backwards in the revised version, and C6 should be omitted or at least positioned after the diode.

If the supplier of the driver chip does not provide a comprehensive application note, one can look for resources provided by suppliers of similar products in the market for guidance. Power Integrations is one such supplier, and I might suggest their AN-82 as a document worth study.