Lower voltage in auxiliary winding of SMPS transformer than what is specified in datasheet - can it cause problems?

Thread Starter

szabikka

Joined Sep 3, 2014
92
Hello fellow electronics enthusiasts!

I'm planning to build my first SMPS using this transformer : Myrra 74030 and this controller IC: THX 203 .
I have noticed that the voltage range of the auxiliary winding of the 74030 is given as 7 - 14.5 volts. I wanted to use the secondary marked as S3 to create a voltage of 12 V and my input to the primary would be about 230 V AC. If I'm correct under these circumstances the voltage generated on the auxiliary would have been around 10 to 11 V, which is not OK since the ICs internal regulation kicks in at 9.6 V and stops the switching until the voltage drops under this specific voltage.
After considering this, I decided to connect S2 and S3 in series to create a larger secondary winding and using this to create the 12 volts. If I'm correct, taking the newly created turns ratio into account the auxiliary voltage will be 5,3 volts which is OK for the IC, but is lower than the voltage range specified in the 74030's datasheet for the auxiliary winding. Can it cause any problems if I drive the auxiliary winding at a lower voltage than what is recommended in the datasheet?
 

Thread Starter

szabikka

Joined Sep 3, 2014
92
If I understand your question, you are asking whether 5.3 volts is enough for Vcc. The datasheet indicates that this is fine.

View attachment 193939
Thank you for your answer Dick! Actually, my question concerns the transformer. I also linked its datasheet previously. The datasheet of the transformer says that its auxiliary voltage is between 7 and 14,5 volts, but I don't know if it would be working with my 5.3 volts, since its lower than the smaller end of the given range.
 

ronsimpson

Joined Oct 7, 2019
710
I think the transformer will make 12V, 12V, 5V, 10.8auxV.
I think the 10.8 is too high and the IC will shut down. (will not damage the IC)
Transformer---diode---cap---3.3VZener---cap--IC This will reduce the voltage by 3.3V.
The idea is to make the 10.8V and then drop the voltage by 3.3V to a 7.5V supply. (cap1 on 10.8 and cap2 on 7.5V)
Also could use a 5.1V Zener. 10.8-5.1= 5.7V

It is also possible to use small diodes to make a low voltage Zener. 1D=0.7V, 2D=1.4V, 3D=2.1, 4D=2.8V …...
1N400x or 1N4148 or ...

The transformer was designed to use a IC driving a MOSFET. The supply would need to be higher for a MOSFET compared to a transistor.
 

Thread Starter

szabikka

Joined Sep 3, 2014
92
I think the transformer will make 12V, 12V, 5V, 10.8auxV.
I think the 10.8 is too high and the IC will shut down. (will not damage the IC)
Transformer---diode---cap---3.3VZener---cap--IC This will reduce the voltage by 3.3V.
The idea is to make the 10.8V and then drop the voltage by 3.3V to a 7.5V supply. (cap1 on 10.8 and cap2 on 7.5V)
Also could use a 5.1V Zener. 10.8-5.1= 5.7V

It is also possible to use small diodes to make a low voltage Zener. 1D=0.7V, 2D=1.4V, 3D=2.1, 4D=2.8V …...
1N400x or 1N4148 or ...

The transformer was designed to use a IC driving a MOSFET. The supply would need to be higher for a MOSFET compared to a transistor.
Thanks ronsimpson! I was thinking about the same thing stabilizing the ICs voltage at a lower value witha zener, but then I thought it must have a reason to have that lock out at higher auxiliary voltages, probably to protect the circuit on the secondary, but since both of us had the same idea it might worth a try ;)
 

ronsimpson

Joined Oct 7, 2019
710
I thought it must have a reason to have that lock out at higher auxiliary voltages
There are two power supply types.
One type uses a MOSFET. It needs 10 to 20 volts of drive voltage and almost no current.
This type used a NPN power transistor in the IC. The Base of this power transistor needs 0.7V and 20mA to turn it on. This 0.7V comes from the aux supply of 9V. (9x0.7=180mWatts.)
This IC uses a high voltage transistor and a low voltage transistor and must turn on both transistors. So two Bases must be turned on. (probably more like 300m Watts of power lose just in turning on the transistors)
If the supply gets too high there will be too much heat in the IC.
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