The output voltage of the MB6S is 1.4 times over the input voltage

[xiangpei]

Hi, everyone!


In this circuit, the output voltage of the MB6S is 1.4 times more than the input voltage during the no-load time. Where is the problem?




Normal time: the output voltage is 12v when the load is empty, the input is not measured.


When the short circuit appears in the behind side while using, the output voltage is up to more than 19v, sometimes even more than 20v during the no-load time. And this time the input voltage is about 11v. What is this going to be?


Of the three capacitors, C2 and C3 are the electrolytic capacitors of 330uf and the resistance is 25V. C1 is the patch capacitance, the capacity and pressure are unknown. C2, C3 are all taken down and measured in a multimeter. The output voltage of the bridge in the absence of capacitance is 10v, which corresponds to the input voltage of 0.9 times. There seems to be no problem.


What went wrong, or was there no problem at all?


Please give me your comments! Thank you very much~

 

[GoldDigger]

Your AC voltage is described by the RMS amplitude of the sine wave.
A capacitor input filter with no load will charge up to the peak voltage of the AC waveform.
The peak value is 1.414 (sqrt(2)) times the RMS value.
Under load, and depending on the source impedance of transformer and bridge rectifier, the DC voltage may be anything less than that.

Sent from my XT1585 using Tapatalk

 

[nec_addicted]

Translation of Polarity - MB6S

Translation of Polarity - MB6S

Polarity symbols marked on the body :

3 & 4 Is the AC Input

1(+) & 2(-) is DC Output

The circuit above and the designation or packaging pin-out of the original post is in error. Thus the filtering capacitors (polarity) are non-compliant or unable to perform to specification UL # E258596

The design supports a 35 A IFSM rating and 5.0 A (sq)Sec I(sq)T Sec rating

 

[gar]

170903-2228 EDT
xiangpei:

This website has major problems with editing and I don't know how to overcome them.
So plots won't display as they should. Found a partial work around.


Your pofile does not provide much information on your background, but I expect you are just getting started with electrical work.

GoldDigger gave you a good general outline. I am going to provide some more specifics with some actual waveforms.

My circuit is similar to yours, but does not use a bridge rectifier. My circuit is called a full wave center tapped rectifier with a capacitor input filter. This uses two diodes instead of four to produce a full wave rectified output. Thus, there is only one diode voltage drop per half cycle compared to two diode drops with a bridge rectifier.

By using this circuit I have an invariant common reference point between the source voltage and the output voltage. The capacitor voltage vs time is the same for both circuits, except for the two diode voltage drops in the bridge circuit.

For illustration purposes this allows for a direct comparison of the source voltage with the capacitor voltage.

For very low currents the forward diode voltage drop is near zero. For most silicone rectifiers under normal load current conditions the drop is more in the 0.5 to 1.5 V range.


Plot 1:

View attachment 18420



For no load on the rectifiers.

The blue curve is the AC voltage between the center tap and the anode of one rectifier. The red curve is the output of both rectifiers. Note: the two curves for the positive half cycle are akmost identical, very little voltage difference.

Ideally the RMS value of a pure sine wave is approximately 0.707 Vpeak, and full wave rectified average value is approximately 0.636 Vpeak, and for a half wave 1/2 that or 0.318 Vpeak. Any graduate EE should know how to derive these values, but I would not expect an electrician to have that background.

I measured the RMS source voltage at 13.0 V, ideally this calculates to 18.4 V, but we have distorted peaks, they are clipped.

The calculated full wave rectified average DC voltage from 18.4 is 11.7 V. I measured 11.44 V.



Plot 2:

View attachment 18421



For 5 ohm resistive load.

Here you see the out[ut voltage lower than the input voltage by the diode voltage drop from the load current thru the diode.



Plot 3.





With a 1000 ufd capacitor load. The capacitor voltage read 17.8 V DC. No vissible ripple.




Plot 4:




With an approximate 1000 ufd capacitor shunted with a 5 ohm resistor. The RC time constant is about 5 milliseconds. The time between the peaks for 60 Hz is 1/120 seconds = 8.3 milliseconds.

The average DC output voltage can be eyeballed at 12.5 V, and I measured a value of 12.9 V. Note the very large amount of ripple on the DC output.


.

 

[jimo144]

Full Wave Bridge Rectifier

Full Wave Bridge Rectifier

The MB6S is a Full Wave Bridge Rectifier. Without any regulation or load, it will have a nominal DC output of approximately 1.4 * the AC Input. The AC Input Voltage is an RMS Measurement and the Bridge Rectifier will Conduct the full Peak to Peak value. If your desired DC Output is approximately 12 VDC, you would need an AC Input of right at 9 Volts. 9 * 1.4 = 12.6 VDC Output. As others have replied, the Capacitors will charge and hold the DC right up to the Peak. There's really nothing wrong with the Circuit. If you are trying to get a relatively constant unregulated Output at a given Voltage, adjust the Input Voltage and you have it. If you are going to use a DC Voltage Regulator, the circuit is ready for it.
JimO