# Need some help with a rectifying circuit.

#### Robot626

Joined Nov 10, 2008
4
Sorry in advance for the length here! I am not really high level electrically so bear with me.

In the last week I was working with an old salt that fixes plasma cutters and we were going over a really simple circuit I put together for a filtration device. On this circuit I run stepped down voltage 12V through a bridge rectifier and use the DC signal (10VDC) unfiltered into a DC motor. It works pretty well except for I lose a little voltage across the bridge which I expected. The problem I am having is that the motor drives a fan and a fuel pump in tandem. In order for me to get the right burn ratio I really need to be up around 13.5V on the motor.

After talking to this guy he is telling me that a "bridge will boost voltage by 1.4" and I am not doing it right. He is throwing out these arbitrary numbers with out know the voltages or currents and the problem is he talks directly to my boss who doesn't know much about elex. Not that I know alot either but I have some basic knowledge. He sent my boss this drawing and I am curious how a resistor and two caps wired up this way would boost voltages. I know ohms law is ohms law so if the current stays the same the Voltage would go up if the resistance does but like I said my rectifier is not filtered. I am feeling a little stupid here, and this drawing isn't making sense to me in that components aren't called out. Is it possible to boost the voltage with a couple caps and resistor?

Any help would be appreciated.

#### Attachments

• 14.7 KB Views: 20

#### bertus

Joined Apr 5, 2008
21,850
Hello,

The capacitor in the schematic you are showing is used as a buffer for the time no voltage is coming from the rectifier.
Take a look at these 2 applets on the electronics demonstrations site:
1) without a capacitor, the voltage will go down to zero every time.
2) with a capacitor, for the time the signal would go to zero in example 1) the voltage is buffered by the capacitor.

There are much more examples of other circuit to look at over there:

Greetings,
Bertus

#### Robot626

Joined Nov 10, 2008
4
Thanks for the replies guys. I have used the Falstad link and understand the filter process but is it possible to actually boost voltage? I know the cap will act like a "surge tank" and supply voltage when the AC wave crosses zero. But the guy I mentioned is telling me this after I referred him to the rectifier circuit applet (which I think is awesome by the way):

"Oh cool that was neat....Now for reality...again typically the rise from AC to DC is aprox. 1.4 times again if it is done correctly. I am sending you a sample drawing...buy the way this section of the power supply is the same on the Servo Drive system on your plasma now. It runs at 55Vdc @ 12amps supplied by aprox. 40 volts AC across the bridge" There has to be more to it right?

Thanks again! I will check out the textbook link.

#### bertus

Joined Apr 5, 2008
21,850
Hello,

You will have to look at the average voltage.
In example 1) there is no capacitor, the average voltage is about the transformers voltage.
In example 2) there is a capacitor that "fills" the gaps between the tops of the rectified signal.
The top of the signal is √2 * the average voltage without the capacitor.
The larger the capacitor is the more it will reach the theoretical maximum of √2 the transformer voltage.
Since √2 is about 1.4 they say that the rectified signal with a large capacitor is 1.4 * the transformer voltage.

Greetings,
Bertus

#### Robot626

Joined Nov 10, 2008
4
I understand now. so they aren't really multiplying voltage they are just getting closer to the theoretical maximum of the AC wave coming off the transformer. So in theory if I put a capacitor into this circuit I may be able to get to 13.5V off of the 12V secondary of the transformer...

Thanks again for your response and patience.

#### bertus

Joined Apr 5, 2008
21,850
Hello,

Yes, the larger the capacitor the more you will reach 12 * 1.4 = 16.8 Volts.
An other thing that lowers the voltage are the diodes in the rectifierbridge.
There are always 2 diodes active, so they will drop your voltage with about 2 * 0.7 Volts = 1.4 Volts.
You will end up with 16.8 - 1.4 Volts = 15.4 volts maximum
The load will let the voltage drop the voltage on the capacitor, as shown in the example 2).

Greetings,
Bertus