# Converting AC to DC

1. Apr 3, 2007

### propergomper

hi, as part of a project me and my friends are undertaking we are trying to measure the change in resistance of an electromagnet when a coin is placed near near the electromagnet. i have been trying to use a bridge rectifier to convert the system to DC but i have been encountering a few problems,

the voltage across the bridge circuit seems to be charging to -15V and does the same even if i change the direction of the capacitor in the bridge rectifier

if anyone could shed any light onto why this is happening or knows how to help then it would be greatly appreciated

many thanks

2. Apr 3, 2007

### junkfunkydude

It sounds like you need to change the direction of your diodes. The capacitor is only a storage device to even out ripples in the current.

3. Apr 3, 2007

### Staff: Mentor

What is your AC source? If it is 120Vrms or 230Vrms AC mains power, please be *very* careful working with it. Hopefully you are not messing with wall power.

If it is from a low-voltage AC source, then yes, you can use a bridge rectifier and storage capacitor to convert the input sine wave to a DC component with ripple. The bridge rectifier is described on this wikipedia.org page, for example:

http://en.wikipedia.org/wiki/Bridge_rectifier

Keep in mind that if you output smoothing capacitor is polar, you must be sure to get the polarity of the bridge output and the polarity of the capacitor matching. If you connect a polar electrolytic capacitor backwards, you will generally blow up the capacitor (bad).

4. Apr 3, 2007

### Bystander

Give us a circuit diagram or at least a verbal description of what you've got and what you're trying to measure. Sounds like you're just looking at your power supply.

5. Apr 3, 2007

### propergomper

heres what ive got so far,

ive got one connection of a signal generator (100KHz) connected to a coil (electromagnet) which is connected in series with a 1 ohm resistor which is connected back to the signal generator. ive been measuring the voltage change across the resistor when different metals were placed in close proximity to the electromagnet. then ive connected either side of the resistor to a bridge rectifier circuit as described in the wikipedia site, ive also tried reversing two diagonally oppposite arms of the bridge circuit as described on another website but to no avail.

thanks

6. Apr 3, 2007

### Staff: Mentor

Whew, I'm relieved to hear you're not using AC mains power.

Just use an oscilloscope or a DVM with 100kHz frequency response to measure the AC voltage across the resistor. There's no need to try to rectify it -- that will introduce errors of its own (the diode voltage drops, for example). Can you just use an AC measurement?

7. Apr 3, 2007

### propergomper

i could but i was needing to use a peak detector to store the peak voltage and have that value stored by a computer program

and i know this sounds lazy but the DC peak detector circuit looked much easier to construct on a protoboard than a AC peak detector

8. Apr 3, 2007

### Staff: Mentor

Okay, but make your peak detector with an opamp to get rid of the diode voltage loss. Do you know how to make an opamp-based peak detector? wikipedia.org may show one, or google definitely would. Or it should just be in your opamp textbook. 100kHz is easily within the range of an opamp-based peak detector.

9. Apr 3, 2007

### Bystander

100 kHz? You say, "electromagnet;" which usually means inductance enough that your entire signal is across the resistor. You're going to want to operate at a lower frequency or wind ten or twenty turns around a cardboard or plastic coil form to get a low enough inductance to see anything.

Edit: Oh, wow, did I ever get that exactly sideways. 'Tany rate, large inductance means the big voltage drop is across the inductor. If you're getting the signal generator's rms across the resistor, you need to check ground and common for the sg, and for whatever you're using to read the bridge output.

Last edited: Apr 3, 2007