Hi there!

I recently built a simple full-wave rectifier circuit/unregulated DC supply utilizing the Analog Discovery 2 just to test the system out. I noticed something odd regarding output.

Currently I'm rectifying 5Vpp AC (using the waveform generator of the Analog Discovery 2) with four 1N4007 diodes, a smoothing capacitor rated at 1000uF in parallel to the output, and a 330 Ohm output resistor. Without the smoothing capacitor, I get the expected peak output of ~3.7 VDC. When I connect the capacitor (or any other capacitor), the Waveforms oscilloscope tells me that I have a significant voltage drop at both the input and the output of the circuit of a ~1V. I simulated the circuit in LTSPICE and didn't see that voltage drop occur. I've tried it with a number of the capacitors I have, all give me the same output. I tested it again by rectifying 2Vpp AC, and didn't see that significant drop occur at both AC input and DC output.

I'm not sure how a capacitor would significantly load a circuit such to create a voltage loss at both the input and the output at any voltage higher than ~2Vpp. Why would this be? Is there something I'm missing?

Thanks!

 

Rather than words, post a schematic of your setup

 

Are you taking the forward drop of the two 1N4007 rectifiers that are on at any given time into account?

A quick schematic would speed this discussion up.

 

Do you have the cap polarity correct?

 

Do you have a load on the output?
If so, what is it?
What secondary current is the transformer rated for?

Keith

 

It's a very basic circuit. I'm using the 1N914 in simulation because it and the 1N4007 have similar foward bias voltages. I tried manually updating my LTSPICE to include 1N4007 and for whatever reason, I didn't have permissions to? (Even though it's my own machine and I'm admin on it?).

Here it the circuit input:

Easy enough.

Here's the output without the 1000uF capacitor in parallel with the 330 ohm load:


As expected, there's a forward drop due to the diodes of about ~1.4V.

Here's what it looks like with the 1000uF capacitor in parallel with the 330 ohm load:


Notice how even the input signal is clipping. I can't figure out why that is.

Thanks!

 

Rather than words, post a schematic of your setup

I posted a reply. Let me know what you think!

 

Running the full wave diode bridge adds 2 x 0.7V diode forward voltage drops into the equation so the output looks ok to me.
And this drop will go up a little as the current does.
One point 'tho, your circuit has 2 different earth connections so D4 is shorted out and it will nor work correctly.

 

The 330 Ohm resistor is drawing a current of almost an Amp from the capacitor. The transformer has to replace that energy when the voltage of the waveform exceeds the voltage on the capacitor. During that time, there will be a drop in the secondary voltage because of the resistance of the secondary winding. That is why you see a flattening at the tips of the waveform. That is also why I asked you what current the transformer secondary is rated for. The thicker the wire on the secondary, the less this loss will be.
Regards,
Keith

 

Those little 1N914 diodes being driven with high peak currents is dropping more than a 600 mv or 700 mv diode drop.


(This is from a Philips datasheet)