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I have recently gotten back into electronics and have access to a lab with a signal generator and oscilloscope. I have been trying to teach myself how to use this equipment by building small and simple circuits.

Last night I wired a full-bridge rectifier on a breadboard. The diodes I used were EM518 diodes (those were the only ones that I could find in the lab). What I was trying to do was set up Channel 1 for the input (5VAC-RMS @ 60Hz). Then I wanted to set up Channel 2 to show the output ( ??VDC @ 120Hz). Below is a visual of how I set up the probes:

enter image description here

What I ended up with was a half-wave at the input and a half-wave of smaller magnitude at the output.

Why is this happening? Am I connecting the probe grounds the wrong way? After searching for how to properly connect ground leads, I am now more confused than I was before.

How is this done when working with circuits not connected to the mains? (I'll stay away from mains powered circuits till I understand what the hell I am doing).

Thanks!

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    \$\begingroup\$ Does your AC signal have a DC offset? \$\endgroup\$ – Shamtam Apr 10 '18 at 17:40
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    \$\begingroup\$ Sig gen GND and scope GND are probably the same (definitely the same if you are probing both input and output with the scope), so you shorted D2 \$\endgroup\$ – Sredni Vashtar Apr 10 '18 at 17:41
  • \$\begingroup\$ DC offset was off when doing this. \$\endgroup\$ – James85 Apr 10 '18 at 17:46
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    \$\begingroup\$ "Ch1 Probe Gnd" == "Ch 2 Probe Gnd". They're internally connected in the 'scope, So you're shorting out your D2 in your bridge. \$\endgroup\$ – brhans Apr 10 '18 at 18:18
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    \$\begingroup\$ You are confusing the scope. Connect one scope GND clip to Function Generator GND. The first GND will work as reference for the second scope probe. \$\endgroup\$ – StainlessSteelRat Apr 10 '18 at 19:13
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If this truly is your wiring:

enter image description here

you are shorting out your DC- rail via your scopeprobe. This will short the supply as well. The local DClink is +- about the return of the supply.

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  • \$\begingroup\$ Sorry but I don't understand your explanation. If my probe grounds are separated, how am I shorting anything? \$\endgroup\$ – James85 Apr 10 '18 at 17:59
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    \$\begingroup\$ They are not separated. They are the same GND, actually, your home ground. You might want to see this: youtube.com/watch?v=xaELqAo4kkQ \$\endgroup\$ – Sredni Vashtar Apr 10 '18 at 18:01
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) The whole circuit. (b) What's conducting on positive half-cycles. (c) What's conducting on negative half-cycles.

You can't connect your scope probes to two different potentials. Fortunately your power supply had limited current drive so you don't appear to have damaged your equipment.

With a more powerful supply you could draw very large current (red arrows) and destroy the earth traces in your test equipment.

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  • \$\begingroup\$ Or blow fuses... \$\endgroup\$ – StainlessSteelRat Apr 10 '18 at 17:54
  • \$\begingroup\$ That is why I am using low voltages on the signal generator. \$\endgroup\$ – James85 Apr 10 '18 at 18:07
  • \$\begingroup\$ @James85 - Low voltage is irrelevant. The short circuit will FORCE the voltage low. The question is, how much current will the supply provide? You might want to look up the specs on arc welders. \$\endgroup\$ – WhatRoughBeast Apr 10 '18 at 18:48
  • \$\begingroup\$ I think James is just saying that he is being cautious and learning with a safe, severely current limited signal generator before he moves onto bigger power. Smart guy. He'll live long and will save by not having to replace test equipment. \$\endgroup\$ – Transistor Apr 10 '18 at 18:53
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Firstly, you need to know: 1. How diode works 2. What happens to the diodes connected in a fullwave bridge rectifier configuration

To address 1, diode turns ON and conduct when the positive terminal of the diode reaches certain positive voltage (usually around 0.7V). This is known as forward bias. Any voltage below the 0.7V applied to the positive terminal will not turn ON the diode. If 0.7V or greater is applied to the negative terminal, it is considered reverse bias and diode will not turn ON as well.

Now that we have explained how diode works, let us look as the fullwave bridge configuration. When you connect a function generator with a sine wave as input to the fullwave bridge assuming that first half of the period, T, is positive and goes from 0V to peak of 5V. From 0V to 0.7V, your diode, D1, will not turn ON. D1 will only turn ON and conduct when voltage is between 0.7V to 5V. Diodes D3 and D4 will be in reverse bias and remain OFF at all times. Similarly, D2 will turn ON to complete the circuit.

Your problem comes in during the second half of the period, T. During this time, your sine wave voltage is negative with reference to GND. Positive terminal of D1 is at a lower potential than its negative terminal - hence reverse bias. D3 will not conduct as well since the voltage drop between both its terminals are close to 0V.

Based on the explanation above, this most likely results in half wave rectification.

http://www.circuitstoday.com/full-wave-bridge-rectifier

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    \$\begingroup\$ Welcome to EE.SE, Kenny. The OP (original poster) seemed to understand the operation of the bridge rectifier. I think that you've missed that the problem is that D2 is shorted out by scope and signal generator ground connections. That means that when the signal goes negative that D4 is connected forward biased across the signal thus creating a short-circuit. Tip: "Positive terminal of D1 is at a lower potential ..." Then it's the negative terminal. Use "anode" and "cathode" to avoid confusion. \$\endgroup\$ – Transistor Jul 9 '19 at 6:39

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