Help determining where does current flow through diodes

In summary, the circuit in the picture is a voltage limiter with an inverting op amp and zener diodes in the feedback path. There is also a possible feedback path through the diodes D1, D2, D3, and D4 under certain conditions. This circuit can be further analyzed using the principle of full wave rectification.
  • #1
oujea
12
0
Hello

Can anybody help me determine where does current flow in the circuit on the picture below:

9qwd2p.png


How it goes when Uul is >0, and how when Uul is negative (<0). There is inverting OP AMP, and Uop will be >0 when Uul is <0, right? And how it goes then through diodes? Zener diodes here are not important for me, because they are only limiting output voltage.

Thank you in advance
 
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  • #2
the zeners are important as they are you only path for feedback. looks like a comparator and rectifier, why would you want to do that?
look up full wave rectification to answer your question.
where did the circuit come from, what does it do?
 
  • #3
oujea said:
Hello

Can anybody help me determine where does current flow in the circuit on the picture below:

9qwd2p.png


How it goes when Uul is >0, and how when Uul is negative (<0). There is inverting OP AMP, and Uop will be >0 when Uul is <0, right? And how it goes then through diodes? Zener diodes here are not important for me, because they are only limiting output voltage.

Thank you in advance

A couple clarifications about the circuit. There should be an explicit connection dot above the resistor Rp connecting it to Uizl. And the polarities shown for +/-Um are backwards...
 
  • #4
I have to analyse how it works for different voltage Uul values. I have to find output voltage for all possibilities of the Uul voltage
 
  • #5
hi oujea, what have you tried to do, and where are you getting stuck?
did you look up rectification? plenty of websites have good explanations of those four diodes in that arrangement.
 
  • #6
Unfortunately I didnt find anything helpful (maybe because my english is bad, so I can't search for the right thing). I tried to analyse this, and I found that all diodes (D1, D2, D3, D4) are conducting when Uul is negative. but I'm not sure if I'm right.
 
  • #7
the circuit on the picture is actually op amp limiter. here is it's characteristic:

5u3eya.png
 
  • #8
oujea said:
Unfortunately I didnt find anything helpful (maybe because my english is bad, so I can't search for the right thing). I tried to analyse this, and I found that all diodes (D1, D2, D3, D4) are conducting when Uul is negative. but I'm not sure if I'm right.

hi Oujea, here is a simple site explaining that diode arrangement.
http://www.allaboutcircuits.com/vol_3/chpt_3/4.html

it was the second hit on google (search full wave rectifier circuit) below wikipedia.

are you supposed to have anything connected to the ±Um nodes? what analysis methods are you supposed to use?
the only way all four diodes would be forward biased is if you had something active there.
perhaps if you show step by step what you have tried, it will be easier to point you in the right direction
 
  • #9
This is not rectifier, and link you posted did not help me because it shows 4 diodes in Graetz circuit (and it is rectifier), but circuit on my pic is voltage limiter.
 
  • #10
oujea said:
This is not rectifier, and link you posted did not help me because it shows 4 diodes in Graetz circuit (and it is rectifier), but circuit on my pic is voltage limiter.

ok, well the principle is the same for those diodes.
lets back track a second, I'm making some assumptions here.

the first thing you do with op-amp circuits is look at the feedback path, which should give you an idea of what's going on. the feedback path in this case is from the output, through those zeners and to the inverting input (negative feedback).

there is another possible path, but only under certain conditions which are
1. there is (as Berkeman says) a connection between Uizl and the top of Rp. this seems likely, as the dot is missing from several other places obviously connected, but the drawing is a little ambiguous
2. there is some current path between -Um and Um, which also seems likely as they have the same Ident, and are drawn like outputs.

are these two conditions true?

if so, then another feedback path exists, but changes its route based on the difference between VUop and VRp.
its nature is described by the rectifier circuit if you model the unseen connection between Um and -Um as the DC load, and the source as VUop to VRp.

does what I'm saying make sense?
 

FAQ: Help determining where does current flow through diodes

How do diodes work?

Diodes are electronic devices that allow current to flow in one direction, while blocking it in the opposite direction. This is due to their structure, which includes a positively doped (P-type) and a negatively doped (N-type) semiconductor material.

How do I determine the direction of current flow through a diode?

The direction of current flow through a diode is determined by the direction of the applied voltage. If the voltage is applied in the forward direction (anode to cathode), the diode will allow current to flow. If the voltage is applied in the reverse direction (cathode to anode), the diode will block current flow.

What is the purpose of a diode in a circuit?

Diodes are commonly used in circuits to regulate the flow of current and to protect other components from voltage spikes. They can also be used to convert AC current to DC current.

How do I calculate the voltage drop across a diode?

The voltage drop across a diode depends on the type of diode and the current flowing through it. For a silicon diode, the voltage drop is typically around 0.7 volts. For a germanium diode, it is closer to 0.3 volts. You can use Ohm's law (V=IR) to calculate the voltage drop at a specific current.

Can diodes be used to amplify signals?

No, diodes are not designed for amplification. They are used for rectification, switching, and protection in electronic circuits. However, they can be combined with other components, such as transistors, to create amplifiers.

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