Can a diode exist with parallel resistors?

Thread Starter

JLoundy

Joined Sep 15, 2022
4
The diode has an "on voltage" of 0.8V is has to overcome in order to turn on, I was thinking since the diode is placed inside a parallel and both resistors have the same resistance that the diode would never have enough voltage to "turn on" and essentially act as an open circuit. Although unsure since it is placed with the op-amp, assuming an ideal op-amp.
Please let me know your thoughts and thank you in advance.

IMG_6884.jpg
 

Sensacell

Joined Jun 19, 2012
3,064
The diode branch will have no effect for negative input currents (Vout = positive)
In the other direction, the diode will start to conduct when Vout = -0.8 volts, from there, the volts/amp of the circuit will be reduced because of the additional feedback.

It will create a "gain breakpoint" where the gain is reduced.
 

crutschow

Joined Mar 14, 2008
30,791
Your thinking is not correct.
Since it is a parallel circuit, whatever voltage is across the top resistor, is also across the bottom resistor and diode.
So the top resistor has no effect on what the bottom resistor and diode do.
Thus whenever the positive voltage from the left side across them is greater than the forward drop of the diode the diode will start to conduct.

It would seem you need to further study circuit theory for parallel and series elements.
 

Thread Starter

JLoundy

Joined Sep 15, 2022
4
Your thinking is not correct.
Since it is a parallel circuit, whatever voltage is across the top resistor, is also across the bottom resistor and diode.
So the top resistor has no effect on what the bottom resistor and diode do.
Thus whenever the positive voltage from the left side across them is greater than the forward drop of the diode the diode will start to conduct.

It would seem you need to further study circuit theory for parallel and series elements.
My understanding was that a diode ideally has zero resistance and so theoretically the current going into the top branch and bottom branch of the parallel would be the same since the resistance over both branches is the same (using a current divider).

And knowing that the voltage across the top resistor and bottom (resistor + diode) is the same I am confused on how there could ever be a situation where the diode could have enough voltage to turn on and keep true to basic circuit theory. For example:

IMG_6887.jpg
How could the top branch have a voltage drop of 2k V and the bottom branch a drop of 2,000.8 V?
 

Ian0

Joined Aug 7, 2020
6,294
Easiest to think of it as follows:
For negative voltages the diode will not conduct, and the gain depends only on one resistor.
For positive voltages the diode conducts and the gain depends on the two resistors in parallel.
That works for perfect diodes.

For real diodes, small positive voltages (<0.7V) don't make the diode conduct, and the gain depends only on one resistor, just the same as the negative voltage case.
 

crutschow

Joined Mar 14, 2008
30,791
And knowing that the voltage across the top resistor and bottom (resistor + diode) is the same I am confused on how there could ever be a situation where the diode could have enough voltage to turn on and keep true to basic circuit theory
Then you are confused about basic circuit theory.
True, the voltage across the top resistor and the bottom resistor-diode are the same, but your error was, making the current through the bottom resistor the same as the top resistor, and not having any voltage across the diode.
(Another way to look at it: for a voltage drop across the resistor there must be a current, but if there's no diode current, then there is no resistor current, obviously a contradiction).
Since there is a voltage drop across the diode then the voltage drop is not the same across both resistors.
That's the failure in your application of circuit theory.

So the answer is that the bottom resistor will have a drop equal to the total minus the diode drop (Vt = Vr -Vd), thus its current will be less than the top resistor.
That makes everybody happy including the circuit theory.
 

Thread Starter

JLoundy

Joined Sep 15, 2022
4
True, the voltage across the top resistor and the bottom resistor-diode are the same, but your error was, making the current through the bottom resistor the same as the top resistor, and not having any voltage across the diode.
So would the current divider formula I2 = Iin * ( R1 / R1 + R2) not be accurate then? Or would the diode simply never be on as there is never enough voltage supplied?
Since the ideal diode has no resistance and resistor 1 and 2 are equal how would the current not be evenly divided? per my example, 1A Iin is divided into 0.5 A in both branches? and therefore an equal voltage drop across both resistors.
 

WBahn

Joined Mar 31, 2012
27,395
Minor point before I continue: Most silicon diodes will drop about 0.6 V to 0.7 V when forward biased under normal currents (though the drop can get up into the ~1 V range with heavy currents). Most design and analysis that assumes a fixed voltage will use 0.7 V unless there's a good reason to do otherwise.

Forget about diodes for a moment. Consider the following circuit:

1663311346312.png

If you know what Io is, what is Vo?
 

WBahn

Joined Mar 31, 2012
27,395
Vo would be -0.7v I believe.
So you are saying that the Vo would be -0.7 V regardless of what Io is?

You REALLY need to step back and learn basic circuit analysis skills.

I'm serious about that. If you lack the fundamental skills needed to work even problems as simple as the one I gave you, then you will not be able to progress to harder problems. If you somehow manage to muddle through this course without rectifying this situation, you will just be even less prepared to handle the next course. You will be digging an ever deeper hole and eventually you will fail out of the program. So go back to the very beginning of your circuit analysis courses and work your way through, being as slow and deliberate as needed to get those skills down cold.
 
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DickCappels

Joined Aug 21, 2008
9,317
So you are saying that the Vo would be -0.7 V regardless of what Io is?

You REALLY need to step back and learn basic circuit analysis skills.

(some text removed for clarity)
How real-life diodes work is a little bit beyond beginner level. Most likely @JLoundy came here for assistance in learning that. I applaud his effort to learn more of the detailed operation of this kind of circuit.
 

WBahn

Joined Mar 31, 2012
27,395
How real-life diodes work is a little bit beyond beginner level. Most likely @JLoundy came here for assistance in learning that. I applaud his effort to learn more of the detailed operation of this kind of circuit.
His questions have never touched on real-life diodes (unless I missed something), but rather nothing more than a diode that exhibits a fixed voltage drop when forward biased.

The underlying problem is that his basic circuit analysis and reasoning skills are not up to even tackling the simpler problem that was given, which comes well before even looking at any model of a diode. Unless he gets a handle on that, he his setting himself up for major problems that will only become harder and harder to overcome and catch up from the longer they go unaddressed.
 

crutschow

Joined Mar 14, 2008
30,791
So would the current divider formula I2 = Iin * ( R1 / R1 + R2) not be accurate then?
Yes, because it does not include the diode voltage drop.
Why do you insist on ignoring that?
Since the ideal diode has no resistance and resistor 1 and 2 are equal how would the current not be evenly divided?
How could it?
Again, even though it has no theoretical resistance, it does have a theoretical voltage drop which you can't ignore.
All voltage drops in a circuit must be represented, not just those caused by resistors.

Why is that so hard to understand?
 
Last edited:

panic mode

Joined Oct 10, 2011
2,301
maybe try to work it out the other way... what if you look at circuit in post #8 and assume Vo is known, could you work out what the three currents are? do you still think I1=I2?
 
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