# Why is diode D2 affected by the current of Vin and the choice of R1?

I was wondering why D2 will not always forward voltage drop .6 volts? Ground is at 0. What happens if R1 is incorrectly chosen? Could too much current through D1 push back on D2 and prevent flow from ground through it? I would think you will always get current through D2, just not as much under some circumstances, and compared to D1. Since you get current through D2, you should always be .6V at A... so .. why the extra concern with R1.. Thanks for any help

• Horowitz & Hill, 2nd edition, page 52. And the question is utterly useless without the previous page so that info should have been in there. – Brian Drummond Apr 2 '17 at 18:07

First you have to understand that real diodes are not switches. They are actually complex, non-linear, voltage controlled resistors.

Lets look at this in steps...

When Vin = 0V, if R1 is small enough there is enough current through D2 to allow it to pass the knee in the diode V/I characteristic so it "turns-on". That is, it presents itself as a low resistance. This can be measured as somewhere around 0.6-0.7V drop across D2, if it is a silicon diode. What that voltage actually is depends on how much current you pass through the diode, and is shown as the Vin = OV bias point on the curve below.

If you raise V1 more current will flow through D1, this will raise the voltage across R1. Which in turn drops the voltage across D2. The bias point indicated on the image above therefore moves down on that curve.

D1 on the other hand is biased below the knee when Vin=0 due to the current limiting of R. When Vin=0V the voltage from Vin to the bottom of D1 will also be the voltage across D2. However because R is in that circuit, the current flowing through D1 is a lot less than that passing through D2.

As Vin increases the effective resistance of D1 falls exponentially as Vin increases. This draws more current through R and attenuates the signal.

At some point, the voltage across D2 will fall low enough that the bias point will move below the knee in the diode's I/V characteristic and it will begin to show a higher and higher resistance. At that point the circuit no longer works as intended.

For this circuit to work as intended it is required that D2 be "biased" far enough up that forward current curve to give it sufficient margin to allow the current to drop and still keep it in the on-state up to the expected maximum signal voltage (Vin).

This is done by selecting the appropriate value of R1.

• Very good answer...just gotta drop the shouty OVER-emphasis here and there and it'll be perfect, your words work fine as they are - HONEST!!! :-) – TonyM Apr 2 '17 at 16:22
• Thanks @TonyM. I only saw one Over-Emph. Did u see another? Feel free to edit if you do. – Trevor_G Apr 2 '17 at 16:25
• I'm not editing it...be like tippexing Oliver Twist in front of Dickens, it's solid as a rock :-) – TonyM Apr 2 '17 at 16:28
• Unfortunately, the circuit is not used with the bias enough to be in the linear delta Vf range of D2 as shown by @Trevor. The image shows that the circuit is used to provide logarithmic attenuation of Vin. To do this D2 would be biased at just the knee of it's characteristics ...and Vin would use from the knee and below the characteristics in D2. Reading this may help with the math and the Vf explanations, though this uses an Op-Amp for higher accuracy: kennethkuhn.com/students/ee431/logarithmic_circuits.pdf – Jack Creasey Apr 2 '17 at 16:35

I was wondering why D2 will not always forward voltage drop .6 volts?

Suppose Vout were at +3 V.

Then if D2 were still forward biased, there would be 3.6 V forward bias on D1, indicating a contradiction in your model.

What actually happens is if a high enough voltage is applied at Vin, D1 becomes forward biased, with its current also flowing through R1. By Ohm's law, at some point the potential across R1 becomes enough to raise node 'A' above -0.6 V, and D2 is no longer forward biased.

• According to the text on the previous page, Vout could never be at +3V. It is near ground. The statement that continues across from the previous page is ... (making Iin accurately proportional to Vin, incidentally). This is because B is near ground. – Jeffrey Edward Messikian Apr 2 '17 at 15:45
• @Jeffrey, that's not true based on the information you've given us. For example, if R is 1 ohm and R1 is 1 kohm, and you apply 3 V at Vin, then Vout will be very near 3 V. If R is much higher valued than R1, then the output voltage would be limited to not rise much above 0 V, but you never specified the R values so there's no way we could know that. – The Photon Apr 2 '17 at 15:54
• Sorry, I though I gave all the relevant info. The previous page stipulated that Vin has to be much larger than .6V So... assume that Vin is 100 and V is at 5... So, because -5 is less than 0, you should always have current and a voltage drop through D2? Under what circumstance can changing the value of R and R1 affect the current through D2 . I guess I would also assume that the voltage is stiff. – Jeffrey Edward Messikian Apr 2 '17 at 16:25
• If R << R1, then Vin at 100 V results in Vout ~ 100 V. If R >> R1, then Vout is limited to not much above 0 V. This is basically what they mean when they say "R1 should be chosen so that...", but it isn't true of the circuit in general, until you choose R1 properly. – The Photon Apr 2 '17 at 16:37
• Lets say R was a wire, 99.999V at point B will forward bias D1. D1 is now basically a wire with a little resistance. Now at point A you have 99.999V - .6V. The 99.399 volts at A has to get to -v through R1. R1 might get hot, but all that would be dropped across the resistor. For D2, ground is at 0. Through D2, ground must get to -V. So, when ground goes through D2 it becomes -.6V. What wins out at point A, the 99.399 volts or the .6V? Is there some cirtcuits law that would block D2 that I should know? – Jeffrey Edward Messikian Apr 2 '17 at 18:33