0
\$\begingroup\$

I was a little confused with the voltage divider circuit with a diode in parallel to the upper resistor, so I simulated the circuit where the resistance R1 is varied from zero to 10k:

enter image description here

(Green plot is the voltage across the diode and R1 with respect to zero to 10k resistance.)

Normally in this scenario, the diode drop is always expected to be around 0.7V.

At first I thought the voltage across the diode will always dictate the voltage across R1. But as you see below a certain resistor value it is not the case and the voltage across the resistor and the diode starts dropping drastically. How can this be explained mathematically or from a circuit theory point of view?

\$\endgroup\$
3
  • \$\begingroup\$ How can a diode drop 0.7 volts across zero ohms - it would have to produce infinite current. The current through the diode dictates the diode voltage, non-linearly admittedly. \$\endgroup\$
    – Andy aka
    Mar 11, 2020 at 9:42
  • \$\begingroup\$ I have a problem analysing and reaching to a conclusion step by step. Where to start? \$\endgroup\$
    – user1245
    Mar 11, 2020 at 9:57
  • \$\begingroup\$ When I remove the diode and calculate the R1 value for where the voltage across R1 as 0.7V, I found out that is the value where the diode voltage starts to drop drastically. That is not superposition method but strangely works. \$\endgroup\$
    – user1245
    Mar 11, 2020 at 10:33

5 Answers 5

4
\$\begingroup\$

As long as the diode is not conducting, the voltage VR1 is determined by the voltage divider R1 over R2. Increasing R1 will increase VR1, as soon as VR1 reached the diode forward voltage( approx 0.7V) the diode start conducting and will clip the voltage to Vf.

mathematically: value of R1 @ which the voltage will be clipped:

R1/(R1+R2) * Vcc = Vf So R1 = (Vf*R2)/(Vcc - Vf) = (approx) 620 ohm

Of course this is the practical diode model, which is a good approximation. The complete diode model will be more accurate but will also introduce some more math

\$\endgroup\$
1
\$\begingroup\$

How can this be explained mathematically or from a circuit theory point of view?

Consider the ubiquitous 1N4148. The extract below is from the ON semi data sheet: -

enter image description here

Look at the top left graph - it shows a linear-log relationship between forward voltage and current. Basically this part of the characteristic boils down to the approximate relationship of \$V_F\$ rising 100 mV for every ten-fold increase in \$I_F\$.

This continues in the top right graph where \$V_F\$ approximately rises from 500 mV to 600 mV for an \$I_F\$ increase of 100 uA to 1 mA.

Then, there is temperature to take into account (and how that might reshape the characteristic). Or maybe consider using the diode equation: -

enter image description here

Better still, go to your preferred simulator and plot V against I. All formula, sims, models and graphs are just approximations - what you might see on one diode might not be too close to what you see with another diode from the same batch.

\$\endgroup\$
1
\$\begingroup\$

Don't know exactly what supply voltage you are working with but assuming it is 12V, you can explain the behavior as follows.
The potential drop across the diode is 0.7V ONLY if it is turned on. It needs a potential drop of 0.7V between the p and the n junction to turn on. This can be seen in the figure below which shows the piece-wise linear model of Diode (green). The x-axis is the voltage drop across the diode and the y-axis is the current through it. Clearly, for voltages less than Vt= 0.7 current is zero implying the diode is off. Thus, if the resistor is too small the potential drop across the diode falls down and it turns off resulting that all the current flows from the resistor R1.

enter image description here
Calculate the value of R1 which will make the diode off as follows: $$I = \frac{12}{R_1 + 10k}$$ $$V_d = IR_1 = \frac{12R_1}{R_1+10k} = 0.7$$ $$R_1 = 0.62k \Omega $$ Here, \$V_d\$ is the voltage drop across the diode which is set to 0.7 below which it turns off.

\$\endgroup\$
0
\$\begingroup\$

You can use a simplest diode model for analysing this. If there is less than 0.7V over the diode because of value of R1, the diode is off. If there would be over 0.7V over the diode due to R1 value, the diode is on, and it will have 0.7V over it.

\$\endgroup\$
2
  • \$\begingroup\$ Can you mathematically show and derive for what estimate value of R1 diode voltage starts dropping drastically? \$\endgroup\$
    – user1245
    Mar 11, 2020 at 9:58
  • \$\begingroup\$ When I remove the diode and calculate the R1 value for where the voltage across R1 as 0.7V, I found out that is the value where the diode voltage starts to drop drastically. That is not superposition method but strangely works \$\endgroup\$
    – user1245
    Mar 11, 2020 at 10:35
0
\$\begingroup\$

Notice from the plots in Andy's answer, the change is diode voltage is about 110 milliVolts per 10:1 change in diode current, over a 10,000:1 range.

In his leftmost plot, from 1uA to 100uA, the diode voltage increases form 275mV to 500mV or 2 decades increase in current. That is 225mV/2 or 112 milliVolt/decade.

The diode-equation has term of Q/NKT,

which becomes 1/0.026 volts if N = 1.

The plot(s) from ONNN on their 1N4148 show their "N" is nearing TWO.

Summary: from 1uA to 10,000uA, the plots show :

275 to 730mV, or 455 millivolt over 4 decades of current.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.