Diode as Regulator Doubt

[elecLear78]

I was trying to analyze the diode circuit as regulator. I have drawn a circuit consisting of 5V- DC power source, the diode and the Load resistance. I replaced the diode with the ideal model, and since the diode is forward biased i replaced it with a constant voltage source of 0.7V. Now the problem is I always get a current of 0.7/RL irrespective of the 5V supply or the load resistance. Is it the correct way of analyzing what am i missing?

 

[MikeMl]

You must add a resistor between the positive pole of the battery and the anode of D1. The series resistor provides a voltage drop of Vbat-Vd. This is how the diode/resistor regulates; the excess voltage appears across the series resistor...

Note that you have built a thermometer.

 

[steveB]

You placed two ideal voltage sources in parallel. That is not allowed.

Also, 5 volts directly across a more realistic diode model will create an impractically large current through the diode.

That is also not a good way to make a voltage regulator for stability reasons. The diode voltage depends too much on temperature.

 

[elecLear78]

Suppose if i modify the circuit considering then the current through the Load resistor is still the same 0.7/RL and the output voltage is always 0.7V. Will this circuit ever fail or the diode will keep supplying the current to load? That is when the diode will fail and stop regulating. Please help.

 

[MikeMl]

Suppose if i modify the circuit considering then the current through the Load resistor is still the same 0.7/RL and the output voltage is always 0.7V. Will this circuit ever fail or the diode will keep supplying the current to load? That is when the diode will fail and stop regulating. Please help.

The diode does not "supply current to the load". It shunts current around the load. It, in effect, wastes some current to achieve the regulation.


Potential failure modes are power dissipation in the series resistor (P=IE) exceeding the resistor rating, and the same for the diode. The series resistor (your R4) passes the sum of the load current and the diode current. The ratio of load current to diode current should be set about 1:1.

As pointed out; this is a very poor means of "voltage regulation". There are much better alternatives.

 

[elecLear78]

But whenever I see standard text books they are replacing the diodes with a fixed voltage of 0.7v but not considering diode resistance. So based on circuit should I change the diode model keeping the practical things on mind. Am I correct in this? please help. I was learning about diodes and I came across this circuit and wanted to know about it. I am not going to use this circuit for practical purposes. But I am learning new things from your valuable suggestions.

 

[ronv]

Suppose if i modify the circuit considering then the current through the Load resistor is still the same 0.7/RL and the output voltage is always 0.7V. Will this circuit ever fail or the diode will keep supplying the current to load? That is when the diode will fail and stop regulating. Please help.View attachment 93115

The diode will stop regulating when the voltage across it falls below 0.7 volts. That will happen when the battery voltage is to low or the load current is to high.

 

[steveB]

But whenever I see standard text books they are replacing the diodes with a fixed voltage of 0.7v but not considering diode resistance. So based on circuit should I change the diode model keeping the practical things on mind. Am I correct in this? .

Yes, you are correct. You can add a series resistor to the 0.7 V voltage source to make it behave more like a real diode. An even better model is to use the Shockley diode model for the ideal diode, and then add a resistor in series with that.

https://en.wikipedia.org/wiki/Diode_modelling

 

[Tony Stewart]

Remember that fixed Voltage active devices like batteries and Diodes and power sources must all have an Effective Series Resistance (ESR). We can't always ignore it and use Ohm's Law to determine the current flow when including these devices are in parallel and must add external series R's to limit the current.

To compare passive (diode) vs linear active ( feedback loop) performance for regulators, there are 2 basic attributes. Source/ Load regulation error and ESR

In all regulator methods, the" load regulation error" on source voltage is important as the range of output voltage over range of output current. THis is also the ESR of the source.
The % of load regulation error is defined as the ratio of Source ESR/Load Impedance.
If the load is static, then the error is DC only, if the load is dynamic, then the impedance ratio induces a % of output DC as AC drop or ripple we call Load Regulation Error.

All Passive diode regulators must draw more current than the load to remain saturated. This can cause all power consumed by load to be dissipated in Diode when there is no load and thus a potential thermal and efficiency issue.

In Active Bandgap Regulators the idle current is only the minimum required to saturate the diode and bias the buffer, which for FETs can be very low current and much lower dropout than bipolar Darlington types with 2V drop.

The ESR of the Zener, LED or silicon Diode can be predicted as the inverse of the Max continuous power rating of the device. ESR= 1/Watts is a good approximation for If>=/10% of max current.

Bandgap Reference Diodes are very temperature stable by design but must be buffered since they can only supply <=10 mA

To use a White LED's as a 3 Voltage Regulator, assuming your bias current is < max rating and greater than MCU load. Then LED will dim when it consumes more current at max clock rate or after waking up. Depending on size, this can range from 20mA 5mm types to 1A 3Watt types.