Design Diode Voltage Regulator for 1.5V 150 Ohm Load

[sandy.bridge]

Homework Statement

Design a diode voltage regulator to supply 1.5 V to a 150 Ohm load. Use two diodes specified to have a 0.7 V drop at a current of 10 mA. The diodes are to be connected to a +5V battery and a resistor R. Specify the value for R. What is the diode current with the load connected?

The Attempt at a Solution

I'm assuming that we determine R with merely the voltage source, resistor R, and the 2 diodes in series.

Thus, R=(+5V-1.4V)/10mA=360 Ohms
However, this puts the output voltage at 1.4 V

Thus, after this point, we attach the load resistor in parallel to the network and the voltage across it will be 1.5V as desired.
The load resistor will draw a current I=1.5V/150=10mA, thus the current through the diode is decreased by this much and thus I=0 when attached to the load.

Does that seem right?

 

[rude man]

'Fraid it doesn't.

First, the current thru the load has to be 1.5V/150 ohms = 10 mA, right?

What if you string the two diodes in series from the 5V supply, then what would you have to do to get 1.5V across your 150 ohm load?

 

[sandy.bridge]

I was under the impression that diodes were placed in parallel to the voltage source when used as regulators. At least any of the examples in my textbook use them in that manner.

 

[sandy.bridge]

Okay, I think I may have an idea. If the load resistor requires 10mA to give a 1.5V drop, and the diodes also require a total of 10mA to provide ~1.5V drop, then the source current would be 20mA total and thus the resistance R=175.

 

[rude man]

I think we're in a semantics trouble-zone here.

You certainly can't put diodes in parallel with a voltage source. You would burn out the diodes in a jiffy!

Can you draw your proposed diagram, scan it into a PDF and attach it, or use some other app?

 

[sandy.bridge]

 

[rude man]

Well, you didn't make 1.5V across your load, did you? You fell 0.1V short.
Plus, your diode current is close to 5V/0.18 ohms = 28A which is a PRETTY big amount of current thru those poor little diodes.

Do this please: connect your diodes in series as they are now, except connect the anode to the 5V supply's + end, the connect a resistor R from the diodes' cathode end and connect the other end of R to your load. Complete the circuit by connecting the other (low) side of your 150 ohm load to the 5V - terminal. Call this terminal zero volts.

Now, how would you calculate R such that i = 10mA? Would i = 10mA satisfy the requirement for the load voltage (+1.5000V)? Would the diode current be close to the level at which the diode drop (0.7V) is specified?

 

[sandy.bridge]

Is it going to matter if the question states to attack the problem considering "The small-signal model"?

 

[sandy.bridge]

For example, if I were to take the question one step further: what is the increase in output voltage when the load resistor is disconnected? What changes occur when the load resistance is decreased to 100-ohms, 75ohms, etc.

When I did as you told me to, I got R=210. However, disconnecting the load doesn't increase the output voltage, it stays constant at 1.5 V.

 

[rude man]

Then you didn't wire up as I suggested.

You got the value of R right. When you remove the load, the output voltage will be 5V. That's because without the load, i = 0 so there is no voltage drop possible across either R or the two diodes.

Don't even try to calculate the output voltage for any other condition than i = 10mA or i = 0. That's because you don't know what the voltage drop across the diodes is for a given current. (As a practical matter, that drop will be close to 1.4V for a wide reange of current, say i = 5mA to 50 mA but that depends on the particular diode type.)

P.S. this has nothing to do with "small signal analysis". That only applies to ac situations. This is a dc problem.

 

[sandy.bridge]

Thats weird. I wonder why it would even say anything regarding the small signal analysis...

 

[sandy.bridge]

Here, I'll quote the question for you. "What is the diode current with the load connected? What change results if the load resistance is reduced to 100-Ohm, To 75-Ohm, or to 50-Ohm? (Hint: use the small signal diode model to calculate all changes in output voltage.)

 

[rude man]

OK, I see. They must have given you a linear model for the diode's i-V characteristic, a plot of i vs. V, or whatever. Please let me know what they gave you since there is no one standard "small-signal" model for a diode. Then we can finish up.

P.S. if I were to do it, I would use i = i0exp(qV/kT),
q = electronic charge, k = Boltzmann constant, T = Kelvin temperature. Diodes all have different i0 values. The value kT/q is about 26 mV at room temperature. Since we know that at V = 0.7V, i = 0.01A, we can deduce i0 = i/exp(V/26) = 0.01/exp(700/26) = 2e-14A. The loop equation would then be 5 = 0.052ln(i/i0) + 210i + RLi where RL is your load resistance. This is a transcendental equation which needs Newton-Raphson or whatever to solve, so that's why they give you a "small-signal" model for the diode, to avoid such an equation.

If you don't understand the last paragraph, don't worry, just send me the diode model they gave you.

 

[sandy.bridge]

"Design a diode voltage regulator to supply 1.5 V to a 150-Ohm load. Use two diodes specified to have a 0.7V drop at 10mA. The diodes are to be connected to a +5V supply through resistor R. Specify the value for R. What is the diode current with the load connected? What is the increase resulting in the output voltage when the load is connected? What change results when the load resistance is reduced to 100-Ohm? To 75-Ohm? To 50-Ohm? (Hint: use the small signal diode model to calculate all changes in output voltage.)"

That's quoted directly from my textbook. They doing supply any more information, and they don't provide an image.

 

[rude man]

They say "use THE small-signal diode model ...". Sure sounds to me like they have one in mind & that it must be given to you. Looking thru that chapter in your textbook, what do they say about diodes?