Problem understanding how a resistor limits current

franceB

I am having problems understanding how a resistor resists or limits current flow (or electron flow).



In circuit 1, if I understand correctly the current at Point 1, Point 2 and Point 3 are all the same – that is if you measure the current at each of these three points you will get the same measurement.

Now to circuit number 2.
I got this circuit from
Scherz, Paul. (2000). Practical electronics for inventors. McGraw-Hill/TAB Electronics. Page 132
The caption that goes with the circuit is the following:

This circuit will supply a steady output voltage equal to the sum of the forward-biasing voltages of the diodes. For example, if D1, D2 and D3 are silicon diodes, the voltage drop across each one will be 0.6V; the voltage drop across all three is then 1.8 V. This means that the voltage applied to the load (Vout) will remain at 1.8V. R1 is desinged to prevent the diodes from “frying” if the resistance of the load gets very large or the load is removed.

Now my question is how does the resistor prevent the diodes from “frying” (or breaking down)? If Rload is removed from the circuit, I am under the impression that if you were to measure the current at Point 1, Point 2, Point 3, Point 4 and Point 5 that you will get the same measurement for current at each of these points. If this is so, how is the resistor preventing the diodes from frying?

f95toli

Diodes more or less act like "short cuts" when used as in figure 2 This means that if you remove R1 the current flowing could be very large; potentially large enough to exceed the maxmum current rating of the diodes and fry them.

Okefenokee

An ideal diode in forward bias mode is a short circuit. In other words it's like a wire. It doesn't have any resistance by itself to limit the current. If you you removed Rload then shorted pins 1 & 2, you would have a voltage (Vin) going through a wire with no resistance. I = V/R(0) = infinity. With R1 where it's supposed to be and Rload still removed, you would have a voltage (Vin) going through a resistor (R1). I = V/R(R1) = a finite value.

vk6kro

R1 is chosen so that the maximum current that could flow through the diodes does not exceed their maximum ratings.
This is never arranged to include the load if the load can be removed, because this would risk destroying the diodes.

Adding a load causes a reduction in this current through the diodes.

If the load causes the voltage at the junction of the two resistors to drop below the 1.8 volts required to turn on the diodes, then the diodes will not conduct. So, this sets a minimum value for the load resistance.

Diodes used as a voltage reference like this are not normally run with anything like their maximum current because the heating of the diodes will affect their terminal voltage.

For example, if the supply was 12 volts and R1 was 100 ohms, the maximum current that could flow would be 120 mA without the diodes. So, if the diodes were 1 amp diodes and they were put in series as shown, it would not matter what the load was, the diodes would not be in any danger.

If the load was 500 ohms, it would have 1.8 volts across it so it would be drawing 3.6 mA.
(1.8 volts / 500 ohms).
But the resistor R1 has 10.2 volts across it (12 - 1.8) so it is conducting 102 mA. (10.2 /100)

So there is 98.4 mA flowing in the diodes and 3.6 mA flowing in the load resistor.
If you removed the load resistor, the diode current would rise to 102 mA.

At no stage would the diodes be at any risk.

franceB

Thanks for the responses everyone.

I have a question about this:
How can/does the load cause the voltage at the junction to drop below 1.8 volts?

vk6kro

How can/does the load cause the voltage at the junction to drop below 1.8 volts?

In the above example, if the resistor R1 was 100 ohms and the supply was 12 volts, a load of 10 ohms would have 1.09 volts across it, by voltage divider action.

(ie (10 divided by (10 plus 100)) times 12 )

1.09 volts is not enough to make the diodes start conducting because this requires 1.8 volts, so the output voltage would be 1.09 volts.