Problem understanding how a resistor limits current

[franceB]

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

**broken link removed**

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?

 

[IanP]

First of all, if you remove R[load} the current will flow through D1+D2+D3 ..
True..

From the characteristic of a diode you can learn that if one tries to increase the voltage across a diode beyond the V[forward] – current will rise to almost infinity [in theory] ..
In practice, any real diode has its maximum current, if you force higher current to flow – you will just fry it .. keep in mind the power converted into heat in a diode is roughly equal to I[current] x V[forward] ..

What the resistor R1 is doing, it is dropping the voltage from Vin to 3xV[forward] and thus allowing only certain current to flow ..
This current is:

{ V[in] – 3*V[forward] } / R1

Of course, by dropping the voltage from V[in} to 3*V[forward} it heat itself up, so in the real life it has to be rated for certain power, otherwise it will fry itself ..

Rgds,
IanP

 

[franceB]

Thanks for the response.

If I understand correctly why then not just get rid of the resistor and use a lower Vin?

Thanks

 

[s_cihan_tek]

If you use a voltage nearly equal to Vout at the input, what will be the point of using this circuit anyway?

 

[payala]

The point of this circuit is that many times you have a limited selection of supply voltages in your board, maybe +12, +5, +3.3.

So, what do you do if you need to give 8V to some load/circuit?, this is where a circuit like this fits.

Of course, the ideal solution is always to have a supply available of the desired voltage (8V in this case), but more often than not, it is not justified to add another voltage regulator just because some particular load requires that special voltage.

Another reason to do this is that maybe Vin is not fixed... maybe it comes from some sort of output (for example from an output pin of a microcontroller) an the voltage level is too high for the load, you need to reduce the voltage level and still allow that signal to activate/deactivate at will.

Anyhow, if the desired voltage output is higher than 3-5 V, it is usually better to use one single zener diode of the required voltage instead of placing tens of diodes. But that is another topic.