# Dont understand simple diodes

Discussion in 'General Electronics Chat' started by brucester, Jan 15, 2013.

1. ### brucester Thread Starter New Member

Dec 24, 2012
19
0
Hi,

I totally dont understand this circuit.

Can someone please tell me why the output voltage is always 740mV?

Is this a clamper circuit?

If someone could explain in simple terms how this works, it would be greatly appreciated.

Thanks

Last edited: Jan 15, 2013
2. ### MrChips Moderator

Oct 2, 2009
18,027
5,644
A diode is a highly non-linear device with a exponential V-I behaviour.
As the voltage increases, the current sky rockets just around 700mV.
This tends to keep the voltage across the diode to around 0.7V, more or less.

3. ### brucester Thread Starter New Member

Dec 24, 2012
19
0
Thanks, but i totally dont understand that

Is it because the diode and the resistor create a voltage divider leaving the forward voltage .74 free to roam past the correct facing diode?

4. ### Audioguru Expert

Dec 20, 2007
10,803
1,219
The voltage across the conducting diode is about 740mV. The remainder of the 9V (8.26V) is across the 90 ohm current-limiting resistor.

5. ### bestos New Member

Jan 7, 2013
1
0
well........... To xplain the simple operation of a diode, it goes thus like this.

A diode is a semiconductor material made up of p type and n type material sandwiched 2gether to form a pn junction.
Think of a diode now as a pn junction.... So what does d pn junction do u may ask?
well its just a barrier dat makes current flow in one direction and prevents current from coming in to the second direction whenever voltage is applied accross it. Just think of a diode as a mechanical valve dat allows for water(current) to flow thru and then prevent water from passing from the other direction and dats all. So wenever current flows and sees an obstruction which is the pn junction barrier then current force reduces. The force of water reduces whenever it comes in contact with a valve or an obstruction.
Hence the mass dat reduced the force traps some force to it self. dats y the forcewuld reduce slightly and hence same with the electric circuit.
whenever a force(voltage is applied accross the diode the diode takes some force to it self due to the barrier it met on the way....) dats why 740mv is accross the diode...... dis is the drop in force due to the barrier acting on the volttage of in the diode
Hope dis makes u to understand diode operation.

6. ### MrChips Moderator

Oct 2, 2009
18,027
5,644
dat is 1 of da wurs xplanashun of a diode i hv evr red

7. ### Audioguru Expert

Dec 20, 2007
10,803
1,219
I hate it wen peepl type with a very strong accent.

8. ### toffee_pie Senior Member

Oct 31, 2009
212
8
christ, text speak is cringeworthy..

9. ### toffee_pie Senior Member

Oct 31, 2009
212
8
another variation of this and very common is a zener diode, this has different ratings ie 4v7.

it will only pass this voltage at whatever current rating it operates on., assuming its forward biased.

Last edited: Jan 15, 2013
10. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
reverse biased.. forward biased, a zener acts as a regular diode...

11. ### toffee_pie Senior Member

Oct 31, 2009
212
8
this is what i mean to explain here

if this zener is reversed in polarity.....?

12. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
It is referred to as the reverse-breakdown voltage...
AAC book

13. ### WBahn Moderator

Mar 31, 2012
23,730
7,295
Let's get back to the OPs quandry.

I have no idea what is meant by,"leaving the forward voltage .74 free to roam past the correct facing diode". I truly have no idea what concept you have in mind and are trying to communicate. So you might try rephrasing it.

Let's think about a resistor voltage divider for a little bit. Let's use easy numbers. I have a 10V supply and I am using a 4kΩ resistor (the top resistor) and a 6kΩ resistor (the bottom resistor).

Hopefully you see that ther output of the voltage divider is 6V (measured from the junction of the two devices to the bottom, which is the negative terminal of the supply).

But why?

Simply put, the two devices together serve to establish a current in the circuit. That current then produces a voltage across the bottom device and the output of the circuit is the voltage across that bottom device.

But how much current is established?

Instead of answering in terms of resistor formulas, let's answer it in terms of concepts that will translate to the resistor-diode circuit. First, because the devices are in series, we know that they have to have the same current flowing in them. Let's imagine that, somehow, there were a current of 0.5mA flowing in them. We can calculate the voltage at the bottom of the top resistor by noting that it is 10V-0.5mA*4kΩ, which works out to 8V. We can also calculate the voltage at the top of the bottom resistor by noting that it is simply 0.5mA*6kΩ, which is 3V. Connecting these two points is a wire (read -- really low value resistor!) and, if this were the actual current flowing, we would have 5V across this wire and it would want a very large current to flow, not just the 0.5mA. So we know that more current is flowing than 0.5mA, but we don't know by how much.

Now imagine that we have a current of 1.5mA flowing. The voltage at the bottom of the top resistor would be 4V and the voltage at the top of the bottom resistor would be 9V. So now we have a current flowing backward in the wire between the two resistors. So we know that less current than this is flowing than 1.5mA but, again, we don't know by how much.

We could keep guessing other possible currents (somewhere between 0.5mA and 1.5mA) and keep narrowing it down and, eventually, we would try 1mA and find that the voltage at the bottom of the top resistor and the top of the bottom resistor are both 6V, which is what is required if we have an ideal piece of wire connecting them.

We can do the exact same thing with a resistor and a diode. The only difference is that we don't use Ohm's Law to compute the voltage that appears across the diode for a given current through it. Instead, we have a logarithmic relationship of the form:

v_d(i_d) = Vo + v_t*ln(i_d/Io)

Where Vo and Io are simply the voltage at a particular reference current and v_t is the "thermal voltage" which is about 26mV at room temperature. Some playing around with this equation reveals that, pretty closely, if the current increases by a factor of ten, that the voltage will only increase by about 60mV.

So let's say that this particular diode has 0.7V across it when there is 10mA flowing through it. Then we would expect 0.76V across it when there is 100mA and 0.82V when there is 1A and 0.88V when there is 10A (and if it is a small signal diode either 1A or 10A would likely smoke the diode). Going the other way, there would be 0.64V if there is only 1mA and there would still be 0.56V across it when there was only 100μA. So you can see that, as a quite useful approximation, we can just assume that if the thing is forward biased and conducting any current at all, that it has about 0.7V across it (and understand that it might vary from that by 100mV or so).