In virtually every practical de circuit containing a diode, there is one simplifying assumption we can make when the diode current is beyond the knee. We have seen (Figure 3-2, for example) that the /- V curve is essentially a vertical line above the knee. The implication of a vertical line on an /- V characteristic is that the voltage across the device remains constant, regardless of the current that flows through it. Thus the voltage drop across a diode remains substantially constant for all current values above the knee. This fact is responsible for several interesting applications of diodes. For present purposes, it suggests that the diode is equivalent to another
familiar device that has this same property of maintaining a constant voltage. independent of current: a voltage source! Indeed, OUI’ first simplified equivalent circuit of a diode is a voltage source having a potential equal to the (essentially) constant drop across it when the current is above the knee.
For a silicon diode, depending on small manufacturing variations and on the actual current flowing in it, the voltage drop above the knee is around 0.6 to 0.7 V. In practice, it is usually assumed to be 0.7 V. For germanium diodes, the drop is assumed to be 0.3 V. Therefore, for analysis purposes, we can replace the diode in a circuit by either a 0.7-V or a 0.3-V voltage source whenever the diode has a forward-biased current above the knee. Of course, the diode does not store energy and cannot produce current like a true voltage source, but the voltages and currents in the rest of a circuit containing the forward-biased diode are exactly the same as they would De if the diode were replaced by a voltage source. (The substitution theorem in network theory justifies this result.) Figure 3-4 illustrates these ideas.
In Figure 3-4(a), we assume that a f rwar=-biased silicon diode has sufficient current to bias it above its knee and that i’ therefore has a voltage drop of 0.7 V. Applying Kirchhoff’s voltage law ~rcund the ‘oop, w have
Our assumption that the voltage drop across the diode is constant above the knee is usually accompanied by the additional assumption that the current through the diode is zero for all lesser voltages. For analyzing the dc voltage and current in a circuit containing a diode, we therefore, in effect, replace the /- V characteristic by one of those shown in Figure 3-5.
The idealized characteristic curves in Figure 3-5 imply that the. diode is an open circuit (infinite resistance, zero current) for all voltages less than 0.3 V or 0.7 V and becomes a short circuit (zero resistance) when one of those voltage values is reached. These approximations are quite valid in most real situations. Note that
it IS not possible to have, say, 5 V across a forward-biased diode. If a diode were connected directly across a +5- V source, it would act like a short circuit and damage either the source, the diode, or both ..When troubleshooting a circuit that contains a diode that is supposed to be forward biased, a diode voltage measurement greater than 0.3 V or 0.7 V means that the diode has falled and is in fact open
Assume that the silicon diode in Figure 3-6 requires a minimum current of 1 mA to be’ above the’ knee of its 1- V characteristic.
1. What should be the value of R to establish 5 mA in the circuit?
2. With the value of R calculated in (1), what is the minimum value to which the voltage E could be reduced and still maintain diode current above the knee?
1. If I is to equal 5 mA, we know that the voltage across the diode will be 0.7 V. Therefore, solving equation 3-7 for R
2. In order to maintain the diode current above the knee, ! mustbe at least 1’mA. Thus,
Therefore, since R = 860
In some de circuits, the voltage drop across a forward-biased diode may be so small in comparison to other de voltages in the circuit that it can be neglected entirely. For example, suppose a circuit consists of a 25-V source in series with a l-kD. resistor in series with a germanium diode. Then / = (25 – 0.3)/(1 kD.) = 24.7 mA. Neglecting the drop across the diode, we would calculate J = 25/(1 kD.) = 25 mA, a result that in most practical situations would be considered close enough to 24.7 mA to be accurate.