# Replacing zener diode with a resistor in current limiter circuit

Can we use a resistor to replace the zener diode in this current limiting circuit ? Why do you think using a simple resistance is not enough to trigger the base ?

• Can you limit the current going through the resistor? Commented May 19, 2016 at 17:19
• I thought zener's sole mission is to provide stable base voltage, didnt know that it helps limiting the current. Can you eleborate ? Commented May 19, 2016 at 17:27
• You can't get a stable voltage from a resistor without limiting the current going through it. Commented May 19, 2016 at 17:30
• If you replace D1 with a resistor (call it Rz) then the nominal output (for low currents) will be Vin(R2/R1+R2)-0.7. In other words, the nominal output will not be fixed voltage, but rather a fixed fraction of the input voltage. This is not entirely useless, since it effectively creates a variable ratio pot with high very low output impedance but no corresponding low total resistance. So, yes, you can do it, if the result is what you want. But it won't affect the current limiter portion, and I don't know what "trigger the base" means. Commented May 19, 2016 at 18:23
• Triggering the base is something we say in Turkish for BJT base biasing :) , i translated it without thinking if it's appropriate. Commented May 19, 2016 at 19:15

Q1 is wired as an emitter follower.

• The voltage on its emitter will be 0.7 V below that on the base.
• R1 and D1 provide a maximum voltage for the base.
• Maximum voltage from this part of the circuit will be $V_{ZD} - 0.7$.

Q2 is wired as a current limiter.

• For low currents (high values of R3) the output voltage of the circuit will be that of the voltage regulator less the voltage drop across R2: $V_{ZD} - 0.7 - IR_2$.
• When the voltage drop across R2 increases to 0.7 V or so Q2's b-e junction is turned on. The collector will divert the base current from Q1 causing the voltage at the base of Q1 to drop which, in turn, will cause the emitter voltage to drop.

In short: at currents up to $\frac {0.7}{R_2}$ the circuit will act as an almost constant voltage source (drooping by 0.7 V between 0 and max current). After that it will behave as a constant current source and the voltage will drop with decreasing values of R3. Replacing the Zener diode with a resistor would ruin the constant voltage characteristics of the supply.

But the zener is supplying the base with a steady voltage , how can the base voltage of Q1 drop?

simulate this circuit – Schematic created using CircuitLab

Figures 1, 2 and 3. Failure of Zener regulation with increased current draw.

• Figure 1 shows the situation when there is no load. 5 mA runs through the Zener and none through the output. Voltage is regulated.
• In Figure 2 the current through the Zener has reduced to zero. It is no longer conducting and therefore no longer regulating.
• As the load resistance, R5, decreases further the voltage drops well below the Zener breakdown voltage. Voltage regulation is completely lost.
• But the zener is supplying the base with a steady voltage , how can the base voltage of Q1 drop ? Commented May 19, 2016 at 19:54
• See the update where I explain it in terms of a simple resistor - Zener regulator. Commented May 19, 2016 at 20:35

This is not simply a current limiting circuit. It's a simple voltage regulator or voltage limiter (D1, Q1, R1) with a current limiter (R2 Q2)

If you don't need the voltage limiting function you can just leave out the zener entirely, with no other changes.

Series voltage regulator with current limit.

$R_1$ is sized to turn zener on. If the zener turns on: $V_{OUT} = V_Z - V_{BE} - V_2$. Zener is the reference voltage for $V_{OUT}$.

Current limit comes from $R_2$ and $Q_2$. If load current $I_L$ ($I_3$) is large enough, there will be 0.7V across $V_{BE}$ of $Q_2$, which should decrease base voltage a little, which should decrease $V_{OUT}$, which should decrease $V_{BE}$ of $Q_2$, which turns $Q_2$ off (more likely decreases impact of $Q_2$), which should increase $V_{OUT}$. Limits current to a specific maximum.

If you replace zener with resistor ($R_4$), you have a voltage divider as reference. $V_{OUT} = \frac {R_4}{R_1+R_4} V_{IN} - V_{BE} - V_2$.

• Shouldn't we substract Vbe instead ? Commented May 19, 2016 at 20:02
• Yes. You are correct. Commented May 20, 2016 at 1:19