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I want to construct a simple voltage regulator like this one:

schematic

simulate this circuit – Schematic created using CircuitLab

However there is something peculiar: the voltage on the emitter is forced to be some value by Vbe and diodes total voltage drop. If then the load changes the voltage will be the same: the current should change to accomodate for the new load. Does this not contradict current gain principle? How can a transistor both supply an amplified current to the emitter and a fixed voltage?

This 'paradox' could be avoided if the load current could change the base current, of course with hFe taken into account. Is this were the answer lies, on the fact that the current on the diodes can change depending on emitter load thus providing a way for changing the base current?

This may seem a strange question to some but I fundamentally know that the collector current can be changed by the base current by hFE amplification when the transistor in in the active region. If it were true for the collector current to change the base then everything would in vain.

Question #2 Why does this configuration(common collector) has high input impedance (reffering to the base of the transistor Q1 in the circuit above)? I imagine because the current needed to drive the base is very small and equivalent (taking into account the base voltage) to a very large resistor. Is this correct?

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    \$\begingroup\$ First, do you understand how a Zener diode shunt regulator works in the first place? \$\endgroup\$
    – G36
    Apr 14, 2020 at 10:26
  • \$\begingroup\$ How can a transistor both supply a fixed current to the emitter Is that happening? I think not. How is the emitter getting a fixed current? What fixes the current? \$\endgroup\$ Apr 14, 2020 at 10:27
  • \$\begingroup\$ The resistor I choose once it is calculated it will provided a current to the transistor which will be amplified by hFE. Thus the emitter will have the collector and base currents added \$\endgroup\$ Apr 14, 2020 at 10:29
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    \$\begingroup\$ There is no paradox - you are misled. \$\endgroup\$
    – Andy aka
    Apr 14, 2020 at 10:31
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    \$\begingroup\$ Uhm in the picture it says "Zener diode". Is your picture wrong then? Then use a proper schematic. Use the schematic drawing tool. Also it does not matter if you use a zener or a stack of diodes as they will behave in a very similar way. \$\endgroup\$ Apr 14, 2020 at 10:33

2 Answers 2

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It is known that this configuration amplifies the current. How can this happen when it also provides a ....'fixed' voltage on the emitter?

Because..... it doesn't quite provide a fixed (or constant) voltage on the emitter.

As the load current increases....

  • The emitter voltage drops slightly and
  • That slightly increases the base-emitter voltage and
  • That, in turn causes more current to flow into the base and
  • That causes more current to flow in the emitter load and,
  • That has the effect of regulating the emitter at a roughly (not perfect) constant value.

It's a reasonable voltage regulator but not perfect. It relies on negative feedback also.

The "system" is driven by slight changes in \$V_{BE}\$. Those slight changes can cause quite significant changes in base current.

Question #2 Why does this configuration has high input impedance? I imagine because the current and voltage needed to drive the base are very small and equivalent to a very large resistor. Is this correct?

And

sorry. bjt base input. that should be high impedance – sergiu reznicencu

If the BJT can avoid saturation by ensuring that the collector voltage is a few volts higher than the emitter voltage then, the current into the base is largely determined by emitter load current ÷ hFE. The impact of this is that the impedance looking into the base is approximately hFE x emitter resistance. "High impedance" doesn't really mean anything but it's certainly hundreds of times higher in impedance than the emitter load resistor.

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  • \$\begingroup\$ Perfect. That's exactly what I thought was happening. Question 2? \$\endgroup\$ Apr 14, 2020 at 10:46
  • \$\begingroup\$ If you can specify where the input is I will try and explain @sergiureznicencu. \$\endgroup\$
    – Andy aka
    Apr 14, 2020 at 10:49
  • \$\begingroup\$ sorry. bjt base input. that should be high impedance \$\endgroup\$ Apr 14, 2020 at 10:50
  • \$\begingroup\$ You mean (emitter load current)/hFE on the base. Might not be obvious for some people \$\endgroup\$ Apr 14, 2020 at 11:01
  • \$\begingroup\$ Is it obvious to you @sergiureznicencu - yes I shall fix that error d'oh! \$\endgroup\$
    – Andy aka
    Apr 14, 2020 at 11:07
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In short:

To understand how this circuit work first you must understand the limitations of a simply Zener regulator

enter image description here

As you can see the Zener current is equal to

$$I_Z = I_S - I_L$$

So as \$I_L\$ varies the Zener current also varies. And this will cause the voltage across the Zener and the load voltage also varies. As \$I_L\$ increases the \$I_Z\$ decreases and Zener voltage \$V_Z\$ will also slight changes(decreases).

Thus the max load current that can flow through that load can not be greater than:

$$ I_{Lmax} = I_S - I_{Zmin}$$

The numerical example should help here.

Lets us assume that we have a \$5.1V\$ Zener diode and we pick \$R_S = 1k\Omega\$. And the input voltage is equal to \$V_{IN} = 10V\$. So, for no-load condition (\$I_L = 0A\$) we have a this situation

$$I_S = I_Z = \frac{10V - 5.1V}{1K\Omega} = 4.9mA$$

But now if we connect the load resistor \$RL = 2k\Omega\$.

The Zenner current will drop to the new value:

$$I_Z = I_S - I_L = 4.9mA - \frac{5.1V}{2k\Omega} = 2.35mA$$

If we further increase the load current (by decreasing the load resistance) the Zener current will further decrease.

And when the load current reaches \$4.9mA\$ there is no current left for the Zener diode. For \$RL < 1k\Omega\$ we have \$I_Z = 0A\$

Thus to reduce the loading on the Zener we add a current amplifier.

enter image description here

$$V_{OUT} = V_Z - V_{BE}$$

The BJT provides a current gain equal to

$$I_L = I_E = I_B + I_C = I_B + βI_B = (β+1)I_B$$

So, now any change in the load current will cause (β+1) smaller change in the Zener current. Because now:

$$I_Z = I_S - I_B = I_S - \frac{I_L}{β+1}$$

And this allows us to increase the maximum load current that we can draw from the circuit.

And again the numerical example:

Let us assume that \$β = 99\$ and \$V_{BE} = 0.6V\$

And because the BJT added the Vbe voltage drop we need to increase the Zener diode voltage to \$5.7V\$ to get \$5.1V\$ at the output.

Now lest as connect \$RL = 2k\Omega\$

The Zenner current will only drop to:

$$I_Z = I_S - I_B = 4.9mA - \frac{\frac{5.1V}{2k\Omega}}{(β+1)} = 4.9mA - 22.5μA ≈ 4.87mA$$

So now thanks to BJT we can increase the max load current \$(β+1)\$ times.

From \$4.9mA\$ (without BJT) to \$490mA\$ (with BJT)

Thus the minimum load resistance is now equal to

$$R_{Lmin} = \frac{5.1V}{0.49A} ≈ 10.5Ω$$

Without BJT

$$R_{Lmin} = 1k\Omega$$

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  • \$\begingroup\$ Uhm in the picture it says "diode". Is your picture wrong then? - Bimpelrekkie This applies to you too. It also makes my question look wrong. \$\endgroup\$ Apr 14, 2020 at 11:07
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    \$\begingroup\$ It also makes my question look wrong Yep, that's because it was inconsistent, the text says diode, the schematic shows a zener. Instead of saying we need to read carefully, you need to make your question consistent. It is you who wants an answer, remember that. Also, now your question shows only one diode in forward, like that the output voltage will be close to zero. Either draw more diodes in series or use the zener. \$\endgroup\$ Apr 14, 2020 at 11:10
  • \$\begingroup\$ The Zener diode will have much "stiffer" voltage than the ordinary diode.This is why Zener diode fits better in this circuit. \$\endgroup\$
    – G36
    Apr 14, 2020 at 11:14

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