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In the figures below, we see two methods of biasing a bipolar transistor: the first uses a voltage divider, and the second employs a zener diode regulator.

Since the voltage regulator appears far more frequently, I'm wondering if there exists some scenario or reason that would compel me to use a zener diode regulator instead.

Biasing using a voltage divider

Biasing using a zener regulator

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  • \$\begingroup\$ I think you should have asked also: Why is a transistor required at all? Why not just use a series resistor for the load? Obviously the supply voltage is assumed not to be stable or the load has variable resistance and the transistor circuit is supposed to provide stable current for the load. For further explanation see my answer below. \$\endgroup\$ – Curd Jul 20 '15 at 14:37
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Based upon the schematics you post most answers here have a good amount to them concerning resistor versus zener.

However, to offer a fair and balanced view towards your specific drawings I'll add a little bit of extra information towards current sinking.

What happens in your circuits is that the transistor is set up as an emitter follower. It means it will want to keep the voltage on its emitter at 0.7V below the base voltage if it possibly can. It will throw its saturation/amplification curves into workhorsing that to work.

If you put a fixed voltage across a fixed resistor, this resistor will want a current to flow. So as long as your load and transistor can support the "required" current, the base voltage of the transistor will determine the current through the load.

Given a set range of supply voltage the current into the base will be quite predictable, so you can set it with both a resistor divider and a zener diode with the proper maths.

Why would you choose one over another?

Well, if you use two resistors, the base voltage will be related to the input voltage. If the resistors are both the same value (low enough to ignore the base current), the base voltage will be 5V at 10V supply, but 6V at 12V supply. That sounds like a problem, but in many cases, by choosing the right balance between the resistor divider it can give a desired effect of limiting the current in the load at power-up of a low-power circuit. It can also give a response to incoming voltages that you want, if you have a control voltage of 6V to 60V, for example, you can turn that into a current curve using just a couple of resistors a transistor and a resistor in the emitter path.

Of course having 50 values of resistors laying around is a very common thing, which adds to the usability of the resistor-only circuit, while usually you have to happen to have the right zener diode.

If, in stead, you have a wobbly power supply, but need a stable current, you should probably use a zener diode.

The zener diode does change voltage over different currents, but if you select a zener diode that is specified to be 5.1V at 5mA, you can assume in your calculations that over 3mA to 6mA it will be relatively stable. You can also calculate the voltage drift using the differential resistance mentioned in the datasheet, but I think that's for a different answer at a different time.

So, if you want the base voltage to stay at 5.1V, you select the 5.1V at 5mAzener, if then the supply is 12V, you select a resistor so that 5mA will go through it:

R = V/I = (12V - 5.1V) / 0.005A = 1.38kOhm.

Say that you could get this value exactly (you can round off to 1.2 or 1.5k because of the relative stability of the zener near the chosen set-point), the voltage can go from:

V = 5.1V + (R * Imin) = 5.1V + (1380 * 0.004) = 5.1V + 5.52V = 10.62V

to

V = 5.1V + (R * Imax) = 5.1V + (1380 * 0.006) = 13.38V

Before you have to start thinking about checking what the zener does at lower or higher current levels, so it adds a lot of stability with respect to the supply voltage, which then makes the current through the load much more stable.

An alternative (that's even more stable) would be simply two standard diodes, like 1N4148, in series to create a fixed voltage of 1.2V to 1.4V (depending on the current range you are looking at). Using those in the forward direction could give you good stability from 0.1mA to 5mA or from 1mA to 10mA, etc. But it's a bit of a black art to many beginning designers to get to the right set-point calculations. Especially since diode datasheets don't always mention all that data, like forward voltage vs low current.

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  • \$\begingroup\$ I have some problems to follow your arguments. Certainly, in your example the assumed base voltage (5.1V) is the result of another assumption for Vbe (0.6...0.7V). Because this is a very rough assumption, it does not make much sense to argue with exact resistor values and corresponding tolerances. These problems/uncertainties can be (and are, in practice) drastically reduced applying dc feedback (emitter resistor Re). \$\endgroup\$ – LvW Jul 20 '15 at 12:48
  • \$\begingroup\$ @LvW If you use the proper datasheets properly, the base to emitter drop can be calculated using the proper formulae down to the milivolt, I don't see why that would be a problem at all. Feedback is a very modern invention by comparison and not at all always a feasible solution, especially when you start to account for reflected impedance and loop phase margins that might even cause unstable behaviour in transistor systems. \$\endgroup\$ – Asmyldof Jul 20 '15 at 13:21
  • \$\begingroup\$ No - I don`t think that a data sheet can give you reliable information on the Vbe value which is necessary for a certain current Ic. More than that, I would not classify "feedback" as a "very modern invention". The principle of feedback was introduced by H. Black already in the 30th of the last century - and it is applied in each transistor amplifier. Otherwise, it would be destroyed due to thermal feedback (self-heating). \$\endgroup\$ – LvW Jul 20 '15 at 14:36
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A zener diode bias means that the bias voltage is almost independent of the supply voltage. This could be useful in any circuit with an unregulated power supply. By contrast a resistor divider produces a bias voltage proportional to the supply voltage.

The dynamic impedance of a zener diode is small (the voltage across the diode does not vary much, even with large changes in current through it). This makes the zener voltage divider very 'stiff', meaning that the input impedance of the amplifier is low. This is generally undesirable, which may be one reason why the zener diode bias is unusual. (Another reason may be that zeners are typically more expensive than resistors.)

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For biasing a BJT based amplifier stage it is common to use a resistive voltage divider. Usually, this divider is chosen to be low-resistance to provide a "stiff" bias voltage. The reason is that the produced base voltage should be as independent as possible of the base current of the BJT which has very large tolerances. On the other hand, these resistors reduce the resulting input resistance of the stage (an undesired effect). For this reason, a trade-off is necessary - resulting, for example, in a current through this divider that is chosen to be approximately ten times larger than the base current.

As an alternative, the desired "stiffness" of the base voltage can be realized with a Z-diode which is able to produce a very constant dc voltage - nearly independent of the current through the upper resistor (between base and supply voltage). It depends on the specific application if the resulting dynamic input resistance is acceptable or not. If not, the bootstrap method could be applied.

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  • \$\begingroup\$ You say that for an amplifier stage (i.e. for AC signal), "it is common to use a resistive voltage divider". Would you agree that it is more correct to say that it is necessary to use a voltage divider? See Rol's answer below stating that for "... AC applications (capacitively coupled) you cannot use a zener, ...". \$\endgroup\$ – seertaak Jul 20 '15 at 10:29
  • \$\begingroup\$ No, I don`t think it is "necessary" because there are other alternatives - for example just a single resistor between base and supply voltage (or collector node providing dc feedback). \$\endgroup\$ – LvW Jul 20 '15 at 12:41
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It is practically easier to find a pair of resistors of determined values than a zener diode of the reverse voltage you might need.

Moreover, you have to count with slightly higher power dissipation at the diode, because the current is higher than when you use a resistive voltage divider.

For AC applications (capacitively coupled) you cannot use a zener, since the voltage will be almost fix at the base.

You want to use a zener to have a stiff voltage reference for DC operation.

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  • \$\begingroup\$ This doesn't explain why you'd ever use a zener-based bias. \$\endgroup\$ – Nick Johnson Jul 20 '15 at 8:34
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Obviously in this particular application the transistor is used to form a constant current source. I.e. its purpose is to provide stable current through the load even if the main voltage supply (20V or 12V) is not stable. This is achived by negative feedback by the voltage across the emitter resistor given that the base voltage is constant.

Therefore a bias voltage directly derived from the main supply voltage (voltage divider) doesn't work. A voltage reference (Z-diode) independent of the main voltage supply is required.

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  • \$\begingroup\$ Good point (usage not as an amplifier, but as a current source). \$\endgroup\$ – LvW Jul 20 '15 at 14:38

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