Transistor biasing, Large Resisor values

Question

schematic

^{simulate this circuit – Schematic created using CircuitLab}

In above circuit the base of Q1 acts as a level-shifter from 3V-50V at it's base to 2.8V. The base of Q1 is driven by PNP open-collector outputs thus R2 holds the Q1 in it's off-state when the input signal is not asserted. The output pull-up can be rather large (22k) since speed is not too important but power-consumption is.

This gives a collector current of:

$$ I_{C(min)} = \frac{2.8\mathrm{V}}{22\mathrm{k\Omega}} = 0.13\mathrm{mA} $$

This leads to a small required base current, even when underestimating the current gain and providing a 30% error budget:

$$ h_{FE(min)} = 100 \\ V_{BE(sat)} = 0.75\mathrm{V} \\ I_{B(min)} = 1.3 \frac{I_{C(min)}}{h_{FE(min)}} = 0.0017\mathrm{mA} $$ As noted above the input voltage at the base relative to ground is in the range of \$V_{in(min)}=3\mathrm{V}\$ to \$V_{in(max)}=50.4\mathrm{V}\$ which imposes the following restrictions on \$R_B\$:

$$ R_{1(max)} = \frac{V_{in(min)} - V_{BE(sat)}}{I_{B(min)}} = 1.3\mathrm{M\Omega} $$

According to those calculations \$R_1=1\mathrm{M\Omega}\$ and \$R_2=10\mathrm{M\Omega}\$ would be sufficient.

However, I wonder whether there are any drawbacks choosing resistors in the Mega-Ohm range for transistor biasing.

Unfortunately the BC846S datasheet by NXP does not list a base capacitance. The collector capacitance is given as max. 1.5pF. I assume the base capacitance is in the same range? — Arne, Nov 12 '13 at 10:41

score 1 · Accepted Answer · edited Apr 13 '17 at 12:32

Two drawbacks come to mind:

Firstly, noise. Two kinds really, thermal noise and shot noise. However, if the transistor is being driven into saturation, noise performance is probably not in your list of considerations. If you were making a linear amplifier however, it could be.

Secondly, switching speed. The base is not without capacitance. The Miller effect effectively amplifies this capacitance. Driving the transistor hard into saturation results in storage delay. There are various methods to reduce these slownesses, but all of them are made more effective by driving the base with a lower impedance, which a resistor in the megaohm range isn't. However, slow may not be a problem for your application if slow is fast enough.

score 0 · Answer 2 · answered Nov 12 '13 at 20:06

You did not ask to verify your calculations, but I think these are erroneous:

Assuming h_FE, min=100 seems too optimistic.
I can't understand why R2=10MOhm follows from R1=1MOhm.

Now, to your question. There are few drawbacks in using this high resistances:

Q1 is very sensitive to noise and leakage currents from Vin
Long turn ON/OFF times
Due to reverse bias leakage current, Base-to-Emitter junction does not have an infinite resistance. This resistance is in parallel to R2. Usually it can be neglected, but if you take R2 in \$M\Omega\$ range you should be careful and verify that this resistance may be neglected (or take it into account otherwise).

R2 does not follow from calculations but simply a rule of thumb R2=10*R1. — Arne, Nov 13 '13 at 08:10

score 0 · Answer 3 · answered Nov 12 '13 at 20:16

You can use even lower values than that because with a 1M ohm resistance you already have enough base current to saturate the transistor with any of your inputs.

I'm currently using multiple BC337 transistors in a project and some of them have a 1M ohm resistance at the base (biasing). It's a telecom project so I'm also dealing with sinusoidal signals (not only operating as switch). As my frequency range does not exceed 400 Hertz and very high speed is not necessary, the high resistors are doing fine.

As you said speed is not so important to you either, so little capacitances should not be a problem. And since you are operating in a switch way, also noise should bother you.

Transistor biasing, Large Resisor values

3 Answers3