# Increasing amplifier input impedance with inductors

Discussion in 'General Electronics Chat' started by Eduard Munteanu, Sep 5, 2007.

1. ### Eduard Munteanu Thread Starter Active Member

Sep 1, 2007
86
0
Consider, for example, a common-emitter BJT amplifier. Adding an inductor in series with each of the base biasing voltage divider resistors will, as far as I can see, increase its input impedance a bit. This is likely to allow a higher biasing DC current without hurting the input signal. Given the inductance is high enough, the input impedance will be almost equal to emitter AC-to-ground path's impedance, which is a bit higher than without inductors.

But I'm not sure how to protect against turn-on/off voltage spikes. Adding diodes in parallel with those inductances would hurt the AC signal. Of course, I could calculate for the best trade-off between AC reactance and turn-on/off voltage (according to max. VBE and VCE for that transistor).

But is this a good idea at all, is it worth the trouble? Or it can't be used because of what I just anticipated?

2. ### techroomt Senior Member

May 19, 2004
198
1
an idea of your input signal would help us knowing your intentions. initial thinking is that the inductor and the base resistor (to ground) would act as a passive high pass filter, so i am guessing you have a respectively high signal frequency?!?

what is the purpose of increasing the input impedance? maybe an intermediate coupling amplifier could match your impedances for you.

3. ### Eduard Munteanu Thread Starter Active Member

Sep 1, 2007
86
0
Let's say it's for an audio amplifier. Having a higher input impedance is good for almost anything (except RF, as Horowitz & Hill say, where you need to match impedances). Those inductors will increase each of those resistors' impedances by $Z = 2 \pi f L$. If you set L to be high enough for the lowest frequency of interest, those inductors resemble open-circuits for AC. Such a base bias divider would look like:
+Vcc --- resistor --- inductor --- BJT base + coupled input signal --- inductor --- resistor --- ground

I don't see any passive high-pass filter if the impedance of those inductors is high enough.

But my concern is whether or not such a high inductance will blow out the transistor when I turn the device on/off, and whether this is useful at all.

4. ### BeeBop Member

Apr 25, 2006
17
0

The above poster was on the right track; you should really use a preamp made from an op amp to increase input impedance, or make your input stage with JFETs.

It is a capacitor, not an inductor, which looks like a short to AC. An inductor looks like a short to DC.

An inductor in series with the input, and a resistor to ground is a LOW pass filter, not a high pass. The larger the inductor, the less bandwidth you will have; it will provide a higher impedance for high frequencies than low frequencies. The low frequencies will have a very low input impedance, and the high frequencies a high input impedance.

The reason designers went to a DC (direct coupled) amplifiers is that the capacitors used for coupling limit the low frequency response of the amp.

Op amps work down to DC, so there is a flatter frequency response.

5. ### techroomt Senior Member

May 19, 2004
198
1
yeah, i misread the series info and read it as in series with the signal. thanks for catching that.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Your theory is sound, but you need to investigate the practicalities of implementing it. Input bias resistors are generally on the order of 10's of kohms. Calculate the value of an inductor with an impedance of 10 kohms (or even 1 kohm) at 100Hz. If you can find one, check the size and the price.
This exact method is used in some RF amplifiers, where the relative bandwidth (bandwidth/center frequency) is fairly low, meaning the impedance will be fairly constant over the frequency range. High frequencies allows the use of small inductors, which are relatively inexpensive, and, of course, small.