Bandwidth problem

Feb 21, 2016 4 Replies
B

I built a 36Khz LC parallel tuned circuit driven by a transistor biased at



2mA. It was for use in detecting IR signals from IR remote controls. It works well enough with a measured bandwidth of about 4Khz at the -3dB points which would be a Q of about 9. As I remember, the unloaded Q of a parallel LC circuit is XL / R. The LC values are 2700pF and 6.8 mH with about 60 ohms of resistance in the inductor wire which equates to about 1500 / 60 =
  1. which is more than twice the measured Q. I found a Q calculator at
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    which indicates .the loaded Q is 9 if the generator impedance is about 22K. So, my circuit must have an impedance of

22K from the transistor driving the tuned circuit. I don't know how they get that from a transistor biased at 2mA and operating from a 9 volt supply. So, how do I equate a transistor biased a 2mA to be a generator of 22K? How do you do the math to find the bandwidth of a parallel LC circuit driven by a transistor? Is there some simple formula for that?










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T

You don't give details on the circuit, but 22k-ohm sounds plausible for the collector impedance of a transistor that's running with a grounded or well-bypassed emitter.

You wish for a higher Q?

www.wescottdesign.com
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B

No, the Q is probably high enough at 9. I read somewhere that a LC tank requires about Q cycles to gain full amplitude. So, if the IR transmitter only sends say 30 cycles per bit, the Q would have to be a lot less than 30. But my problem is understanding how to obtain 22k from a simple grounded emitter transistor driving the tank from a 9 volt battery. The base bias resistor is 330k and connects the base to the 9 volt battery. So, the collector current depends on the hFE of the transistor. I get about 2.5mA of collector current , so the hFE gain must be about 100.. I tried using LTSpice, but the .program assumes the hFE gain at a much higher figure of

310 or so. But it does agree fairly close as to Q if I use a 1.2 meg bias resistor instead of the 330k. So, it appears that the internal resistance of the transistor is a function of DC collector current. As the bias current goes higher, the gain goes higher, and the Q goes lower.

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T

Yes, the collector impedance of a transistor changes with increasing collector current.

Since you want to quote hybrid parameters, the one you're failing to take into account is h_ox (or h_oe): .

Tim Wescott Wescott Design Services http://www.wescottdesign.com
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T

I forgot: if the source that's driving your transistor base is high- enough impedance, then any collector-base capacitance is going to create feedback that's going to make the output impedance of the resulting amplifier lower than just the transistor's output impedance.

I don't think this effect should be great at 40kHz, but I've never had reason to build a circuit like what you're trying, so I can't be sure.

Tim Wescott Wescott Design Services http://www.wescottdesign.com
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