Bandwidth problem

[Bill Bowden]

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 =
25. which is more than twice the measured Q. I found a Q calculator at
http://www.pronine.ca/qbw.htm. 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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[Tim Wescott]

On Sat, 20 Feb 2016 19:18:08 -0800, Bill Bowden wrote:

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 = 25. which is more than twice the measured Q. I found a
Q calculator at http://www.pronine.ca/qbw.htm. 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?

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

 

[Bill Bowden]

"Tim Wescott" <tim@seemywebsite.com> wrote in message
newssadnS_Uasur6FTLnZ2dnUU7-YPOydjZ@giganews.com...

On Sat, 20 Feb 2016 19:18:08 -0800, Bill Bowden wrote:

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 = 25. which is more than twice the measured Q. I found a
Q calculator at http://www.pronine.ca/qbw.htm. 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?

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

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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[Tim Wescott]

On Sun, 21 Feb 2016 20:15:58 -0800, Bill Bowden wrote:

"Tim Wescott" <tim@seemywebsite.com> wrote in message
newssadnS_Uasur6FTLnZ2dnUU7-YPOydjZ@giganews.com...
On Sat, 20 Feb 2016 19:18:08 -0800, Bill Bowden wrote:

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 = 25. which is more than twice
the measured Q. I found a Q calculator at
http://www.pronine.ca/qbw.htm. 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?

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


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.

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): <https://en.wikipedia.org/wiki/
Bipolar_junction_transistor#h-parameter_model>.

--

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

 

[Tim Wescott]

On Mon, 22 Feb 2016 12:39:22 -0600, Tim Wescott wrote:

On Sun, 21 Feb 2016 20:15:58 -0800, Bill Bowden wrote:

"Tim Wescott" <tim@seemywebsite.com> wrote in message
newssadnS_Uasur6FTLnZ2dnUU7-YPOydjZ@giganews.com...
On Sat, 20 Feb 2016 19:18:08 -0800, Bill Bowden wrote:

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 = 25. which is more
than twice the measured Q. I found a Q calculator at
http://www.pronine.ca/qbw.htm. 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?

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


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.

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): <https://en.wikipedia.org/wiki/
Bipolar_junction_transistor#h-parameter_model>.

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