sine wave voltage amplified in parallel transistor

[noelcucu]

My sine wave output has 30khz frequency and 22vp-p when I input it on the base of the parallel transistor the voltage across the inductor in the collector went up to 38V what is the theory behind why the voltage in parallel NPN transistor amplifies the voltage?

 

[kubeek]

Please show your entire circuit, along with what you measured at which points.

 

[noelcucu]

the output sinewave of the buffer has a frequency of 30khz and 22vp-p. when I measure the voltage across the inductor i measured 38vp-p

 

[Beau Schwabe]

The resonant frequency of your LC combo is 26.902kHz .... 30kHz is enough to cause it to resonate slightly, therefore, the voltage produced will be higher.

Tune your circuit closer to the resonant frequency, and you should see a more dramatic voltage difference.

BTW) You might want to put a resistor on each base of the transistors

Frequency calculator reference:

Web Based Resonant Frequency Calculator

www.ham-radio.com

 

[unclejed613]

when you increase the current, the voltage rises proportionally

 

[noelcucu]

I already put a resistor in the circuit I just forgot to put it in the schematic. So is it normal that my voltage will increase even though my supply is only 12v?

 

[Beau Schwabe]

"So is it normal that my voltage will increase even though my supply is only 12v? " ... Yeah, but at resonant frequency and a high "Q" factor, you could see a lot more than 38V pk-pk. If you aren't looking for that, then change your frequency, cap value, or inductor value. If your design is simply fixed components and you happened to get close the the resonant frequency, then you could be borderline for a mysterious failure if you aren't expecting the higher voltages. Better to understand the potential problem now than to "hunt" for it later.

 

[noelcucu]

Thank you for your answer sir. one more question sir, why do the measurement in oscilloscope and the digital multimeter is not the same. when I measured the sine wave in the receiver coil it shows that it has 16Vp-p, but in the multimeter it shows 27v?

 

[Beau Schwabe]

The meter is probably showing an RMS value and there could be some differences in loading. A scope typically has a higher impedance than a meter.

Reference: (See formula for Sine Wave)

Root mean square - Wikipedia

en.wikipedia.org

 

[noelcucu]

Thank you so much, sir. it really helps a lot.

 

[ronsimpson]

The meter might be measuring the AC voltage + the DC voltage.
If the meter is connected to the (collector/inductor) node and connected to ground then you have 12Vdc and some amount of AC on top of that. I do not know how your meter will respond to that.

 

[audioguru]

Few multimeters can accurately measure the amplitude of a frequency as high as 30kHz. Most are designed to measure only 50Hz or 60Hz electricity, not ultrasound frequencies.

 

[Nigel Goodwin]

The meter is probably showing an RMS value and there could be some differences in loading. A scope typically has a higher impedance than a meter.

Scopes are usually 1 meg impedance, or 10 Meg with a x10 probe, a modern digital meter is 10 Meg - older analogue meters vary according to range, and are higher than 10 Meg on the highest ranges (but considerably less on the lower ones).

As AG has pointed out, you can't usually measure 30KHz using a multimeter, and the result is likely to be either nothing, or reading a lot lower than it should.

Obviously the only way to make any sensible measurements on this bizarre circuit (which seems to be a strange attempt at class C?) would be an oscilloscope.

 

[AnalogKid]

Note that the two output transistors are acting as a half-wave rectifier. They do not conduct during the sinewave negative half-cycle. The resonant tank is being excited by a series of half-sine pulses. These a re probably closer to squarewave pulses because the transistors are operating at their maximum gain. Can you post an image of the collector voltage?

If you put a very small resistor in series with the tank, you can get an image of the collector current. With this, you might be able to eliminate one of the output transistors.

Also, what is the purpose of the circuit?

ak

 

[Nigel Goodwin]

Note that the two output transistors are acting as a half-wave rectifier. They do not conduct during the sinewave negative half-cycle. The resonant tank is being excited by a series of half-sine pulses. These a re probably closer to squarewave pulses because the transistors are operating at their maximum gain.

Like I said, crude attempt at class C