Base Photo Diode

RF detector?does it work fine,the base of the transistor is biased with 1.5V?

RF detector?does it work fine,the base of the transistor is biased with 1.5V,power supply is 9V and i have used a common diode 1N34A please check it and tell me if it works fine " http://www.flickr.com/photos/aj_silverthunder/4211391061/sizes/o/ "
please help me out ,thank u
when the RF out from the detector is connected to the amp ,the 15K and 3K will bias the base to 1.5V

no, as the diode won't conduct until it sees a RF signal in excess of 2 volts. without the transistor, you will get a DC output with RF signals of a few hundred millivolts.

Why do you insist on this circuit, it does not work for this application? Just put the meter across the 3k resistor. Your DC signals will be several volts or more. And that would saturate the transistor.

The transistor collector would swing from 7 volts down to 1 volt or so with increasing RF.

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Sweeper Robot Infrared Communication

transistor as a switch,please help?

friends in this circiut,the objective is to
convert the audio signal to DC and amplify it to 10Volts.
but my problem is ,1)i need the amplifier to work as a switch,
in this circuit a CE configuration is used.
but even there is no input from the base,the collector will
have voltage of about 6V,but i need it to be very very low,
and when i apply 3V to the base ,it should be amplified
to 10Volts..
so i think i should use the amplifier in common base
configuration,am i right?
2)and about the audio input,"the audio is amplified till
it reaches 10Volts AC and then given input to the diode
(1N4007),and the out put from the diode will be atleast
3Volts,again do u think its correct?
here is the schematic " http://www.flickr.com/photos/aj_silverthunder/4227184967/sizes/o/ "

Firstly, you are half wave rectifying the audio signal coming in with the 1N4007 diode, this will cause enormous distortion of the audio signal. A small coupling capacitor in the pico farad range is usually placed where the diode is.

Secondly you have 3V and 1.6Volts pictured at base (the two points you have pictured will be at the same potential). The voltage at that point will be set by voltage divider network of the 10k and 2.2k resistors. 2.2k / 12.2k X 9volts = 1.6 volts at base.

The set up pictured prevents thermal runaway of the transistor. Thermal runaway is where the transistor self destructs with the heat in the component.

Thirdly you can not get out more than you are putting in. The positive rail is at 9volts therefore there is no way you are going to get out 10 volts, this is in the field of perpetual motion. Even at the very most you are going to get out less than 9 volts peak to peak AC. This will mean an RMS voltage of only a few volts.

The gain (amplification) of this transistor can be found on the specs sheet at -

http://www.datasheetcatalog.org/datasheet/philips/BC107_108_109_4.pdf

Finally a coupling capacitor is used on the output to prevent the DC portion from reaching the speakers etc. If dc reaches the speakers you will get a terrible hum or buzz. But if you are planing on using the circuit as a switch you could place the diode on the output.

If you make these changes the circuit should work OK as a pre amplifier. Depending on what speaker you are trying to drive you may need to amplify the signal again through a higher power transistor if you are trying to drive a large speaker.