A dirt cheap “spectrum analyzer” with an RTL-SDR dongle(hansvi.be)
hansvi.be
A dirt cheap “spectrum analyzer” with an RTL-SDR dongle
http://hansvi.be/wordpress/?p=91
1 comments
D1 and D2 are normal diodes in forward direction[1], in series they drop 2⋅0,7V = 1.4V. R2 just pulls them down. So the cathode of D2 will sit 1.4V below supply voltage.
Q1 in this schematic is acting as a current source: If conducting a reasonable amount of current, a transistor needs a base-emitter voltage of Ube = 0.6V. This is a PNP transistor[2], so base needs to be 0.6V below emitter, hence the voltage over R1 will be U(d1d2) - Ube = U(R1) = 0.8V. The current then is of course set (to first order, ignoring the base-current contributing) to be U(R1)/R1.
Why such a "complicated" circuit and not only use R1 to bias D3? Because a current source has a much higher (theoretically: infinite) impedance: Z=dU/dI, but dI=0 (stays constant) whatever the change in voltage (dU > 0) might be.
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1) forward direction: A diode that is operated, like a LED supposed to light, will have a certain voltage drop. For a standard silicon diode, this is 0.7V, a red LED would have 1.6V or so.
2) PNP transistor: A transistor has three terminals, base is the cross-bar, emitter is the diagonal line with the arrow, collector is the other one. a NPN transistor runs on a base current (technical direction + to -) into the base terminal and conducts current from emitter to collector, a PNP transistor runs the reverse.
Edit: emitter/collector mixup
Q1 in this schematic is acting as a current source: If conducting a reasonable amount of current, a transistor needs a base-emitter voltage of Ube = 0.6V. This is a PNP transistor[2], so base needs to be 0.6V below emitter, hence the voltage over R1 will be U(d1d2) - Ube = U(R1) = 0.8V. The current then is of course set (to first order, ignoring the base-current contributing) to be U(R1)/R1.
Why such a "complicated" circuit and not only use R1 to bias D3? Because a current source has a much higher (theoretically: infinite) impedance: Z=dU/dI, but dI=0 (stays constant) whatever the change in voltage (dU > 0) might be.
-------------------
1) forward direction: A diode that is operated, like a LED supposed to light, will have a certain voltage drop. For a standard silicon diode, this is 0.7V, a red LED would have 1.6V or so.
2) PNP transistor: A transistor has three terminals, base is the cross-bar, emitter is the diagonal line with the arrow, collector is the other one. a NPN transistor runs on a base current (technical direction + to -) into the base terminal and conducts current from emitter to collector, a PNP transistor runs the reverse.
Edit: emitter/collector mixup
I think, to put it more simply, those two diodes exist to cause a 1.4 volt drop to the transistor.
The transistor is a PNP transistor, so the base (middle terminal) must be about .6V lower than the emitter (top terminal) for the transistor to turn on.
I'm guessing one diode wouldn't turn on the transistor reliably, so the author used two in series. This also probably allows for a more convenient value of R1.
The transistor is a PNP transistor, so the base (middle terminal) must be about .6V lower than the emitter (top terminal) for the transistor to turn on.
I'm guessing one diode wouldn't turn on the transistor reliably, so the author used two in series. This also probably allows for a more convenient value of R1.
Exactly. One diode will not turn on the transistor. Replace the two diodes with one LED, and the circuit will run the same.
One often finds LEDs used for the prose of simple voltage stabilization (often green ones @2.1V or so..)
One often finds LEDs used for the prose of simple voltage stabilization (often green ones @2.1V or so..)
So why not just use an appropriate resistor?
A resister has a voltage drop related to the current. A diode will always drop the same voltage (within operating parameters).
> collector is the diagonal line with the arrow, emitter is the other one
I suppose you have a typo here. Should be the other way around.
I suppose you have a typo here. Should be the other way around.
Yes, of course, fixed. Thanks!
Googling on "diodes in series", I found some discussion about how you can do this if you need a reverse breakdown voltage that is higher than that of one diode, but aren't D1 and D2 always forward biased in this circuit?