Hello,
I'm learning on my own and got a German book. In the chapter about diodes, the author shows how auxiliary voltage is created with the help of diodes (see attached image). He doesn't really explains what is going on a beginner level and I am not sure if I should go one with reading or try to understand before I move one, that is why I am here. I already learned about the Kirchoff's law.
We have 9 V battery, on the left side 33k R, two diodes and on the right side three diodes.
1) Are the 9 V divided between left and right part of the circuit? E.g. 60% of voltage available on the left and 40% on the right side of the circuit?
2) I can't tell the voltage drop on 33k R, because I don't know how much voltage is used on the right side of the circuits (three diodes), right?
3) Why is the need/auxiliary voltage is between resistor and the two diodes? It feels like the needed voltage of 1.2V should be available after all three components.
Thank you for your time.
The paragraph about the circuit:
An additional lower voltage is often required in an electronic circuit. In our example, an "auxiliary voltage" of approx. 1.2 to 1.5 V can be generated at the two diodes (D1 and D2).
This corresponds to the voltage loss that occurs "automatically" at the two diodes. The resistor R can have a rather high ohmic value, if the current drop is only small. The drawn capacitor C serves for "smoothing" the 1.2 V voltage. At the Diodes D3, D4 and D5 there is a voltage loss of about 3 x 0.6 V (= 1.8 V). The residual voltage is thus approx. 7.2 V. Here no series resistor is necessary and the max. current consumption can be as high as the silicon diodes and the batteries can handle.
I'm learning on my own and got a German book. In the chapter about diodes, the author shows how auxiliary voltage is created with the help of diodes (see attached image). He doesn't really explains what is going on a beginner level and I am not sure if I should go one with reading or try to understand before I move one, that is why I am here. I already learned about the Kirchoff's law.
We have 9 V battery, on the left side 33k R, two diodes and on the right side three diodes.
1) Are the 9 V divided between left and right part of the circuit? E.g. 60% of voltage available on the left and 40% on the right side of the circuit?
2) I can't tell the voltage drop on 33k R, because I don't know how much voltage is used on the right side of the circuits (three diodes), right?
3) Why is the need/auxiliary voltage is between resistor and the two diodes? It feels like the needed voltage of 1.2V should be available after all three components.
Thank you for your time.
The paragraph about the circuit:
An additional lower voltage is often required in an electronic circuit. In our example, an "auxiliary voltage" of approx. 1.2 to 1.5 V can be generated at the two diodes (D1 and D2).
This corresponds to the voltage loss that occurs "automatically" at the two diodes. The resistor R can have a rather high ohmic value, if the current drop is only small. The drawn capacitor C serves for "smoothing" the 1.2 V voltage. At the Diodes D3, D4 and D5 there is a voltage loss of about 3 x 0.6 V (= 1.8 V). The residual voltage is thus approx. 7.2 V. Here no series resistor is necessary and the max. current consumption can be as high as the silicon diodes and the batteries can handle.
