# Can anybody explain to me this circuit

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#### QuickStrike

##### New Member

1) Why connect a 10K resistor and not connect the photodiode directly to the +12V?

2) The 0.1uF cap is acting as bypass cap right?

This is what I really don't understand

3) What is that attenuator for? , Why the T connection of resistors?

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#### stevez

##### Active Member
My limited experience with attenuators suggests that the attenuator does not significantly change the input impedance of the circuit that follows. The "T" configuration is often designed with values that have the same impedance on either side. Simply adding a resistor or resistors without due consideration for this would certainly attenuate energy coming in but would likely affect operations otherwise. ARRL Handbook has a nice section on this.

#### mstechca

##### New Member
1) Why connect a 10K resistor and not connect the photodiode directly to the +12V?
If I remember, a phototransistor would suit that schematic more than a photodiode. the 10K resistor limits the current to 1.2ma. Some devices can only handle a certain amount of current and/or voltage as its maximum and will perform poorly if the maximums are exceeded.

2) The 0.1uF cap is acting as bypass cap right?
This is what I really don't understand
I would say it is with reduced current. Every component is using the 1.2ma given through the resistor.

3) What is that attenuator for? , Why the T connection of resistors?
It seems that the photodiode is actually a signal input. Resistors limit current. I don't think the attenuator is necessary because the resistors are already weakening the strength of the signal. The resistor R3 is a resistor that sets minimum requirements.

Equation wise, R1 and R3 is a voltage divider if the photodiode and the 10K resistor was not present.

If you use the equation R3 / (R3 + R1) * 12V, you will see how many volts the rest of the circuit is getting (after the R1 and R3 pair).

#### JimB

##### Super Moderator
"Damp Rate" is a curious expression, but it just means the attenuation ratio of the attenuator expressed as a percentage.

JimB

#### Russlk

##### New Member
R1, R2 and R3 are a matching network. The amplifier input is 50 ohms and if not driven from a 50 ohm source it will ring or maybe even oscillate. The photo diode is a high impedance source, so I don't think R1 is even needed. In that case, R2 + R3 should add up to 50 ohms. R1 may provide some isolation for capacitance at the photo diode, but at 1.5gHz, the circuit cannot stand much capacitance.

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