What does the diode in this picture mean?

kal9623287

What does the diode in this picture mean? [Copy link]

 

Dear forum teachers, may I ask, what does the diode D1 in this picture mean and what is its function?

PowerAnts

Parallel rectification, output negative half cycle

kal9623287

PowerAnts published on 2022-4-27 08:51 Parallel rectifier, output negative half cycle

Can it be understood as rectification and filtering?

PowerAnts

kal9623287 posted on 2022-4-27 08:53 Can it be understood as rectification and filtering?

Do you see the filtering?

maychang

kal9623287 posted on 2022-4-27 08:53 Can it be understood as rectification and filtering?

"Can it be understood as rectification and filtering?"

Don’t “understand it as”.

This circuit is a rectifier, but there is no filtering in the diagram.

PowerAnts

There are a number of things that landlords need to consider:

1，C2*(R1+RL)<<0.5T

2，C2*(R1+RL) ≈ 0.5T

3, C2*(R1+RL)>>0.5T

Fred_1977

This is a rectifier circuit. The positive half cycle is filtered out by the diode, so the output is the negative half cycle.

PowerAnts

Fred_1977 posted on 2022-4-27 09:09 This is a rectifier circuit. The positive half cycle is filtered out by the diode, so the output is the negative half cycle.

Note item 3 on the 6th floor. In this case, the voltage across C2 is positive at the top and negative at the bottom. The voltage = the peak value of the LC parallel circuit voltage - 0.6. So is the right side as simple as just the negative half cycle?

PowerAnts

If the output is added with one D and one C, it becomes a voltage doubler rectifier.

kal9623287

PowerAnts posted on 2022-4-27 09:04 The OP needs to consider a variety of situations: 1. C2*(R1+RL)<<0.5T 2. C2*(R1+RL) ≈ 0.5T 3. C2*(R1+R ...

Why is C2*(R1+RL) compared with 0.5T?

Which part is RL?

Can you give me some pointers?

kal9623287

maychang posted on 2022-4-27 09:03 "Can it be understood as rectification and filtering?" Don't "understand it as". This circuit is rectification, but there is no filtering in the figure.

What is the function of the inductor in the figure?

Is LC filtering rectification?

se7ens

This circuit is a bit strange. What product is it used for?

maychang

kal9623287 posted on 2022-4-27 09:47 What is the function of the inductor in the picture? Is LC filtering the same as rectification?

You haven't figured out the direction of signal (power) flow.

The signal flows from left to right. L1C1 forms a parallel resonant circuit, and the signal energy flowing into the left end is stored in L1C1. During a small part of the L1C1 oscillation cycle, diode D1 is turned on, charging capacitor C2, with the left side positive and the right side negative. During the rest of the AC cycle, the diode is turned off, and C2 discharges to the right end load (not shown) through R1.

PowerAnts

se7ens posted on 2022-4-27 09:51 This circuit is a bit strange, what product is it used for?

For situations that the poster didn't explain, don't guess that it's just a single application. If you see it, think about it more yourself. It will also improve your abilities.

There are too many possibilities, such as LLC over-current detection, induction cooker power detection, metal detection, coil turn-to-turn short-circuit test, etc.

maychang

kal9623287 posted on 2022-4-27 09:47 What is the function of the inductor in the picture? Is LC filtering the same as rectification?

Therefore, diode D1 is the rectifier, and L1C1 only stores energy, charging C2 through the diode during a part of the AC cycle (of course, before and after the AC reaches its positive peak).

kal9623287

se7ens posted on 2022-4-27 09:51 This circuit is a bit strange, what product is it used for?

FR front-end receiver circuit

PowerAnts

kal9623287 posted on 2022-4-27 10:05 FR front-end receiver circuit

I was fooled by you! Then D1 is a varactor diode, and the negative DC voltage on the right side changes the tuning frequency

maychang

PowerAnts posted on 2022-4-27 10:28 I was fooled by you! Then D1 is a varactor diode, and the negative DC voltage input on the right changes the tuning frequency

"Then D1 is a varactor diode, and the negative DC voltage input on the right changes the tuning frequency."

This is really possible!

jimtien

1. The figure does not indicate that it is a variable capacitance diode. 2. V is not indicated as a negative voltage. Therefore, this diode should be used as a switch tube. When the diode is forward biased, it is connected to C2 to change the oscillation frequency.

kal9623287

It is the function of frequency modulation. This is the circuit diagram that my senior asked me about. But isn’t the symbol of the varactor diode this?