DC component

[noelcucu]

Based on the figure below, if i measure the receiver coil using an oscilloscope and it shows ac signal with dc component. where does the dc component come from transmitter or rectifier?(note: transistor was used in transmitter)

 

[gophert]

Based on the figure below, if i measure the receiver coil using an oscilloscope and it shows ac signal with dc component. where does the dc component come from transmitter or rectifier?(note: transistor was used in transmitter)
View attachment 118497

If you are seeing a DC offset between the + and - then the zero offset on the scope needs to be adjusted to zero.
.or you didn't connect the grounding clip to the (-) test pint on your drawing.
.
.

 

[noelcucu]

I measure it on the receiver coil, not in the output of the rectifier.

 

[gophert]

I measure it on the receiver coil, not in the output of the rectifier.

Well, that's one reason. Each diode that current is flowing through will drop 0.6 to 0.7 volts. Most people use the ( - ) test point as their 0-volt reference.

 

[noelcucu]

This is what i get from the oscilloscope using the circuit in the right side. im confused where did the DC component come from? does it come from the transmitter or the receiver(rectifier)? help me plss.

 

[gophert]

Please - the device with the two coils is called a transformer. The side attached to power is called the Primary, the other side (not connected to a power source) is called the secondary.

The DC offset comes from the 0.1uF capacitor on the secondary side. It holds the charge from the previous cycle and does not allow the signal to drop to zero.

Try removing it and measure again.

 

[AnalogKid]

The 0.1 uF cap is *before* the rectifiers, not after.

A 1N4001 rectifier is not fast enough for most switching power circuits.

We still need to see more of the schematic.

The scope probe capacitance could be affecting the displayed signal.

ak

 

[noelcucu]

This is the whole schematic. I used class C amplifier even though is not a good idea. help me pls

 

[ronsimpson]

where did the DC component come from?

If just the circuit shown in the top left: Note: using a battery but reversing the battery very fast is the same as ac.
Note the meter (or scope) has one wire connected to ground.
No dc!

When you connected the 4 diodes in you caused the DC.
Sorry I tried to make a picture but too hard.
Words: There are now 4 diodes. In this picture I show that the wires on the transformer can not go below -0.7V but they is no limited to go positive. So a 10V p-p signal will drive the wires to -0.7V and +9.3V no matter the phase. (+ or -). Even though you are not taking power from the output of the rectifier birdge, just adding the bridge and connecting it to ground causes the "DC".

 

[1Lynx1]

Hello noelcucu, nice to see some intriguing schematics

What frequency is the oscillator supposed to work with?
Checking the datasheets for the OP amps it seems as though the slew rate leaves a little to ask for, atleast when it comes to handling all too high voltages for the higher frequencies at hand, so if you check the non inverting input on the right hand amplifier, does it meet your demands?

Also as there's 2 TIP31C working in parallel to power the transformer primary, leaving the primary almost without current during the oscillator's negative half cycle (class C amplifier), this also means that what's left to power the primary would be what's stored in the 0.1 uF capacitor, which would discharge fairly quickly, so this could also help explain the wave form in your scope shot.

However, if you fire up the circuit and try charging a battery, does the battery voltage increase over time?

 

[noelcucu]

The oscillator must produced at least 30khz. Yes, it does meet my demand. It produced at least 20Vp-p. Using class C amplifier is not good idea, but it is also working. When i fire up the circuit and measure the voltage across the battery, it shows that the voltage is increasing over a period of time. I already changed class C amplifier so that the negative cycle wont be wasted.