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This box should add DC offset to a resonant circuit. I drew the simulation on LTspice. However, it's not similar to the result in oscilloscope. So, my question is: Is my circuit diagram correct?

#1 sine signal enters, #3 DC voltage enters.

enter image description here

I compensated for oscilloscope input impedance with 1Mohm resistor. and adjusted the simulation connection. still, I cannot get the same result as in the oscilloscope. I need to simulate with ltspice the box with the actual offset of 1V enter image description here.

SamGibson
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Nana
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    DC does not pass through capacitors. – Math Keeps Me Busy Jul 27 '23 at 20:17
  • is your oscilloscope reading correct? ... what reading were you expecting? ... what did you measure? – jsotola Jul 27 '23 at 20:18
  • this is non-polarized capacitors at high frequency around 1Mhz it should work with both AC and DC – Nana Jul 27 '23 at 20:23
  • Is this a followup on https://electronics.stackexchange.com/questions/675470/advice-on-simulating-amplifier-and-power-splitter-circuit-in-ltspice? – Julien Jul 27 '23 at 20:23
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    Point "3" is common R2-C8-C9. High C8-C9 are "ground" and not "Vcc". – Antonio51 Jul 27 '23 at 20:26
  • Please provide a more detail explanation of the problem. It is very unclear what behavior you have at the moment. Please show your simulation results and your test results. Otherwise, we are doing guess work. – Julien Jul 27 '23 at 20:28
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    @Nana A little tip on your LTspice schematics, don't put the ground symbol and labels like Vcc right on a wire, it makes them difficult to see. Bring a short wire out and attach them to that. Also on labels, it can help to set their style as input or output, it makes them stand out and be easier to spot. For example on the supply I would make Vcc an output label, where it connects to the circuit I would make input labels. – GodJihyo Jul 28 '23 at 15:35
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    @Nana - Hi, You removed vital information from your question, which made answers useless as they used the details which you deleted. Remember that part of the Stack Exchange approach is to create a collection of questions & their answers *even after* you have received whatever answer *you* need, to allow for future use. Therefore I have reversed (rolled-back) your destructive edit. Please don't do that to your questions. Thanks. – SamGibson Aug 03 '23 at 13:55

2 Answers2

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Your schematic is wrong. Note that in the photo of the box one of the resistors is connected to jack 3, but that's not done in your schematic. Also the two paralleled capacitors in the photo are connected between point 3 and ground, not in series between 3 and the resistors.

The resistors provide a path for DC from jack 3 to jack 2, the capacitors will not pass DC.

I believe it's supposed to be like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The two parallel caps are to bypass RF at the DC bias jack.

So Bias comes in Jack 3, goes through the resistors to Jack 2. RF comes in Jack 1, goes through C3 to Jack 2. DC is blocked from going to Jack 1, RF is bypassed from going to Jack 3.

In your simulation you will need to take the oscilloscope input impedance into account as the resistors are quite high in value, the scope will see roughly 1/10th the bias voltage if it's input impedance is 1 meg.

GodJihyo
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From the answer previously posted (that shouldn't have been an answer) it was stated that you saw 1V while you were expecting 10V. Essentially, the DC source you are plugging to your biasing port is in series with 10Mohm. If you check the output with an oscilloscope with a 1Mohm input entry (fairly typical) you essentially create a voltage divider with a ratio of 1:10. This is why you are measuring ~1V instead of the 10V.

To confirm this issue you can use several technique. The simplest one is remove your AC signal and use an high impedance multimeter and measure the DC voltage. It should be close to 10V then. An other way it to go probe between your two resistor with an oscilloscope probe. You should see around 2V if you are using a 1Mohm probe.

How to solve your issue. Sadly, this is an easy one, change your design because 10Mohm of source impedance is way too high for 99.9% of applications. Or (and it's not a good idea) increase a lot your voltage. At 100V you should have 10V with a 1M impedance. To drive an amplifier, you might need several kV.

Obsolete: I see that you connect the VCC to the outer conductor of the BNC. Is your scope isolated? Because otherwise your biasing get grounded and could mess up with everything.

Julien
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  • I think the chassis is ground, DC bias is on the center pin. See my diagram. – GodJihyo Jul 27 '23 at 20:46
  • The schematic of OP clearly states VCC from that point. I have to agree with that since it is a standard biasing tee topology. It allows to isolate the DC supply from the AC signal. – Julien Jul 27 '23 at 20:48
  • Well, it's almost a biasing tee. Usually you don't place 10Mohm resistance on your dc supply. Usually you se a lambda/4 transmission line to present yourself as an open, but at 1MHz it is not possible. It might not be a biasing tee because of that. OP would need to comment! – Julien Jul 27 '23 at 20:50
  • Just to be sure, the resistor that is in series with DC power source, do you mean( The 4.7Mohm in the simulation)? And what range of values will be best for the resistors ? Given that it operates around 200V and 1Mhz – Nana Jul 28 '23 at 06:43
  • Sadly, this is something you have to figure out for yourself. In general, it depends how much current you need at the output of your biased tee, the impedance of it and the voltage span. Lower serie resistance will always be better for power delivery but will reduce the idolation of your tee. – Julien Jul 28 '23 at 12:19