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IamSmooth

Wed Mar 03 2010, 04:11AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

I am taking a high voltage sinusoidal wave from my tank capacitor. The voltage can be as high as 600v and the frequency will be 50-150khz.

The voltage goes through a 100k/10w resistor and is clamped at 15v and -0.6v (diode drop). The problem is the output voltage is lagging the input signal. I am guessing this is a shift due to capacitance. The shift is gone if R is very low. The problem is I need a high R to keep the current low due to the high voltage. Is there anything I can do to reduce or eliminate the shift?

otherwise, I can adjust my calculation to account for the shift.

radiotech

Wed Mar 03 2010, 04:48AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546

Use a voltage divider and clamp the low voltage tap.

IamSmooth

Wed Mar 03 2010, 05:16AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

Using a voltage divider, I believe, won't solve the problem. I will still need a large resistance to keep the current down with the divider, bringing me back to where I started.

ScotchTapeLord

Wed Mar 03 2010, 06:40AM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
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Posts: 635

A voltage divider with a variable inductor in series with the parallel resistor?

Dr. Slack

Wed Mar 03 2010, 08:02AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

What are you doing with the signal at the "out" node? Why is phase shift a problem? Why are you clamping the output voltage?

If you are merely taking it to your scope to measure a reduced version of your input voltage, then you need a low value shunt resistor instead of the diodes, making a "voltage divider" with the input resistor. The low value resistor keeps the voltage low, and makes a low impedance to drive whatever stray capacitance you have with low phase shift.

I suspect though that this is a feedback path for your induction heater? What properties do you want it to have? It's easier to design a suitable circuit if we know what it has to do. You might be better off with a pure capacitive divider?

IamSmooth

Wed Mar 03 2010, 12:31PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

Ideally:

no phase shift
output voltage clamped between 0 and 15v
high enough impedance so it does not affect the RLC induction heater tank

signal is going into one of the PLL inputs, which is why I need to clamp the voltage

As I said in the beginning, if the design is too complicated, I can adjust the constant I am using that relates the phase shift of the drive and capacitor tank signal.

I am curious why the shift is more evident with a higher R value?

Sulaiman

Wed Mar 03 2010, 01:17PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

You could try making a high impedance divider with capacitors and resistors,
e.g for 101:1 use 100k//100pF and 1k//10nF

IamSmooth

Wed Mar 03 2010, 01:21PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

What is the purpose of putting the resistor and capacitor in parallel?

Also, Sulaiman, why do I see the shift more with higher resistance values? I would think if I was looking at a vector diagram with R on the horizontal, and C or L on the vertical, the phase shift would be more pronounced when R is smaller.

EDIT:

I have looked up information on phase shift topologies:

If my current voltage divider is the 100K reisistor (R1) and the high imedance of my input device (R2), there is a stray capacitance across my R1 resistor. I am guessing this is more evident at higher R values because the "capacitance" is not being shorted out.

Using this diagram and formula

couldn't I just put a higher C value (translating to a lower Xc) in parallel across my R which will lower ArcTan(R/Xc), and decrease the phase shift?

klugesmith

Wed Mar 03 2010, 03:40PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

The lag in your picture could be created by just the capacitance of your oscilloscope probe.
100 k ohms x 12 pF is 1.2 us. In other words, the waveform is different when you aren't looking at it.

For design suggestions, it would help if we knew what your clamped waveform is for. And which current is the one you want to limit.

Along the lines of previously suggested voltage divider: to your circuit as drawn, suppose you add a 10K resistor to ground. Voltage and RC time constant are reduced by a factor of 11, and no current is significantly larger than what you started with.

[edit]
>>The problem is I need a high R to keep the current low due to the high voltage.
The high voltage is only across the large value series R. In existing circuit with +-600V applied, the clamp diode currents are limited to +- 6 mA and your series R dissipates 1.8 watts. You could reduce the 100K to 22K and not burn it up.

Steve Conner

Wed Mar 03 2010, 03:55PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

The basic solution to the problem is a small capacitor across the 100k, as others have pointed out. However the comment about scope probe capacitance is a valid one too.

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