# BALUN transformer for capacitive loads

Discussion in 'Electronic Basics' started by [email protected], Aug 22, 2006.

1. ### Guest

Hi,
I need to drive 2 capacitive electrodes with out of phase RF (0-100
Vp-p, 10KHz - 1MHz). For this purpose i got myself one of the T&C power
amplifiers (AG1021) with a gain of 53dB upto 1MHz. I planned to use a
BALUN transformer from North hills (75 ohm: 75 ohm, 1KHz - 2MHz) to
unbalance the amplifier RF. I use a function generator as a source.
However, things are not working as i thought.
The setup is as follows: Function generator feeding the signal to the
power amplifier. The output i measure on a oscope is good 100 Vp-p for
small input signal < 1 m volt (10Khz) . As soon as i connect the output
of the amplifier to the BALUN, my reflected power to the amplifier
shoots up. And the 2 unbalanced outputs of the BALUN show merely 12
Vp-p.
I know the power amplifier is rated for 50 ohm, is there an impedance
problem. With the capacitive load at the secondary of the BALUN, the
impedance which the amplifier at the primary is surely not 50 ohms.
How can i overcome this, can i just use a BALUN with more turns on the
secondary? Say 50 ohm : 600 Ohms BALUN to step up the voltage.
I would appreciate the suggestions.

2. ### Guest

can you rephrase that?

3. ### John PopelishGuest

frequency rises. Using a transformer to change an unbalanced source
to a balanced pair of outputs does nothing to solve this mismatch.
The largest capacitance the unbalanced 50 ohm load can drive to
nominal output voltage at 1 MHz is about 3000 pF. If the balun steps
the total voltage up 2:1 then the capacitance each side can drive is
half of that.

So what capacitive load are you trying to drive to 100 v p-p at up to
1 MHz?

4. ### Guest

Hi,
In brief, i have 2 electrodes each with ~20pF capacitance to ground. I
need to drive each of them with an out of phase RF signal i.e. one with
A cos wt and other with A cos(wt-180). The RF amplitude and frequency
range is around 0-100Vp-p and 1KHz to 1MHz.

5. ### John PopelishGuest

So if you need to apply positive 100 volts to one electrode, while
applying -100 volts to the other electrode, and each electrode has 20
pF to common (10 pF in series) the impedance across the differential
output is about 16 k at 1 MHz. Neglecting stray transformer and
wiring capacitance, you should be able to use a transformer with an
impedance ratio of up to about 318:1. Of course, those stray
capacitances may not be at all negligible, compared to 10 pF, so the
real limit is probably about half that or less. 100:1 may be
achievable. I think I would try a toroidal core transformer, with a
center tapped secondary having 10 times as many turns as the primary.
At this frequency range, I doubt that worrying about transmission
line transformers is worth the trouble. This is more in line with
switching power supply components. The core size will be dominated by
the low frequency extreme (the primary will have to have an inductance
of at least 8 mHy to stay above 50 ohms at 1 kHz), and winding
capacitance and leakage inductance will dominate the high frequency
problems. With a 100:1 impedance ratio, the implied minimum secondary
inductance is .8 Hy. Winding a tightly coupled .8 Hy inductor with
low losses at 1 MHz and low capacitance is a fair trick.

6. ### Guest

John,
Thank you, this all makes sense to me. I have started to find the right
torroid for this. I haven't made any transformers before in house,
wondering if there are any practical guidelines.
Do i have to worry about the saturation of the core, keeping in mind
that i have 100 Vp-p at the secondary?

7. ### John PopelishGuest

At the low frequency end, saturation is the prime consideration. This
is because the integral of voltage over time (for each half cycle) is
proportional to the flux swing of the core. So if it will handle 1
kHz at 100 volts, it will not saturate at any higher frequency.
Here are some tutorials on transformer design:
http://www-s.ti.com/sc/techlit/slup123.pdf
http://www-s.ti.com/sc/techlit/slup126.pdf
http://www.mag-inc.com/library.asp
http://www.triodeguy.com/transformer_math.htm

8. ### Guest

John,
So, i got myself a ferrite Core torroid with permeability 5000 and OD =
..87in and ID = .25 inch.
I wound up couple of turns and connected it to the output of the 50 ohm
wavetek. The output dropped from 1 volt to few mVs at 10 KHz.
So i kept on increasing the number of turns untill i get half a volt
which is 12 turns now, this implies 50 ohm impedance at 10 KHz.
Calculating it back, tell me the inductance is arpprox. .8mH. Now, to
get higher impedance at 1KHz i have to wind more turns on the primary
say 20 turns. That will imply around 200 turns for secondary, which i
am not sure i can fit on this small core. I use a 30 Gauge magnet wire
which is already too small. And winding 200 turns on this torroid will
almost cost me a whole day. Is there a better way to get this done?

9. ### Guest

John,
So, i got myself a ferrite Core torroid with permeability 5000 and OD =

..87in and ID = .25 inch.
I wound up couple of turns and connected it to the output of the 50 ohm

wavetek. The output dropped from 1 volt to few mVs at 10 KHz.
So i kept on increasing the number of turns untill i get half a volt
which is 12 turns now, this implies 50 ohm impedance at 10 KHz.
Calculating it back, tell me the inductance is arpprox. .8mH. Now, to
get higher impedance at 1KHz i have to wind more turns on the primary
say 20 turns. That will imply around 200 turns for secondary, which i
am not sure i can fit on this small core. I use a 30 Gauge magnet wire
which is already too small. And winding 200 turns on this torroid will
almost cost me a whole day. Is there a better way to get this done?
Any commercial products, i can buy quickly?

10. ### John PopelishGuest

Something like that. So to get that same effect at 1kHz, you will
have to increase the inductance by at least a factor of 10, so the
turns will have to increase by at least the square root of 10 or 38
turns, minimum. Lets say 40. Then you need about 10 times that many
turns on the secondary, center tapped. Of you stack two of these
cores and cut the turns in half, etc.
Use a larger core. Perhaps a LFB360230-300, \$1.35 from Digikey.
I suspect there might be something you could use, but you would
probably be better off asking the makers than me.

11. ### Tim AutonGuest

10^0.5 != 38
10^0.5 = 3.16

Tim

12. ### John PopelishGuest

Now, multiply that factor times the 12 turns he has tested.

13. ### Tim AutonGuest

Sorry about that. We'll now return you to our usual programming.

Tim

14. ### John PopelishGuest

Shouldn't take an hour. The trick is to wind all the wire you need
and a bit extra over a soda straw, end to end. Notch the ends with
scissors to help the wire stay on. Then you have a neat and thin
object to pass through the core (instead of pulling a long lose piece
of wire), with minimum fuss and tangle.

After you have something that works, you can take the design to a
custom magnetics house for a quote on automated production.

15. ### Guest

Thanks John,
I see the part # LFB360230-300, why not get the one which has the
higher Net impedance at 1 MHz, say LFB43064-000? That will ensure i
have high enough reactance with fewest number of turns (to make my life
easy).

16. ### Guest

Thanks John,
I see the part # LFB360230-300, why not get the one which has the
higher Net impedance at 1 MHz, say LFB43064-000? That will ensure i
have high enough reactance with fewest number of turns (to make my life
easy).
http://dkc3.digikey.com/PDF/T062/1057.pdf

17. ### John PopelishGuest

I picked one with a large cross sectional area, but also a large
window area, in case you needed space for lots of turns, especially if
you go with Kynar or Teflon insulated wire, yo lower the inter winding
capacitance. For all they cost, you might as well get an assortment.

You might also consider placing a pair of the tall, narrow form factor
beads side by side, and winding through both, cutting the number of
turns in half.

18. ### Guest

Works like a charm. I just finished winding and testing.
Thanks a lot John, the core from the digikey was sufficient for the
job.
regards,

19. ### John PopelishGuest

Congratulations!