Trek Application Note
Number 6001
Voltage level shifting
Dr. Maciej A. Noras
Abstract A brief description of voltage shifting circuits.
1
Introduction
2U
U
U
In applications requiring a unipolar AC voltage sig0
0
nal, the signal may be delivered from a bi-polar
t
t
T
voltage amplifier/generator and repositioned rela-U
tive to a zero volt reference (an earth ground) usT
ing a voltage level shifting circuit. In addition, the
voltage level shifting technique is frequently used
to generate high voltage DC outputs from lower
Figure 2: Shifted signal.
voltage AC sources. The maximum voltage (AC
The steady state current drawn from the AC
or DC) produced at the output of a single stage of
source
(amplifier, for example) follows the depenthe level shifting circuit is nominally equal to the
dency:
peak-to-peak value of the AC signal delivered by
C1 · C2 dU
the amplifier/generator.
·
(1)
I=
C1 + C2 dt
For a sinusoidal AC voltage input
2
AC voltage shifting
U = U · sin (2πft)
(2)
the current is:
An example of an AC voltage shifting circuit diaC1 · C2
I=
· 2πfU cos (2πft)
(3)
gram (Villard circuit) is shown in Figure 1. FigC1 + C2
ure 2 shows the input and the output voltage of
The output voltage Uout across the capacitor C2
the circuit. In order to move the voltage signal
is:
toward negative values, it is necessary to reverse
C1
Uout = U ·
(4)
the diode shown on the schematic in Figure 1. AdC1 + C2
ditionally, if the capacitors C1 and C2 are polar,
they also need to be reversed.
3
voltage shifting section
C1
C2
amplifier
(AC source)
load
Figure 1: Villard voltage level shifting circuit.
TREK, INC.
• 190 Walnut Street
Copyright © 2013 TREK, INC.
Consider the following example:
U=1000 · sin (2πft) V;
f=100 Hz;
C1=C2=63 nF.
The maximum current is 20 mA (eq. 3) and the
voltage across the C2 is 1000 V (eq. 4). These
results are shown in Figure 3.
When the capacitor C1 becomes smaller
(C1=31.5 nF), the output voltage value drops (Figure 4) according to equation 4. Current I also decreases, as the total capacitance of the circuit decreased.
• Lockport, NY 14094
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Example
• Tel: (716) 438-7555
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0408/MAN Rev. 2
page 1 of 5
Trek Application Note
Number 6001
Voltage level shifting
input voltage
voltage, V
2000
1000
0
−1000
−2000
voltage across C1
voltage, V
2000
1000
0
−1000
−2000
voltage across C2
voltage, V
2000
1000
0
−1000
−2000
current, mA
current through the load C2
80
60
40
20
0
−20
−40
−60
−80
0
0.01
0.02
0.03
0.04
C1=63 nF, C2=63 nF, f=100 Hz
0.05
time, s
0.06
0.07
0.08
0.09
0.1
Figure 3: Simulation results for the circuit shown in Figure 1, where f=100 Hz, C1=C2=63 nF.
When considering a choice of capacitors for
the voltage shifting circuit, two factors have to
be taken into account: total capacitance of C1
C1·C2
and C2 connected in series( C1+C2
), and the ratio
of a capacitive divider formed by both capacitors
C1
). It is important to pay attention on the volt( C1+C2
age rating of the capacitors.
The frequency of the input voltage signal also influences choice of capacitors. For example, if the
frequency is changed from 100 Hz to 1 kHz, the
current drawn from the source multiplies by the
factor of 10 (Figure 5). Capacitor values have to
be lowered by the factor of 10 to get back to the
20 mA current level.
4
DC voltage generation
By combining several stages of the basic circuit
from Figure 1, a DC high voltage signal can be obtained. Figure 7 presents a two stage DC voltage
multiplier generating positive output DC voltage.
A similar, two stages negative DC voltage circuit
is shown in Figure 8.
+ D.C.
amplifier
(AC source)
The output voltage Uout from the capacitive divider is maximized for C1>>C2. For example, let
C1=12,000 nF and C2=1 nF. Results of a simulation with these C1 and C2 values are shown in
Figure 6.
D
C
D
C
Figure 7: Positive voltage multiplier.
TREK, INC. • 190 Walnut Street • Lockport, NY 14094 • Tel: (716) 438-7555
Call: 1 800 FOR TREK • FAX: (716) 201-1804 •
Copyright © 2013 TREK, INC.
Email: sales@trekinc.com • Web: www.trekinc.com
0408/MAN Rev. 2
page 2 of 5
Trek Application Note
Number 6001
Voltage level shifting
input voltage
voltage, V
2000
1000
0
−1000
−2000
voltage across C1
voltage, V
2000
1000
0
−1000
−2000
voltage across C2
voltage, V
2000
1000
0
−1000
−2000
current, mA
current through the load C2
80
60
40
20
0
−20
−40
−60
−80
0
0.01
0.02
0.03
0.04
C1=31.5 nF, C2=63 nF, f=100 Hz
0.05
time, s
0.06
0.07
0.08
0.09
0.1
Figure 4: Simulation results for the circuit shown in Figure 1, where f=100 Hz, C1=31.5 nF, C2=63 nF.
C
D
- D.C.
amplifier
(AC source)
D
n * 2U
C
2U
Figure 8: Negative voltage multiplier.
U
0
t
-U
T
Figure 9: DC multiplier signal.
The number of stages that can be used in this
kind of design is limited by current capabilities of
TREK, INC. • 190 Walnut Street • Lockport, NY 14094 •
Tel: (716) 438-7555
Call: 1 800 FOR TREK • FAX:(716) 201-1804 • E-mail: sales@trekinc.com •
Copyright © 2013 TREK, INC.
0408/MAN Rev. 2
Web: www.trekinc.com
page 3 of 5
Trek Application Note
Number 6001
Voltage level shifting
input voltage
voltage, V
2000
1000
0
−1000
−2000
voltage across C1
voltage, V
2000
1000
0
−1000
−2000
voltage across C2
voltage, V
2000
1000
0
−1000
−2000
current, mA
current through the load C2
800
600
400
200
0
−200
−400
−600
−800
0
0.001
0.002
0.003
0.004
C1=63 nF, C2=63 nF, f=1000 Hz
0.005
time, s
0.006
0.007
0.008
0.009
0.01
Figure 5: Simulation results for the circuit shown in Figure 1, where f=1000 Hz, C1=C2=63 nF.
the circuit. The DC output voltage has an AC voltage ripple δU (Figure 10) given by equation [1]:
n * 2U
no load
∆U
with load
δU =
I n · (n + 1)
·
,
fC
4
δU
(5)
T=1/f
Figure 10: Voltage ripple and voltage drop.
where I is the load current and n is the number of
stages. A voltage drop ∆U (Figure 10) due to the
load can be calculated using formula [1]:
I
∆U =
fC
2n3 n
3 6
TREK, INC. • 190 Walnut Street •
Lockport, NY 14094
Call: 1 800 FOR TREK • FAX:(716) 201-1804 •
Copyright © 2013 TREK, INC.
(6)
References
t
[1] E. Kuffel, W. S. Zaengl, and J. Kuffel. High
Voltage Engineering: Fundamentals. Newnes,
2 edition, 2000.
•
Tel:
E-mail: sales@trekinc.com •
0408/MAN Rev. 2
(716) 438-7555
Web: www.trekinc.com
page 4 of 5
Trek Application Note
Number 6001
Voltage level shifting
voltage across C1
2000
1000
1000
voltage, V
voltage, V
input voltage
2000
0
−1000
0
−1000
−2000
−2000
voltage across C2
current through the load C2
1000
current, mA
voltage, V
2000
0
−1000
−2000
8
6
4
2
0
−2
−4
−6
−8
current, mA
current through C1
8
6
4
2
0
−2
−4
−6
−8
0
0.02
0.04
0.06
0.08
time, s
C1=12000 nF, C2=1 nF, f=100 Hz
0.1
Figure 6: Simulation results for f=1000 Hz, C1=12000 nF, C2=1 nF.
TREK, INC. • 190 Walnut Street •
Lockport, NY 14094 •T Tel: (716) 438-7555
Call: 1 800 FOR TREK • FAX:(716) 201-1804 • E-mail: sales@trekinc.com •
Copyright © 2013 TREK, INC.
0408/MAN Rev. 2
Web: www.trekinc.com
page 5 of 5