How to linearly scale up and down a voltage?

[Michael Noone]

Hi - I have two problems I'm trying to figure out how to solve:

1. Take a +-200V signal (call it Vin) and scale it down to +-10V (call
that Vout), without drawing much/any current from the input. The output
will not need to sync much current at all - 10 ma would probabaly be
more than adequate. I actually think that this input is high impedance -
so I'm going to have to look into that. (someone I'm working with
promised to e-mail me specs tomorrow morning)

2. Take a +-10V signal and scale it up to +-200V. The input could
probabaly draw 10ma or so, and the output needs to be able to sync 20ma.
Also - I need some sort of way to limit current to 20ma. Anything past
that the circuit should either shut off or just limit the current to
20ma.

For the first problem - the only thing I could really think of was a
voltage divider, as simple as that seems. I believe that the circuit
connected to Vout is high impedance, so I think I could just connect say
R1 to Vout and ground, and R2 to Vout and Vin, with R1=9R2. If it's not
very high impedance (say - Z ohms) I could instead connect a resistor
also Z ohms between Vout and ground, and then connect a resistor of 9Z/2
ohms between Vin and Vot, thus again dividing the input voltage by 10.

I'm thinking for the second problem I should use op-amps in non-
inverting amplifier configurations - input connected to V+, resistor R1
connected between Vo and V- and resistor R2 connected between V- and
GND. Then Vo=Vs(1+R1/R2) - so I'd want 10=1+R1/R2, thus 9R2=R1. But are
there op amps out there that can handle 20ma and 200V?

Any thoughts/suggestions? Oh, and if it matters each of these circuits
will be implemented 10 times on the same board. Thanks for the help!

-Michael Noone

 

[Rene Tschaggelar]

Hi - I have two problems I'm trying to figure out how to solve:

1. Take a +-200V signal (call it Vin) and scale it down to +-10V (call
that Vout), without drawing much/any current from the input. The output
will not need to sync much current at all - 10 ma would probabaly be
more than adequate. I actually think that this input is high impedance -
so I'm going to have to look into that. (someone I'm working with
promised to e-mail me specs tomorrow morning)

2. Take a +-10V signal and scale it up to +-200V. The input could
probabaly draw 10ma or so, and the output needs to be able to sync 20ma.
Also - I need some sort of way to limit current to 20ma. Anything past
that the circuit should either shut off or just limit the current to
20ma.

For the first problem - the only thing I could really think of was a
voltage divider, as simple as that seems. I believe that the circuit
connected to Vout is high impedance, so I think I could just connect say
R1 to Vout and ground, and R2 to Vout and Vin, with R1=9R2. If it's not
very high impedance (say - Z ohms) I could instead connect a resistor
also Z ohms between Vout and ground, and then connect a resistor of 9Z/2
ohms between Vin and Vot, thus again dividing the input voltage by 10.

Yes, you're correct.

I'm thinking for the second problem I should use op-amps in non-
inverting amplifier configurations - input connected to V+, resistor R1
connected between Vo and V- and resistor R2 connected between V- and
GND. Then Vo=Vs(1+R1/R2) - so I'd want 10=1+R1/R2, thus 9R2=R1. But are
there op amps out there that can handle 20ma and 200V?

Correct.
Yes, there are. Have a look at APEX :
http://www.apexmicrotech.com/

Rene

 

[Winfield Hill]

Michael Noone wrote...

Hi - I have two problems I'm trying to figure out how to solve:

1. Take a +-200V signal (call it Vin) and scale it down to +-10V (call
that Vout), without drawing much/any current from the input. The output
will not need to sync much current at all - 10 ma would probabaly be
more than adequate. I actually think that this input is high impedance
- so I'm going to have to look into that. (someone I'm working with
promised to e-mail me specs tomorrow morning)

If you truly mean "high" high impedance, you won't be able to afford
the load from attaching a resistive divider. I had a situation like
that, which I solved with a 400V mosfet voltage follower. But you'll
have a +/-220V power supply available (see below), so you could use a
low-cost APEX PA97 opamp to buffer the divider-resistor signal. See
http://www.apexmicrotech.com

2. Take a +-10V signal and scale it up to +-200V. The input could
probabaly draw 10ma or so, and the output needs to be able to sync 20ma.
Also - I need some sort of way to limit current to 20ma. Anything past
that the circuit should either shut off or just limit the current to
20ma.

200V at 20mA implies 4W of power delivered to the load. If you use a
high-voltage power opamp, such as the APEX PA15, you'll need a bipolar
dual regulated power supply with 220V (+/-5V) 6W each polarity to run
the opamp. You'll also be needing some sizable heat sinks, and perhaps
even a small fan. The PA15 provides for an adjustable current limit.

Finally, remember the rule authorized personnel follow when working on
high-voltage circuitry, "one hand behind your back."

 

[Michael Noone]

Michael Noone wrote...

If you truly mean "high" high impedance, you won't be able to afford
the load from attaching a resistive divider. I had a situation like
that, which I solved with a 400V mosfet voltage follower. But you'll
have a +/-220V power supply available (see below), so you could use a
low-cost APEX PA97 opamp to buffer the divider-resistor signal. See
http://www.apexmicrotech.com

I think buffering it is a very good idea, and would allow me to not have
to worry about what I attach to it's output.

Something doesn't seem right here though - I looked up the PA97 on
Apex's web site - and it is listed as costing $52.61. That seems to be
in quantities of 1 as well... Am I looking at something wrong? Also - I
was unable to find any distributor that carries these parts, (I checked
the 3 usual suspects - Digikey, Mouser, and Newark) besides Apex
themselves.

200V at 20mA implies 4W of power delivered to the load. If you use a
high-voltage power opamp, such as the APEX PA15, you'll need a
bipolar dual regulated power supply with 220V (+/-5V) 6W each
polarity to run the opamp. You'll also be needing some sizable heat
sinks, and perhaps even a small fan. The PA15 provides for an
adjustable current limit.

Finally, remember the rule authorized personnel follow when working
on high-voltage circuitry, "one hand behind your back."

The Apex PA15 does indeed look like a good match for my needs - but
again is that price right? It's listed as over $100 in their online
store, and as far as I can tell that's in quantities of 1!

Thanks so much for the help,

-Michael Noone

 

[Tim Wescott]

I think buffering it is a very good idea, and would allow me to not have
to worry about what I attach to it's output.

Something doesn't seem right here though - I looked up the PA97 on
Apex's web site - and it is listed as costing $52.61. That seems to be
in quantities of 1 as well... Am I looking at something wrong? Also - I
was unable to find any distributor that carries these parts, (I checked
the 3 usual suspects - Digikey, Mouser, and Newark) besides Apex
themselves.





The Apex PA15 does indeed look like a good match for my needs - but
again is that price right? It's listed as over $100 in their online
store, and as far as I can tell that's in quantities of 1!

Thanks so much for the help,

-Michael Noone

Yes those prices are correct, no you can't get them from Digi-Key. Apex
specializes in making stuff that no one else does, and charging you
through the nose for it. They're generally hybrid circuits, and they're
generally bullet-proof.

They make good sense if you're designing just a few pieces and you
charge for your time because making and validating precision high
voltage/high power stuff takes lots of effort, while plunking an Apex
part on a board doesn't take much more time than plunking down a 741.

If you don't need such precision (i.e. if you could load your 200V line
with a 10M resistive divider, and your 200V output could be really slow
and/or have lots of offset) then your buy/design tradeoff may come at a
different production point.

 

[Michael Noone]

Yes those prices are correct, no you can't get them from Digi-Key.
Apex specializes in making stuff that no one else does, and charging
you through the nose for it. They're generally hybrid circuits, and
they're generally bullet-proof.

They make good sense if you're designing just a few pieces and you
charge for your time because making and validating precision high
voltage/high power stuff takes lots of effort, while plunking an Apex
part on a board doesn't take much more time than plunking down a 741.

If you don't need such precision (i.e. if you could load your 200V
line with a 10M resistive divider, and your 200V output could be
really slow and/or have lots of offset) then your buy/design tradeoff
may come at a different production point.

OK well a couple questions: Is it really necessary to use a high voltage
Apex chip for the voltage divider circuit? It seems to me that it will
only be seeing about 10V - so wouldn't a normal op-amp (ie 741) work
just as well?

But for the other circuit - are there any other options? Problem is each
circuit needs to be implemented 10 times - so I'd be looking at $1000 in
chips alone, which would probabaly make my boss rather unhappy. I must
admit I'm not familiar with high voltage linear amplifier design - how
difficult is it to implement such an amplifier without specialized chips
like the Apex chips?

Thanks again,

-Michael Noone

 

[Mac]

Hi - I have two problems I'm trying to figure out how to solve:

1. Take a +-200V signal (call it Vin) and scale it down to +-10V (call
that Vout), without drawing much/any current from the input. The output
will not need to sync much current at all - 10 ma would probabaly be
more than adequate. I actually think that this input is high impedance -
so I'm going to have to look into that. (someone I'm working with
promised to e-mail me specs tomorrow morning)

2. Take a +-10V signal and scale it up to +-200V. The input could
probabaly draw 10ma or so, and the output needs to be able to sync 20ma.
Also - I need some sort of way to limit current to 20ma. Anything past
that the circuit should either shut off or just limit the current to
20ma.

For the first problem - the only thing I could really think of was a
voltage divider, as simple as that seems. I believe that the circuit
connected to Vout is high impedance, so I think I could just connect say
R1 to Vout and ground, and R2 to Vout and Vin, with R1=9R2. If it's not
very high impedance (say - Z ohms) I could instead connect a resistor
also Z ohms between Vout and ground, and then connect a resistor of 9Z/2
ohms between Vin and Vot, thus again dividing the input voltage by 10.

I'm thinking for the second problem I should use op-amps in non-
inverting amplifier configurations - input connected to V+, resistor R1
connected between Vo and V- and resistor R2 connected between V- and
GND. Then Vo=Vs(1+R1/R2) - so I'd want 10=1+R1/R2, thus 9R2=R1. But are
there op amps out there that can handle 20ma and 200V?

Any thoughts/suggestions? Oh, and if it matters each of these circuits
will be implemented 10 times on the same board. Thanks for the help!

-Michael Noone

I've read some of the other replies and your replies to those replies.

I think you have your first problem well in hand. Use a voltage divider
with 1% tolerance resistors, and you should be good to go. Note that 200
Volts is pretty high, and can do a lot of damage if something goes wrong.
Someone could even end up getting hurt. I am not dealing at all with
the safety issues surrounding this thing.

For the second problem, there are other ways to potentially do this that
are a LOT cheaper than using the high-voltage op-amps you priced.

If the signal is AC, maybe you can use a transformer inside the feedback
loop to an ordinary op-amp. What is the bandwidth of the signal? Is there
a DC component?

Or you could build some kind of switching regulator with high Voltage
output transistors and opto-couple the transistor drivers to the
control circuitry. With your relatively low current requirement, this
shouldn't be too hard. In order to sink and source current, you might need
something like an H-bridge.

But a lot depends on how Vin varies over time, and what the load is like.
If it is a slow varying signal, then this approach could work. If it is a
high bandwidth signal, you might have to look at something else.

Good Luck!
--Mac

 

[Michael Noone]

I've read some of the other replies and your replies to those replies.

I think you have your first problem well in hand. Use a voltage
divider with 1% tolerance resistors, and you should be good to go.
Note that 200 Volts is pretty high, and can do a lot of damage if
something goes wrong. Someone could even end up getting hurt. I am not
dealing at all with the safety issues surrounding this thing.

Mr. Hill suggested I use an op-amp to buffer that voltage divider output -
and this makes excellent sense to me, as I would then not even have to
worry about the impedance of the circuit connected to the output of the
voltage divider circuit. This makes great sense to me - but I don't
understand why he suggested an apex op-amp - to me it seems that any normal
op-amp would be perfectly suitable. What do you think?

For the second problem, there are other ways to potentially do this
that are a LOT cheaper than using the high-voltage op-amps you priced.

If the signal is AC, maybe you can use a transformer inside the
feedback loop to an ordinary op-amp. What is the bandwidth of the
signal? Is there a DC component?

I've been told the signal will be changing at about 1KHz. I think it may
actually be an entirely positive signal (never dipping below 0V), but I'm
not positive about that just yet. So it will be entirely DC.

Or you could build some kind of switching regulator with high Voltage
output transistors and opto-couple the transistor drivers to the
control circuitry. With your relatively low current requirement, this
shouldn't be too hard. In order to sink and source current, you might
need something like an H-bridge.

Can you explain this circuit a bit more? I'm having trouble picturing it.

But a lot depends on how Vin varies over time, and what the load is
like. If it is a slow varying signal, then this approach could work.
If it is a high bandwidth signal, you might have to look at something
else.

Good Luck!
--Mac

The load is a bit wierd. It's some sort of polarized fluid. I'm not too
sure on the details of it as they haven't been given to me.

Thanks for all your help,

-Michael

 

[Tim Wescott]

OK well a couple questions: Is it really necessary to use a high voltage
Apex chip for the voltage divider circuit? It seems to me that it will
only be seeing about 10V - so wouldn't a normal op-amp (ie 741) work
just as well?

It all depends on just how much current you can draw. If you can't draw
_any_ current then you need to use a differential FET pair, and the APEX
part is a least-engineering way to do it. If you can draw _some_
current then you can use a resistive divider into an op-amp (pay
attention to input bias current -- you may be best with a JFET op-amp).

But for the other circuit - are there any other options? Problem is each
circuit needs to be implemented 10 times - so I'd be looking at $1000 in
chips alone, which would probabaly make my boss rather unhappy. I must
admit I'm not familiar with high voltage linear amplifier design - how
difficult is it to implement such an amplifier without specialized chips
like the Apex chips?

Not too, if you know what you're doing -- but the fact that you need to
ask suggests that you'll at least have to learn how.

If you take the amount you get paid before taxes and multiply by two or
three thats about how much you cost your company to work there. Assume
that you can build the functionality you need with $10 worth of parts
(this may be doable if you're good and your requirements aren't too
great and your production volume is moderate to high).

Now take that $990 per board difference and multiply it by the number
of boards you expect to produce in a year or two. Call it cost A.

Now estimate how long it'll take you to learn how to build the circuit,
debug it, validate it, train the production staff not to kill themselves
while testing it, etc. Multiply that by how much you cost your company
per hour. Call it cost B.

If A > B then start designing.

If A < B then start buying.

If your boss doesn't understand the reasoning then do whatever he says
or brush up your resume.

 

[Winfield Hill]

Michael Noone wrote...

Mr. Hill suggested I use an op-amp to buffer that voltage divider output
- and this makes excellent sense to me, as I would then not even have to
worry about the impedance of the circuit connected to the output of the
voltage divider circuit. This makes great sense to me - but I don't
understand why he suggested an apex op-amp - to me it seems that any
normal op-amp would be perfectly suitable. What do you think?

Michael, you wrote, "1. Take a +-200V signal (call it Vin) and scale
it down to +-10V (call that Vout), without drawing much/any current
from the input." The only way to not draw "any" current from the
+/-200V input is to avoid directly connecting your voltage divider,
and to instead connect it after an active FET buffering stage. Such
a buffer would have to operate over the full +/-200V input range, so
clearly "any normal op-amp" would NOT be perfectly suitable. If you
want to redefine your problem, and allow for some input loading, then
select a pair of appropriately-high-value resistors and have at it.

 

[Mac]

Mr. Hill suggested I use an op-amp to buffer that voltage divider output -
and this makes excellent sense to me, as I would then not even have to
worry about the impedance of the circuit connected to the output of the
voltage divider circuit. This makes great sense to me - but I don't
understand why he suggested an apex op-amp - to me it seems that any normal
op-amp would be perfectly suitable. What do you think?

Mr. Hill is well-known as one of the authors of an excellent and popular
book on electronics: The Art of Electronics. The third edition will be
out one of these days, but until then, get the second edition.

I wouldn't presume to speak for Mr. Hill, but I believe he was operating
on the assumption that the output impedance of your 200 V signal was too
high for you to put resistor on it. And unless you can tell us how much
current we are allowed to draw when the input is at 200 V, that may be the
best assumption.

I on the other hand saw that you yourself proposed a Voltage divider, so I
assumed, perhaps wrongly, that it was OK to draw a little current. Anyway,
if the output impedance of the 200 V signal is, say, 1000 Ohms or even
100,000 Ohms, you might be able to get away with the divider. Above
that, I'm not sure. Some of the people here would know. Anyway, if the
divider is acceptable, then you will certainly want to buffer the divider
output using some kind of fet (or jfet) input op-amp as suggested by Tim
Wescott, I think it was.

Unfortunately, there are a lot of unanswered questions about your
application, so everybody is making assumptions, and some of them are
bound to be wrong. For example, what Voltages do you have available? Do
you have +/- 200 (or higher) VDC rails? If not, you might want to ask
yourself how you are going to generate them, especially if you are in the
USA where even line Voltage isn't that high.

I've been told the signal will be changing at about 1KHz. I think it may
actually be an entirely positive signal (never dipping below 0V), but I'm
not positive about that just yet. So it will be entirely DC.

It would be good to know whether the signal dips below 0V or not. But what
I was getting at with my question about a DC component was that it might
be possible to output a signal at a much lower voltage and then pass it
through a transformer to get the voltage up to where you need. But this
will only work if there is no DC component, since transformers don't
pass DC. Where is the signal coming from?

Can you explain this circuit a bit more? I'm having trouble picturing it.

An H-bridge is probably too complicated. Especially if the signal is
always positive.

But since you asked, an H-bridge looks like this: (Use courier or similar
for ASCII art schematic):


V+ V+
| |
| |
S1 / S2 /
/ /
| |
| |-----| |
+----|load |-----+
| |-----| |
| |
S3 / S4 /
/ /
| |
| |
V- V-


You turn the switches on in pairs. S2 and S3 can go on together, and S1
and S4 can go on together. Note that the H bridge can apply +/- V+-V-
Volts to the load, which means that it doubles the available voltage
swing, in a sense. For example, if V+ is 200 V, and V- is 0 V, the
H-bridge can apply +/- 200 Volts.

This setup is usually used for inductive loads, which naturally smooth out
the voltage square wave being applied to them.

For the switches, you would use high voltage MOSFET's, and you could use a
small drive circuit referenced to the source of the MOSFET. This small
drive circuit would be turned on and off by an opto-coupled input which
would be generated somewhere else based on the Voltage across the
load.

An opto-coupler chip typically has an LED and a photo-transistor inside of
it. When you pass current through the LED, the photo-transistor turns on.
Since there is no electrical connection between the input and output, they
can be at totally different voltages. They will have an isolation rating
which will tell you how large the difference can be. 200 Volts is no
problem.

What I had in mind was a small drive circuit, powered by line voltage
connected to the smallest transformer you can find which has a suitable
isolation rating between the primary and secondary. This entire drive
circuit would be referenced ("grounded") to the source of the MOSFET, and
turned on and off with the opto-coupler. Some of the clever people around
here might even be able to put together a drive circuit using a capacitive
Voltage divider from line Voltage or something like that instead of a
transformer. But I'm not that clever. ;-)

Anyway, with your slow moving signal, you might be able to somehow
trick a standard switching DC-DC controller into creating the control
signal for a single high voltage FET which you then drive as I outlined.

It looks like this (use courier, again):
V+
|
|
-- D
G____|| N-channel FET
||
--S
|+----------------+
--- |
| | ---
| | inductor /\ diode
| | ----
--- |
|_________________|
| |
--- |
--- capacitor |
| ---
| | |
| | | load
| | |
| ---
GND |
GND

I am assuming that the Voltage is always positive. If not, then replace
GND with -200 (or 220 or whatever) Volts.

You would sense the voltage at the load, and, using a voltage divider,
feed it back to the controller chip (which I haven't shown). You would
drive the gate with a tiny drive circuit referenced to the FET source, as
I said. And the drive would be turned on and off with an opto-coupler.

I don't know for sure that this would work, but I think it has a shot. It
might not be the world's safest circuit, with V+ being over 200 Volts...
What happens if the FET fails as a short circuit?

The load is a bit wierd. It's some sort of polarized fluid. I'm not too
sure on the details of it as they haven't been given to me.

Interesting. Do you think you might be in a little over your head? Well,
sometimes that's how one learns, I guess. Just don't electrocute yourself
or anyone else. ;-)

Thanks for all your help,

-Michael

--Mac

 

[Michael Noone]

Michael, you wrote, "1. Take a +-200V signal (call it Vin) and scale
it down to +-10V (call that Vout), without drawing much/any current
from the input." The only way to not draw "any" current from the
+/-200V input is to avoid directly connecting your voltage divider,
and to instead connect it after an active FET buffering stage. Such
a buffer would have to operate over the full +/-200V input range, so
clearly "any normal op-amp" would NOT be perfectly suitable. If you
want to redefine your problem, and allow for some input loading, then
select a pair of appropriately-high-value resistors and have at it.

Hi - I'm sorry - I misunderstood what you were saying. I thought you
were suggesting having a voltage divider circuit connected to the input,
and then have an op-amp connected to the output of the voltage divider.
I think you can understand why I would think a normal op-amp would work
in that situation! Again - my bad, I need to learn how to read better.
But my misinterpretation also still makes alot of sense to me. So I was
thinking for the voltage divider circuit, use the largest possible
resistors. The largest I've seen is 22M, though I expect larger are
available. Thus at 200V, about 18 micro amps would be drawn by the
voltage divider circuit, which I expect is not enough to affect the
current that is being monitored, as it will be nearly 3 orders of
magnitude greater (I believe about 10ma compared with 20ua). Thus then a
normal op-amp could be used (as, unless I'm being dumb, it would only
see about 10V max if I set up the voltage divider correctly) to buffer
the output from that voltage divider, and the output from the op-amp
would be the output of the whole circuit (which would then be connected
to a +-10V ADC).

So how does that sound? Thanks again for all your help - I really can't
thank all of you enough.

-Michael

 

[Michael Noone]

Michael Noone wrote:

It all depends on just how much current you can draw. If you can't
draw _any_ current then you need to use a differential FET pair, and
the APEX part is a least-engineering way to do it. If you can draw
_some_ current then you can use a resistive divider into an op-amp
(pay attention to input bias current -- you may be best with a JFET
op-amp).

I think a very small amount of current can be drawn. I believe current
through the circuit attached to the voltage divider circuit (or whatever
else i use) would be about 10ma, so I'd like for the voltage divider
circuit to draw at most a couple of orders of magnitude less current.
(so say something along the lines of .1ma

Not too, if you know what you're doing -- but the fact that you need
to ask suggests that you'll at least have to learn how.

If you take the amount you get paid before taxes and multiply by two
or three thats about how much you cost your company to work there.
Assume that you can build the functionality you need with $10 worth of
parts (this may be doable if you're good and your requirements aren't
too great and your production volume is moderate to high).

Now take that $990 per board difference and multiply it by the
number of boards you expect to produce in a year or two. Call it cost
A.

Now estimate how long it'll take you to learn how to build the
circuit, debug it, validate it, train the production staff not to kill
themselves while testing it, etc. Multiply that by how much you cost
your company per hour. Call it cost B.

If A > B then start designing.

If A < B then start buying.

If your boss doesn't understand the reasoning then do whatever he says
or brush up your resume.

Don't worry about my time costing too much - I am but a research
assistant at my university (UIUC). In a full day's work (though I don't
actually work 40 hours a week as I'm a full time student) - I make less
at my job then just one of those op-amps costs My boss is the head of
the MIE nanotechnology department (called nano-CEMMS) - and thus his
knowledge of electronics is at most limited. He pretty much tells me
what he wants and assumes everything is easy and cheap. It's also
wonderful when I ask him questions like "what accuracy do I need?",
"what speed do I need?", "how much current can be drawn?", etc. and he
just tells me as accurate as possible, as fast as possible, as little as
possible, etc. Thus hopefully this accounts for why I am so vague - as
I'm given no information to work with. Thanks for your patience,

-Michael

-Michael

 

[Winfield Hill]

Michael Noone wrote...

... I was thinking for the voltage divider circuit, use the largest
possible resistors. The largest I've seen is 22M, though I expect
larger are available.

100M and 1G are easy to get, such as the Ohmite “Mini-Mox” series,
http://dkc3.digikey.com/PDF/C051/1032.pdf