How to linearly scale up and down a voltage?

• posted

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

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."

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Thanks,
- Win```
• posted

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)

1. 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

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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.

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Tim Wescott
Wescott Design Services```
• posted

Yes, you're correct.

Correct. Yes, there are. Have a look at APEX :

Rene

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• posted

Tim Wescott wrote in news: snipped-for-privacy@corp.supernews.com:

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

• posted

Winfield Hill wrote in news: snipped-for-privacy@drn.newsguy.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.

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

• posted

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

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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).

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.

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Tim Wescott
Wescott Design Services```
• posted

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.

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Thanks,
- Win```
• posted

Mac wrote in news: snipped-for-privacy@bar.net:

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?

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.

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

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

• posted

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.

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?

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?

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. ;-)

--Mac

• posted

Winfield Hill wrote in news: snipped-for-privacy@drn.newsguy.com:

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

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Tim Wescott wrote in news: snipped-for-privacy@corp.supernews.com:

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

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

• posted

100M and 1G are easy to get, such as the Ohmite ?Mini-Mox? series,
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