Relatively high voltage precision DAC amplification

[peter-h]

However, you don't need to do this, in the 21st century. In 1970 etc it was the only way hence HP and Tek etc developed all those fancy circuits, but now you can get HV devices.

[jbb]

… that’s a good point. For a low volume (especially a one-off!) design, just buying a premium high V device is likely cheaper than developing & properly testing a cascode arrangement.

Same argument applies to precision resistor arrays, I guess.

[peter-h]

Yes. I was doing precision analog, also HV, in the 1970s, and back then you had to develop all kinds of weird techniques. So a job got solved with 10x more parts than would be the case today. A lot of stuff used hand-selected / matched components, especially references, and (for HV) feedback resistors, temperature ovens for references, etc. It was an interesting area because those who developed those skills were able to sell products which nobody else could make. And some of it has not got easier because so few people understand analog.

Today you just buy a 1ppm/C ref for a few quid, an op-amp for less, and you are mostly done.

Today's precision is amazing. I am building a product with an ADS1118. All of them have their internal Vref within about 0.01% - many times better than the spec. Even the 10 pence TL431 tends to be within 0.01% or so. Well, they are laser trimming, so why not laser trim to 0.01% even if the spec says 1%.

In this case, building a 1kV op-amp, just throw in some 4kV devices and you are done. Just don't blow yourself up - this stuff is fatal!

[Zero999]

However, you don't need to do this, in the 21st century. In 1970 etc it was the only way hence HP and Tek etc developed all those fancy circuits, but now you can get HV devices.

I looked at the links in your post. The IXTT02N450HV looks good. Would you advise using two of them, or just one and a cascode for the low current level shifting/driver transistors?

https://www.mouser.co.uk/datasheet/2/240/media-3321478.pdf

[peter-h]

But why complicate things.

[Laval]

This circuit cannot output 0v or even near 0v. The transistor on the left will hit saturation before.

It might be easier to stabilise, if you add local feedback to the transistor output stage. This example has a gain of 101. The op-amp IC will need to have a -2V supply for the circuit to output voltages below 70V.

This circuit is puzzling me. It looks almost like Q2 is in a common base configuration but without the base being to ground (and in fact varying with the output). Could you give some explanation as to how it works ?

[Zero999]

This circuit is puzzling me. It looks almost like Q2 is in a common base configuration but without the base being to ground (and in fact varying with the output). Could you give some explanation as to how it works ?

It is common base, of sorts. Just because the base voltage can vary, it doesn't mean it's not common base. The same is true for a common emitter amplifier, without a capacitor, even though the emitter voltage varies, depending on the input. Where it's common base or emitter, depends on which terminal the input is connected to.

Another way to look at it is, the BJT is really a type of differential amplifier. The collector current, depends on the voltage difference between the base and emitter. The collector resistor converts this current into a voltage, which is buffered by another transistor to reduce the impedance, so with now have a type of differential voltage amplifier. The emitter of Q2 is the non-inverting input and the base is the inverting input and Q1's emitter is the output. The output is connected to the inverting input via a potential divider (R2 & R3) which acts as a negative feedback network and the input signal is connected to the non-inverting input. This is similar to that of an op-amp in non-inverting configuration, except it has a massive voltage offset of V_BE (about 0.6V), due to Q2's base-emitter junction. The output voltage is approximately equal to the following formula:
V_OUT = (V_IN+V_BE)*(1+R2/R3).

Here's a graph showing the voltage at Q2's base (oa_out), when the input voltage is swept from 0 to 10V. The U1 corrects the voltage offset, hence a negative voltage is required to get to zero. The voltage drop across R2 due to the base current, means U1's output needs to be a little more negative, than the simple formula above would suggest.

[BrianHG]

This circuit is puzzling me. It looks almost like Q2 is in a common base configuration but without the base being to ground (and in fact varying with the output). Could you give some explanation as to how it works ?

You just figured out why I just 'love' that circuit...

It might be easier to stabilise, if you add local feedback to the transistor output stage. This example has a gain of 101. The op-amp IC will need to have a -2V supply for the circuit to output voltages below 70V.
(Attachment Link)

Damn, I like that circuit....

[peter-h]

This circuit cannot output 0v or even near 0v. The transistor on the left will hit saturation before.

Yes, but this is really basic electronics. You can put a couple of diodes in series with the output transistor's emitter, which will drop 1.2V or so, so the output can swing down to 0.

But if you want the ability to control a zero voltage then of course you also need a negative rail, which can be quite small, say -0.5V.