DC Voltage amplification

hello board,
how can I convert DC voltage range +/-14 V to +/-40V? so that -14V gives -40V and +14V gives +40V.
thank you in advance

You need a simple non inverting op-amp with a gain of 2.xx. Theoratically this just constitutes of an op-amp and 2 resistors. But practically, you have to find a chip that can handle an 80V output swing. I would suggest using an op-amp coupled with a class-B output stage with suitable transistors to handle the voltages.

The design of +/-40V output stage depends on the required output power (current).
Here is an example of 120V (+/-60V) opamp:
https://www.prema.com/Design/120vopamp.html
but only 1.5mA can be drawn from its output ..
Regards,
IanP

thanks IanP and techie. actually the output of this circuit will be fed to the gates of two MOSFETs so I don't need power amplification ; but the main problem is OpAmps which meets those specification are not available for me( I need them rather urgently) or are very expensive. so I decided to do the job with transistors. the main problem that I have is that I want to do this with supply voltage about +/-40 it means output swing voltage will be near the supply voltage. thank you again.

Attached is a diagram of High-Voltage amplifier ..
The gain of this circuit = -R2/R1 ..
The output voltage is almost +Ucc - to - -Uee, and is only limited by the voltage rating of transistors.
In this application you can use any available cheap opamp, including popular types such as LM741, LM358 or similar.
Regards,
IanP

Thank you Very much IanP, I'm really thankful for your time.
I understood the cicuit as follows please correct me if i'm wrong.
two transistors that provide supply voltage for Opamp(lets name them Q1, Q2)
their bases are biased with 2 voltage dividers to give the needed voltage.
Voltages of their collectors(Q1,Q2) are constant so 2 transistors which their
bases are connected to Q1 & Q2 and arranged in a push pull-like configuration
(Let's name them Q3 for PNP, Q4 for NPN) draw a constant current
(I assume their bases are biased with 2 resistors so that they source and sink
the same current). so if there is no input voltage output voltage will be zero.
no problem to this point but how the opAmp works? because
even if the gain R2/R1 ratio will be appropriate -14V input gives +14V
output(Saturation) for OpAmp this voltage will be blocked by High pass filter.
if Q3 and Q4 are supposed to sink and source same amount of current from where
input can sink or source current? the opamp has a high pass filter so how it can
feedback dc voltages? is it possible to change this circuit so it can be used for
DC voltage amplification ? thank you again.

The role of capacitors between the output of the opamd and in parallel with R2 is to provide path for HF compensation.

You didn't mention the function of the resistor connected between the output of the opamp and 0V.
Let's call it RL.
It provides a load to the opamp dc path. Its value can be as low as <500Ω .. (depending on the max current of selected opamp).
DC feed back is provided through R2 .. and the control of the output stage (transistors) is through the supply lines of the opamp ..

So, it is amplifier that will amplify signals from DC to fmax limited by the value of both capacitors ..

Regards,
IanP

Thanks a lot IanP!!! I think now I understood how the circuit works! nice idea!
again please correct me if I'm wrong. Q3 and Q4 do not always sink and source same amount of current as previously I thought! the current they sink or source are proportional to the current which each of supply lines of OpAmp takes. so RL purposefully selected to have a low amount so when there are differences between
inputs of OPAmP the corrosponding power supply of OpAmp takes more current and voltage drop across collector resistance of Q1 or Q2 becomes higher and corresponding Q3 or Q4 sink or source more current and as the amount of current that the other supply line takes is constant so the extra amount of current comes from path R1 & R2 and it continues until inverting input of OpAmp becomes Zero and RL (and supply lines) dosn't take any current and circuit becomes stable! nice idea! thank you!
Well, I also came with an idea, the idea is to use a current mirror as it is shown in
the attachment ; but I'm wondering which one is better ? as you see in the circuit there is a feedback so I'm not worry about voltage drifts or temprature drifts but the most important thing that I need is low noise and precesion. the output voltage may be needed to be precise within a fraction of a millivolt.well I can replace current mirror with your circuit easily but which circuit do you think will be more precise and more stable with lower noise: Current mirror or your circuit nested in the feedback loop In the attachment?
thank you very much again!

Use simple opamp in the non-inverting feedback configuration. Adjust the gain, using external componants.
The accuracy of the depends upon the loop gain and the precision of the external componants used.

Is it drawing error, or should Q1 be a p-channel?
Also, I don't think you need a voltage divider and additional buffer between output and error amplifier ..
Regards,
IanP

IanP said:

Is it drawing error, or should Q1 be a p-channel?
Also, I don't think you need a voltage divider and additional buffer between output and error amplifier ..

Click to expand...

Hello,
I'm Sorry. Q1 should have been mirrored so that source be connected to Vcc and drain to the output.
the reason for the voltage divider is that output voltage isn't compatible with opamp inputs and it may cause damage to the OpAmp. OfCourse R5, R6 also form a voltage divider.
currently I'm testing your circuit and make measurements to see how much stable and low noise it is. the circuit that I've been posted is just an idea and I havn't implemented it yet ; because I like your idea better I decided first implement your circuit. well till now after blowing some BC337 and frying some resistors, which may or may not can be blamed on lack of sleep , I made the prototype and I should say it works. in your circuit no matter of the gain output voltage sign is opposite of the input but it is not very important for me right now.
the most important thing for me is the stability of the design and noise in comparison to other options. as I said till now I concluded that the circuit works and now I'm busy with measuring stability, noise and sensitivity parameters of that. have you used this circuit anywhere? or do you have information if it has been used elsewhere ? have you ever measured sensetivity, stability and noise parameters of the design or know about it ?
I need the output voltage be stable within a few 10 microvolt range(Vcc- Vee ~ 80V). do you think I can acheive that goal with your circuit ? anyway thank you very much for your time and helps!

Well, when you conduct experiments there will be always victums ..
Please bear in mind that all transistors should be rated for voltages above the +/- voltage swing .. and BC337 is a low voltage transistor ..
I used that circuit on several occasions and the main issue is to use appropriate HV transistors ..
On another topic, µVs stability is out of question .. but mVs stability is more realistic ..
Regards,
IanP

i think the optimum solution of ur problem is the boost converter , it's the one used for charging batteries with high voltage using low voltage batteries.

check it out in power electronics books for detailed circuits , also u can find a lot of online schematics covering such topic

hope that would help