integrated power amplifier 700kHz - Page 1
Forum section: electronics / beginners

author: raff5184
author_rank: Regular Contributor
timestamp: January 16, 2017, 10:42:54 pm
content: Hi all,do you know if there exists and where could I find an integrated (or small) power amplifier for signals around 700kHz?I am using a very big/heavy one for a prototype but I want to miniaturize my circuit.Thank you

author: tszaboo
author_rank: Super Contributor
timestamp: January 16, 2017, 11:02:17 pm
content: Can you specify power and voltage?Otherwise, APEX makes power amplifiers.

author: raff5184
author_rank: Regular Contributor
timestamp: January 16, 2017, 11:25:35 pm
content: first of all thank you.I am working with voltage values varying from 1V peak-to-peak up to 2V peak-to-peak. I have no idea what values I can expect in output of such a small amplifier. Tents? hundreds of the input value?

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: January 17, 2017, 12:18:37 am
content: output voltage is meaningless unless you specify the load. a 74HC device will do 5Vp-p easily but that's probably not what you're looking for.We can't help you without more information.

author: raff5184
author_rank: Regular Contributor
timestamp: January 17, 2017, 12:58:59 am
content: the load varies between 500\$\Omega\$ and 2k\$\Omega\$... let's say 1.5 k\$\Omega\$ is a good value

author: raff5184
author_rank: Regular Contributor
timestamp: January 17, 2017, 02:04:32 pm
content: probably 100x voltage gain is a little too high, but around 40-50 would be better, and probably easier to find (?). Also the cost is not a problem

author: tszaboo
author_rank: Super Contributor
timestamp: January 17, 2017, 03:37:04 pm
content: Quote from: raff5184 on January 17, 2017, 02:04:32 pmYou can easily reduce the requirements, if you add a preamplifier to the circuit. Just have a general purpose opamp amplify the 1V signal by 10-15, running from a +/-15V supply, so the power amplifier does not need to be that accurate (offset) or optimize the bandwidth or anything like that.

author: edavid
author_rank: Super Contributor
timestamp: January 17, 2017, 05:18:09 pm
content: Your stated requirements of 50-100V PP, 700kHz, 1.5Kohm don't really make sense.At 700kHz, a capacitance of 200pF has a reactance of 1.1Kohm. There aren't many real loads that can tolerate 50V+ and have capacitance that low.What kind of load are you driving?What is your actual load voltage requirement?

author: raff5184
author_rank: Regular Contributor
timestamp: January 17, 2017, 07:39:08 pm
content: thank you edavid. I am driving a small piezoelectric transducer (PZT) for ultrasounds. I am trying to drive it with a high voltage value because I have a high attenuation of the ultrasonic signalSo basically I have a signal generator to drive the PZT, bust since the maximum amplitude of the signals that my generator supports is too small I need to amplify it before applying it to the PZT.I know that the PZT has a resistance R=1.5k\$\Omega\$

author: edavid
author_rank: Super Contributor
timestamp: January 17, 2017, 08:05:16 pm
content: Quote from: raff5184 on January 17, 2017, 07:39:08 pmThat is unlikely. A PZT is almost a pure capacitance.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: January 17, 2017, 08:33:25 pm
content: Pzts are a complex load that changes with load

author: Marco
author_rank: Super Contributor
timestamp: January 17, 2017, 08:41:34 pm
content: Quote from: raff5184 on January 17, 2017, 07:39:08 pmWhat kind of signal? If it can be semi-square wave you can simply usetotem poles.

author: raff5184
author_rank: Regular Contributor
timestamp: January 17, 2017, 08:59:29 pm
content: What kind of signal? If it can be semi-square wave you can simply usetotem poles.just a sinusoid

author: raff5184
author_rank: Regular Contributor
timestamp: January 17, 2017, 09:02:16 pm
content: Quote from: edavid on January 17, 2017, 08:05:16 pmThan I must be wrong, sorry. This is what I have, I thought that real was the internal resistance

author: qno
author_rank: Frequent Contributor
timestamp: January 17, 2017, 09:21:46 pm
content: Why not use a transformer?

author: Marco
author_rank: Super Contributor
timestamp: January 17, 2017, 09:22:51 pm
content: Quote from: raff5184 on January 17, 2017, 08:59:29 pmThe sonic output will be sinusoid any way, it's a pretty good filter.

author: raff5184
author_rank: Regular Contributor
timestamp: January 17, 2017, 09:31:54 pm
content: Quote from: qno on January 17, 2017, 09:21:46 pmbecause they are too bulky (?)

author: retrolefty
author_rank: Super Contributor
timestamp: January 17, 2017, 09:42:25 pm
content: Quote from: raff5184 on January 17, 2017, 09:02:16 pmNo that verical axis is Z the impedance, mostly due to capacitance reactance, not a DC resistance.

author: raff5184
author_rank: Regular Contributor
timestamp: January 17, 2017, 10:55:18 pm
content: Quote from: retrolefty on January 17, 2017, 09:42:25 pmOk, so from my figure, which is value of the load do I have to consider to evaluate the power that I am transferring to the PZT?

author: edavid
author_rank: Super Contributor
timestamp: January 18, 2017, 02:05:58 am
content: It looks like you need a resonant driver to make it work.

author: Marco
author_rank: Super Contributor
timestamp: January 18, 2017, 04:07:28 am
content: Quote from: edavid on January 18, 2017, 02:05:58 amIsn't the piezo transducer alone enough for that? Simply generate a square wave signal of the same polarity as the measured current through the transducer, you get ZCS and automatic resonance tracking AFAICS (just need to get it going before the control loop can take over).

author: dmills
author_rank: Super Contributor
timestamp: January 18, 2017, 11:38:15 am
content: Ah, driving high frequency sonar transducers, been there done that.The usual thing is to use a tuned transformer (Resonate the secondary winding inductance with the transducer fixed capacitance) as part of a fifth order or so matching network, then drive the whole thing with a square wave, the transducer will do a good job of turning it into a sine in the working fluid. Note that your driver will probably have to deal with a load that is not in fact resistive, active snubbers can be your friends.These transformers can be small as the duty cycle is usually negligible, but the voltages can be huge so kapton tape is a good thing, as is litz wire to keep the Q where you want it.Usually the input match is some form of L network, often with deliberate series resistance to make the bandwidth a little wider, do make sure you work from a G/B plot done in water, they measure **very** differently in air.Are you sure that plot is impedance rather then admittance? I would expect the resonance to be a low impedance as the devices model as series resonant devices with an annoyingly large shunt cap and the acoustic radiation resistance in the series arm.73 Dan.

author: Zero999
author_rank: Super Contributor
timestamp: January 18, 2017, 06:55:11 pm
content: Quote from: raff5184 on January 17, 2017, 09:31:54 pmHow small does it have to be?A 700kHz transformer with a power rating of a few Watts, isn't very big.

author: BrianHG
author_rank: Super Contributor
timestamp: January 18, 2017, 07:11:17 pm
content: Quote from: Hero999 on January 18, 2017, 06:55:11 pmIn fact, how much you want to bet that a cheap 500khz-1Mhz 10 watt 12v wall-mount switching supply's tiny transformer could be used in reverse to drive the PZT.Though I cant say anything about the efficiency...

author: Zero999
author_rank: Super Contributor
timestamp: January 19, 2017, 09:13:55 am
content: Quote from: BrianHG on January 18, 2017, 07:11:17 pmIt should work, but you might have to adjust the input voltage to give the correct output voltage.

author: raff5184
author_rank: Regular Contributor
timestamp: March 09, 2017, 10:40:41 pm
content: Quote from: blueskull on January 17, 2017, 01:46:24 amPlease can you give me more details bout the second part?Bootstrapped opAmp is to reduce the impedance so that I can give a higher voltage in input without distortion?Why a BJT buffer?Someone suggested me a multistage amplifier with the AD825 but I can't simply give the output of the first AD825 in input to the 2nd AD825, assuming I configure both with the same say inverting feedback configuration

author: raff5184
author_rank: Regular Contributor
timestamp: March 10, 2017, 12:09:00 am
content: Quote from: blueskull on March 09, 2017, 11:33:58 pmYes, I meant to configure both in inverting configuration, with two similar -but separate- feedbacks.Thanks for the paper I'll read it and see how to create the cascade.

author: raff5184
author_rank: Regular Contributor
timestamp: November 30, 2017, 09:28:30 pm
content: Quote from: blueskull on January 17, 2017, 01:46:24 amso, I'm re-arising this old post. About the powering of the 2 cascaded op amps, do I have to use 2 independent supplies like -+17.5 V (by the way, why exactly 17.5?) for the first stage and another +-Vsupply for the bootstrapped op amp?Also, I was able to have the second opAmp with a gain of about 6-7 (without bootstrapping network) but when I connect the opAmp to the bootstrapping circuit, it doesn't work any longer or its output is the same of the input. I am using a bootstrapping circuit very similar to the one in Figure 6 on this paper, but with and AD825:http://www.proaudiodesignforum.com/images/pdf/Bootstrapping_Your_Op_Amp_Yields_Wide_Voltage_Swings_King_Watkins_EDN_May_13_1999.pdf

author: raff5184
author_rank: Regular Contributor
timestamp: December 01, 2017, 07:05:36 pm
content: Quote from: blueskull on December 01, 2017, 05:27:34 amYes I got very closeQuote from: blueskull on December 01, 2017, 05:27:34 amDo you get this info from Figure 7 of the datasheet, right? Because I don't see it elsewhere on the datasheet.Quote from: blueskull on December 01, 2017, 05:27:34 amNo I used R2:R1=R4:R3=3:1. What's the criterion to choose of R1,R2,R3,R4?Also I noticed that it is better not to insert RB