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I have made a signal generator based on DDS and amplified its output to 0-20 V pk-pk. Now I want to prepare the output side. My main concern is short circuit protection to prevent damage to my device. I was thinking about isolating the signal on the device side from the output side.

First choice was a transformer. But it is not applicable because finding a good transformer for such wide frequency and amplitude is not easy. Other option was winding the transformer myself, but I have no experience on transformer winding.

Second choice was a coupling capacitor. Using a coupling capacitor may isolate the signal to "some extent" , but does not seem enough ( AC current passes through it , and also it needs to be of a very large capacitance in lower frequencies that leads to problems as current leakage,... ).

Can anyone give me a suggestion please?

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    \$\begingroup\$ a circuit of your output stage would be useful. I'm pleased to here this project is moving along well. \$\endgroup\$
    – Andy aka
    Oct 13, 2013 at 21:12
  • \$\begingroup\$ @Andyaka Thank you!. You were the first one who suggested DDS for my purpose and now it is going to end. Never will forget your favors along this project. \$\endgroup\$
    – Aug
    Oct 13, 2013 at 21:17
  • \$\begingroup\$ I'm pleased for you dude and if you get a chance please email me the final circuit if that's ok with you. My email should be on my profile page and if not follow the link to my website and email from there. I understand if not. \$\endgroup\$
    – Andy aka
    Oct 13, 2013 at 21:47
  • \$\begingroup\$ My pleasure! I will do. \$\endgroup\$
    – Aug
    Oct 13, 2013 at 21:49

2 Answers 2

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If you are making a signal generator, then the 50 ohms in the output is a good start. With a 20Vp-p sinewave and the output shorted the resistor will disipate about 1W so firstly choose this resistor to have the power rating but, what op-amp operating at maybe 30MHz can deliver 1W? Probably none so it might be worth adding a push-pull stage made from PNP and NPN emitter followers.

Because they are emitter followers and, because they can be biased (with a little care) to be on the verge of conduction (it's called class AB) taking the op-amp feedback from the output of the push-pull circuit may work OK. Be aware that it can be easy to cause ringing and oscillations so layout is a little critical but methinks you have learned this so far (did you relay the DDS and get it working better?).

The transistors should have an \$H_{FE}\$ that remains above 1 at 1GHz or higher.

Basically, my idea is to build a little power into the final drive stage and use a +1W 50 ohm resistor on the output. Be also aware that the full 20Vp-p may not be attainable using this method - maybe you can increase the power supply levels a little. Also, your choice of op-amp is worth disclosing.

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  • \$\begingroup\$ OK, please let me know if I got it well: by this method we are putting the short circuit pressure on the power transistors which can bear it well in the push-pull design. In fact we will not need a short circuit protection design at all. Is that correct? \$\endgroup\$
    – Aug
    Oct 13, 2013 at 22:27
  • \$\begingroup\$ There is no secret about the op-amp. It is AD811. It works fine for the design. It has 100mA output current which with 50 ohm, it can tolerate up to 5V ( x2 vpp =10v) I tested it with 50 ohms in 50MHz and voltage dropped less about 40% ( it is better than the standard of 50%!). with such op-amp do I need a push-pull output? I have bad experience with them because their cross-over distortion is really difficult to manage. \$\endgroup\$
    – Aug
    Oct 13, 2013 at 22:43
  • \$\begingroup\$ I wouldn't run the AD811 at lower than 50 ohm load despite the output current drive being 100mA. The cross over distortion can be managed by biasing the output transistors to have some collector current when not driving a signal and the Rfb of the AD811 taken from the push-pull output - this then should work better. \$\endgroup\$
    – Andy aka
    Oct 14, 2013 at 7:16
  • \$\begingroup\$ I tried it right now. I dont know why negative feedback AD811 combined to any kind of emitter follower starts oscillating !? ( it is much worse with positive feedback). I cant see any 0 degree phase shift positive feedback loop which is necessary for oscillation. I am struggling with it now! \$\endgroup\$
    – Aug
    Oct 14, 2013 at 7:41
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    \$\begingroup\$ AD8016 looks like it could do the job - should get 20Vp-p from +/-12V supplies - what are your supplies. AD815 (see fig 50 in pdf document) can really drive out some current. These are both probably better than 811 at driving. 815 looks the best though. \$\endgroup\$
    – Andy aka
    Oct 14, 2013 at 8:49
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If you do not already have a resistor in the path(s) that provide current to the output then add one. This can be a relatively small value (i.e. 1 ohm or less). Then design an opamp circuit that can amplify the voltage across this resistor to a reasonable working voltage level. This voltage will be proportional to the current that is being delivered to the output via the path that the resistor is placed in. During a short circuit this detected voltage will increase to a level that would be deemed detrimental to the life span of the critical components in your signal generator output. Now compare the amplified voltage to a fixed reference so that then it exceeds a safe level for the health of the unit cause the output transistors to be shut off or gate off the power to the output stages via an appropriate MOSFET.

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  • \$\begingroup\$ Should this Op-Amp be one with high bandwidth product? As I see, this idea really seems reasonable but I am thinking it should not be dependent on the output frequency. A LM741 would suffice? \$\endgroup\$
    – Aug
    Oct 13, 2013 at 21:23
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    \$\begingroup\$ In almost all circuit design cases the current through the sensing resistor will have a frequency component to it that will follow the output frequency of your generator. So - yes you need to have an op-amp that has a gain-bandwidth product and slew rate capable of tracking the current waveform in the sense resistor. \$\endgroup\$ Oct 13, 2013 at 21:29
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    \$\begingroup\$ Op-Amp selection will also depend highly in the circuit topology and where the sense resistor is located. It may very well be that you may need a device with rail to rail inputs, a device whose inputs can support a common mode range that includes the negative rail voltage, or even a device that is able to support the "over the top" capability one vendor touts which is the capability to support common mode voltages that go above the positive supply rail of the op-amp. \$\endgroup\$ Oct 13, 2013 at 21:33
  • \$\begingroup\$ I think there may be a problems: most high bandwidth Op-Amps ( CFO/VFO) have a smaller allowed input differential voltage. In my application, during a short circuit, the voltage sensed along the resistor reaches 20 v. \$\endgroup\$
    – Aug
    Oct 13, 2013 at 21:39
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    \$\begingroup\$ Just as a comment - there are 100's (if not 1000's) of different Op-Amp types from many vendors. It seems likely that at least one suitable one can be found. \$\endgroup\$ Oct 13, 2013 at 21:48

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