# Generating a rippled voltage supply

1. Aug 9, 2007

### taleebe

So i'm trying to do some regulator PSRR tests and for this i'll need a regular DC source with a ripple on it (say 200mv pk-pk) so i can compare this to the ripple at the output and someone suggested this circuit set up to me. My question is about the capacitor, i was told "a signal generator might not be able to handle DC on its output so a capacitor may be needed".First of all i'm not sure if this is true and second of all i'm not sure what it means or how a capacitor would solve this problem,and if it does how to figure out the capacitance to use.
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To the administrators:Sorry for the double post! I figured this would be a better forum to ask

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2. Aug 9, 2007

### chroot

Staff Emeritus
If you directly connect the output of two signal generators, and they are producing different voltages, then very large currents will flow from one to the other. Of course, their limiting circuitry will kick in, and they will essentially turn themselves off.

Instead, you use a capacitor to couple one signal to the other. The capacitor blocks DC current, but permits the passage of AC current. You can use almost any size of capacitor; you just need to make sure that whatever RC filters are created do not significantly attentuate the AC signal you wish to pass.

- Warren

Last edited: Aug 9, 2007
3. Aug 9, 2007

### Staff: Mentor

How much power are we talking here? Many signal generators will put out an AC signal with a DC offset. That's what I typically use for PSRR measurements on chips.

4. Aug 9, 2007

### taleebe

hmm! thanks chroot that makes alot more sense now, berkeman my DUT sinks up to 1A and i want to test PSRR up to 20Mhz. do you think i could just connect my signal generator to my DUT? :surprised

5. Aug 9, 2007

### Staff: Mentor

Not at that current. Just use a power BJT as a follower (that's what we do for higher power) -- 20MHz is pretty fast, but could probably be done.

6. Aug 10, 2007

### taleebe

this has been suggested to me before, but i never understood the operating conditions(biasing etc) or connections (besides driving the base with the signal generator)or what specifications to look at in a BJT that would suggest it could work at the Mhz range

7. Aug 10, 2007

### Staff: Mentor

I admit that 20MHz is pretty dang fast for a power follower, but it should be do-able. If I have time, I'll do a couple SPICE runs on typical power BJT candidates, and see what their 200mV banwidth looks like across frequency.

EDIT -- What are the DUT's characteristics? You mentioned up to 1A of current -- what is the supply voltage that you want, and how much does the DUT current vary? You'll be removing any decoupling capacitance from the DUT for this test, correct? Without the extra decoupling capacitance, how much capacitance will the supply circuit see across the DUT?

Last edited: Aug 10, 2007
8. Aug 14, 2007

### taleebe

a supply voltage of around 3.3 V, DUT current varies from 600mA-1A.the capacitance seen by the supply circuit were not given but before all of that could you please explain rudimentary power follower circuits

9. Aug 14, 2007

### Staff: Mentor

An emitter follower is when you use an NPN transistor to give you current gain with approximately unity voltage gain (the emitter voltage "follows" the base voltage, minus a Vbe drop):

http://www.ece.osu.edu/ee327/Figures/emitter_follower_schem.gif [Broken]

The signal you put into the base shows up at the emitter, but you have about Beta times more current available from the NPN transistor than from the source driving its base. It has relatively good bandwidth, but you can see it is uni-directional in supplying current, so the load or extra emitter resistors to ground would be what pulls the voltage back down as the base signal is going down. That's why you want very little load capacitance, and a knowledge of the load current.

So that would be the type of cirtcuit that I would simulate first (but I honestly don't know if I have time), using some real-world power NPN BJTs that would be good starting candidates. If the low-going bandwidth were a problem, I'd look at doing more of a class-C power amp arrangement, with two opposing BJTs and appropriate biasing.

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10. Aug 14, 2007

### Odin42

The capacitor should block the DC offset, that's kind of what caps do(block dc).
It depends on the frequency you are looking to do, but you'll probably want to use a very small capacitance value, otherwise you'll start dropping the ac output across the cap.

Also in your other thread I noticed you were asking about the second connection from the signal generator. Several signal generator I've seen have bnc(coax) output port, in this case the shell is the reference and the center is the signal.

Another option to a transistor may be an op-amp, just make sure you check the gain bandwidth product(I can't remember typical values but I think you should be okay).

11. Aug 14, 2007

### taleebe

Ok,thanks alot for responding

so if the center is the signal can i still slide it in my configuration somehow?

Ive been looking everywhere for an op amp that is unity gain stable up to 20MHz and can source up to 1A.

12. Aug 14, 2007

### Odin42

I'd say put the center contact to the cap and the bnc shell to the resistor (even though this is all moot once you add an amplified(be it transistor or op-amp) to the circuit)

13. Aug 15, 2007

### taleebe

will this work

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14. Aug 15, 2007

### Odin42

Probably, measure resistence from the center and the shell to the posts. If the shell goes to one and the center goes to the other then that should work. But as previously mentioned, I doubt the signal generator can source 1A.

15. Aug 15, 2007

### Staff: Mentor

Yeah, that's not an opamp. That' a low RF power amp.