Comments on: Injecting AC into the DC Power Supply Rail

By: Adrian S. Nastase

Adrian S. Nastase — Sun, 28 Apr 2019 02:27:18 +0000

In reply to Eric Gracka. Simply add the two signals.

By: Eric Gracka

Eric Gracka — Sat, 23 Mar 2019 16:08:50 +0000

I would like to superimpose a sine wave on top of a square wave.

By: Xiong Rui

Xiong Rui — Thu, 24 May 2018 08:58:46 +0000

Thank you very much. It is a great article. Could you recommend a book which describes this method?

By: Adrian S. Nastase

Adrian S. Nastase — Sun, 07 Aug 2016 02:24:06 +0000

In reply to CC. No. Without the inductor the power supply will short-circuit the AC signal generator.

By: CC

CC — Tue, 19 Jul 2016 08:56:07 +0000

Can you inject an AC pulse without the inductors?

By: Adrian S. Nastase

Adrian S. Nastase — Wed, 08 Apr 2015 02:39:31 +0000

In reply to Steve. I would guess that your inductor together with the capacitive load create a tank with a self resonant frequency of 5 Hz. Or, it may be that your 12 V power supply is overloaded and goes into protection mode (it folds) oscillating between on and off. Good supplies should stay folded until you power cycle them.

By: Steve

Steve — Thu, 02 Apr 2015 19:14:53 +0000

I need to add a 1Vp-p, 50 Hz sine wave on 12VDC so I determined that I needed a 1.0H inductor (very big and heavy) and some value of capacitor (I forget exactly). Anyways, I built it with a 48 ohm power resistor (actual load needs 250mA) and it worked very well. I ended up having to increase the DC voltage a little to over come the internal DC resistance of my inductor to get my required 12VDC at my load, but it worked until I plugged in my actual load. The DC output oscillates like a square wave around 5 Hz! I am not sure why this is happening. The load has about 20uF capacitance and internal DC-DC regulators and switchers. The best answer I can think of is that maybe the internal DC-DC switcher/regulators shut down perhaps due to an over voltage condition??? Or maybe the DC input rise time is too slow and causes some weird condition inside??? I have now removed the AC and capacitor and just have my benchtop DC voltage generator to this large inductor to my load. Any thoughts of what could be happening?

By: Adrian S. Nastase

Adrian S. Nastase — Tue, 01 Oct 2013 20:38:05 +0000

In reply to Lloyd. What is the load in ohms? If the voltage is 5V, with 10 mA load current, then your load is 500 ohms. You want your inductor impedance to be 10 times higher, at 5 kohm. Therefore, L1 = 5 kohm / (2 * PI * 15 kHz) = 53 mH.

By: Lloyd

Lloyd — Mon, 30 Sep 2013 18:18:36 +0000

I am trying to work the formula for L1 and the numbers just don’t add up. My circuit in somewhat different, in that my frequency is 15kHz and my load is 10mA, but the final desired outcome is the same. When I try to calculate the values for L1 (for example) as stated, I don’t come up with 1.58, my calculation is 3.1830 (I assume this is Henry’s). Is there an error on the website or a misunderstanding on my part? What am I doing wrong?