Combining a Power Supply and a Function Generator - Page 1
Forum section: electronics / beginners

author: eev_carl
author_rank: Regular Contributor
timestamp: November 10, 2017, 03:06:20 pm
content: Hi,I was using the DC waveform on my function generator to see how a BJT reacts to changes in voltage. My circuit lights an LED if the transistor is forward biased by a function generator. However, if I shut off my function generator, I get a small light emitting from the LED. I measured a voltage of -13V across the LED. I figure that this is coming from a difference between the ground I'm using from my power supply and the negative terminal positioned of the function generator in the circuit. I don't get this behavior if I swap out the function generator for a battery and a switch.In the schematic below, I have the function generator - marked DG1022 -- providing a Vbe. When the DG1022 is off, I measured -13V between point A and the ground of the power supply.Is there a way to bridge the ground difference or is this a bogus configuration?Thanks in advance,Carl

author: Zero999
author_rank: Super Contributor
timestamp: November 10, 2017, 03:21:29 pm
content: It's probably capacitive coupling between the mains and the earth of the signal generator, lighting the LED, which doesn't take much current for modern, high efficiency LEDs.

author: eev_carl
author_rank: Regular Contributor
timestamp: November 10, 2017, 03:46:34 pm
content: I put a 22k resistor in parallel with the LED and the light now goes off when the function generator is off. Also, I no longer see the -13V difference (it's 0V).Did this fix the capacitive coupling? I just did this through trial-and-error because I saw it on other schematics.

author: Damianos
author_rank: Frequent Contributor
timestamp: November 10, 2017, 07:22:05 pm
content: Quote from: eev_carl on November 10, 2017, 03:06:20 pmThe DP832 has floating output (which means it is not grounded).For the DG1022, I guess, that the output is grounded, as it is in almost all the function generators.If both were grounded, then, during the experiment, the LED would not lit, due to short circuit.Both instruments have couplings (mostly capacitive) to the ground and the environment. Think of it, as there are "small" AC sources between connections.In the presented diagram, grounding (from the generator) is connected somewhere in the middle of the circuit. This creates a somewhat complex situation...The measured voltage is the result of interference rectification, from the LED. This is from the reverse bias portion; the forward polarity is when the LED glows. The value is the average...For your experiment, I think it is better to have one reference point. Assuming you experiment with common emitter transistor topology, connect the emitter and sources to common (0-volt, common, reference, ground or whatever someone calls it!). So, move the LED from there. In this way, while the voltmeter or oscilloscope is connected, to measure the VBEor VCE, less interference is injected.

author: IanB
author_rank: Super Contributor
timestamp: November 10, 2017, 07:28:10 pm
content: I would give the same answer as Damianos, but in fewer words. I think you should move the LED up to where the 220 ohm resistor is on the collector side, and give the power supply and the signal generator a common ground at the emitter.

author: eev_carl
author_rank: Regular Contributor
timestamp: November 10, 2017, 07:49:31 pm
content: Thanks to both of you. I wasn't using the chassis ground terminal on the DP832. Moving the LED up in the circuit and the putting the supplies to a common ground at the emitter did the trick. Here's the revised schematic.

author: bson
author_rank: Unknown
timestamp: November 12, 2017, 09:18:09 pm
content: Quote from: eev_carl on November 10, 2017, 07:49:31 pmConsider moving the resistor to the emitter instead. This makes the emitter current (Vin-Vbe)/Re. While Vbe is somewhat variable, relative to the voltage across Re it's small, unlike hfe which can vary by integer factors. Hence its impact on the emitter current is minor. The resistance looking into the base is hfe*Re, so with a high-hfe BJT Ie will be very close to Ic. The other advantage of this is it can handle more than 5V on the input without blowing up the BJT, or even wear it down with noise riding on top of a 5V signal.