Question about TVS (Transient Voltage Suppression) circuit - Page 1
Forum section: electronics / projects

author: bkish
author_rank: Newbie
timestamp: October 18, 2016, 06:27:01 pm
content: I need help identifying the functionality of this TVS diode that a previous engineer used in his circuit. I know that TVS are designed to "Load Dump" voltages that are over their threshold, but the previous engineer connected to the TVS in a way that does not make sense to me and seems to nullify its functionality.The datasheet (See link below) diagram on page 1 specifies that the TVS should have the ground connected to the pin labeled "G", and that the signal line going to the "Line to be protected" pin. However, when I review the PCB layout, the the previous engineer has the signal going into the "NC" pin and the "Line to be protected" pin is not connected. (See part number and datasheet below)It looks to be connected correctly in the schematic, but it is connected in the way I described on the PCB board.The signal coming in is a small +/- 1V AC signal coming in from BNC (J5) Pin 1. The signal is generated by a crystal.What is the reason for this?I work for a small company, who's previous engineer has passed away, so I am unable to ask him directly. That is why I am asking on here, as this community seems to be very knowledgeable.I can't place the full schematic on here for IP reasons, but I have attached a small snippet of the circuit concerned.part: DLP05LC-7-FDatasheet:http://www.diodes.com/_files/datasheets/ds30283.pdfIs there a specific reason for connecting to the TVS in this way?Any information is appreciated.

author: madires
author_rank: Super Contributor
timestamp: October 18, 2016, 07:17:13 pm
content: A TVS has a high capacitance and for high frequency or high data rate applications it has to be hidden by a diode bridge to reduce the capacitance. That's what the DLP05LC does. The wiring you've described uses the DLP05LC a little bit differently. It limits the positive part of signal to Vf of the diode, and the negative part to Vf of the other diode plus the 6V of the TVS. The typical Vf of 0.7V for a silicon diode would be to low for a +1V signal, and the 6.7V to high for a -1V signal. Maybe the idea was to clip the input signal at +0.7V while also having an input protection for the negative part?

author: bkish
author_rank: Newbie
timestamp: October 18, 2016, 10:00:01 pm
content: Thank you for this! The frequency of the signal is 6Mhz. So your assumption about it being a fast signal is correct. I am a little confused by how the capacitance of the TVS plays in? So do you mean because the signal is so fast, that the TVS would essentially hold the signal higher, causing a slow drop off on the sinusoidal signal or simply cause it to stay high? (I am picturing a low pass filter effect from trying to smooth out an AC signal using a full wave rectifier where the signal stays high with a slight ripple. Is that what you were driving at?I have attached an image of how I BELIEVE the crystal signal flows through the TVS with the way the previous engineer has wired it.Does it matter which way the TVS is facing? I believe that the TVS is polar correct? It is only when you have two of them back to back that it is bipolar?If I understand correctly the Single diode on the right side would clip the sign wave in half, so you would only see the positive voltage on the Crystal Signal wire but you would see a -0.7v from the forward voltage drop. However, when it flows though the left path with the TVS and the Diode, I don't know how the TVS would affect the signal since it appears to be going though the TVS in the reverse direction.If you remove the TVS out of the "Circuit Redrawn" doesn't the two diodes in parallel in opposite directions cause the signal wave to clip like a diode limiter? So you would get a voltage of +/- 0.7v due to the forward drop?Something similar to this?http://www.electronics-tutorials.ws/diode/diode62.gif?x98918So how would the TVS play into the circuit?Again thank you so much for your your help.P.S. I just noticed after posting this that the only thing that really changes in the circuit is the direction of the TVS from the Datasheet specified typical application connection, vs the previous engineer's connection. But I can't seem to find anything explaining how reversing the TVS affects the signal. And also, does reversing the TVS prevent the capacitance on the signal that you were referring to? Or is there another reason that I am missing for reversing the TVS?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 19, 2016, 12:57:45 am
content: If the schematic is correct, then the layout is incorrect (and footprints?).Also... if the schematic is correct, and the signal is connected as you describe, then surely the unit doesn't workat allbecause the signal path is wrong?Tim

author: bkish
author_rank: Newbie
timestamp: October 19, 2016, 03:04:26 am
content: Thanks Tim.The footprint is in fact laid out as described, where the signal is connected to the "NC" pin and ground is the "G" pin. The circuit is quite complex, and I just wanted to make sure that if I re-design it, I am not going to make a mistake.

author: madires
author_rank: Super Contributor
timestamp: October 19, 2016, 01:18:59 pm
content: Yes, a directly connected TVS would be a low-pass filter (capacitance might be up to a few nF). Most standard TVSs are available as unidirectional and bidirectional types. The unidirectional is like a zener (for DC) and the bidirectional like two zeners back-to-back (for AC). You can compare the way the DLP05LC is wired with the clipping circuit in the picture (two anti-parallel diodes). Instead of both directions limited to +/- 0.7V the circuit of the former engineer limits the signal to -6.7V and +0.7V. I got no idea if that was done intentionally or if it's some mistake. Usually you would connect the TVS to Vcc/signal and Gnd, while the input goes to the connection of both diodes (bot diodes in series and this parallel to the TVS). This setup limits the input to -0.7V below ground and +0.7V above V_BR of the TVS. The DLP05LC does the same, but the datasheet shows a different wiring in the application examples, actually increasing the total capacitance by paralleling the diodes.I think, instead of trying to guess what was meant, it would be better to re-define the specs for the input protection and replace the current mess.

author: bkish
author_rank: Newbie
timestamp: October 19, 2016, 05:31:53 pm
content: Thank you so much madires! This is a huge help. What you are saying makes a lot of sense. I am going to go back and re-examine the circuit. Like you suggest, I might have to re-build the circuit, but this information is invaluable. Thanks again!