I can't clip the voltage with Diodes unless a bias is present

[LoveLaika]

I'm trying to build a voltage clipping circuit that will clip my signal voltage using diodes. Essentially, if the signal gets too high or too low, the diodes will turn on accordingly and clip the voltage down to its turn-on value. This is to suppress excess voltage spikes from environmental noise. In my case, I would like to clip the signal voltage to +/-2 volt range.

Now, looking online, the concept seems pretty simple. A pair of diodes will clip the signal voltage according to its forward voltage. This can be done with or without a DC voltage bias. I am trying to do it without a voltage bias because I don't have one available. So, I'm depending on the forward voltage of the diode.

However, I'm having issues with my circuit simulation. I've attached it below for LTspice as well as a picture so you can see my test circuit. I've also included the SPICE model of the diode I'm using to try and get this circuit right. The diode is the BY203-20STR, with a max forward voltage of 2.4 volts at 200 mA. From the circuit, node IN2 should clip my 5-volt sinusoidal voltage from V2 after passing my filter to a max/min of 2.4 and -2.4 volts respectively. However, after passing the said filter, the voltage gets clipped to -0.73 and 0.73 volts, much less than what I was expecting. Is it possible that I built my circuit incorrectly? I don't really understand why this is not working the way it should be. I've tried this with several other SPICE models, but it doesn't seem to work properly.

For reference, the SPICE macromodel consists of two diodes in parallel (from my understanding), with one model being the forward bias and the other model being the reverse bias. I copied the .MODEL statements and put them together accordingly to skip the need to make a model of it. I've also had .MODEL statements of other diodes I tried to use in my schematic, and they seemed to work, though I'm taking those with a grain of salt, as I had to extract the diode model information from the datasheet, so it may not be completely accurate.

[TimFox]

2.4 V at 200 mA is a maximum specification from the manufacturer:  they guarantee that the voltage will not exceed 2.4 V at that current.
Diodes have an approximately exponential I(V curve, so the actual forward voltage at lower current is smaller for two reasons:
A.  The spec is a maximum voltage.
B.  You don't have 200 mA through the diode, and the voltage is approximately logarithmic in the current.  5 V peak, less the diode voltage, through 100 ohms is < 50 mA peak current.

[Andy Watson]

200 MEG

The datasheet for the BY203 give a capacitance of 16pF with zero reverse bias. I suspect this is swamping any diode-action at 200MHz.

[TimFox]

Andy is right:  at 200 MHz, the reactance of one diode is only 50 ohms, shorting out most of the current from the 100 ohm source.
By the way, in that Spice circuit model, is the 5 V the peak, peak-to-peak, or rms value?  I assume it’s the peak value.

[Benta]

A high voltage rectifier with a trr of 300 ns at 200 MHz?

Is this a joke? We're quite some time away from April 1st.

[TimFox]

I should have looked up the datasheet on that diode.  Even a 1N4148 has a 4 ns trr, which is far from ideal at 200 MHz (5 ns period).
At that frequency, Schottky diodes are needed.

[LoveLaika]

Thanks for the replies. Looking at all of these replies, they've been really helpful. For what its worth, the 5-volts in my signal is peak-to-peak.

Putting aside my choice of diode for the moment, the real problem is the current. With my load at the end, the 9k resistor and everything past that, a max of 2 volts results in 0.22 mA at best. Is there even a diode that turns on at 2 volts given that current? Going by the max spec, if the diode will hold (say for example 2 volts at 1 amp), the turn-on voltage will be lower than that given such a tiny current. Thus, it stands to reason that we need to have a higher forward voltage than 2 volts (so that, for a small current, the forward voltage will be 2 volts). At this frequency, it seems nigh impossible unless my logic is wrong.

[LoveLaika]

Looking at another way to achieve the same result, it appears that using back-to-back Zener diodes in series to clip the voltage also works too. Going this route, what are the advantages between Zener diodes vs finding a suitable Schottky diode? Zeners are built to hold the voltage, so are they more naturally suited for this type of purpose, or am I still facing the same issues with my low current draw?

https://www.electronics-tutorials.ws/diode/diode-clipping-circuits.html

[TimFox]

Look at the data sheet I-V curves for any diode:  there is no "turn-on" voltage threshold, as found in Zener diodes, for a signal diode.
A good graph shows the voltage on a linear scale vs. the current on a logarithmic scale.  As I stated above, it is better to treat the current as the independent variable and the voltage as the dependent variable when computing quantitative values.  See  https://datasheet.octopart.com/1N4148-ON-Semiconductor-datasheet-42765246.pdf   for the Fairchild 1N4148 (which, at 4 ns trr, is not fast enough for your 200 MHz case).
Figures 3 to 6 show this for different conditions (current ranges and temperature).  In Fig 4, at 25 C, 200 uA corresponds to about 535 mV.  If you want less voltage at that current, you need a Schottky diode (e.g., 1N5711), which is also faster.  By the way, in your Spice calculation, is 5 V truly the peak-to-peak in your specification of input voltage?  In regular Spice, the voltage parameter in an independent voltage source "V" is the amplitude, which is the peak voltage.  This is important in a .TRAN calculation for non-linear devices, as opposed to a linearized .AC calculation for linear passive devices and small-signal behavior of active devices.
 

[TimFox]

Zener diodes are used in low-frequency circuits for clipping, but they have huge capacitance at zero bias, and are not designed to be fast.  A 1N746A 3.3 V Zener has about 150 pF capacitance at zero volts.
In clipping circuits where you have power supplies available, a good solution is to DC bias two Zener diodes, and connect low-voltage diodes from the input to each Zener.  Also, low-voltage (true-Zener) diodes (say, 3.3 V) have relatively "soft" curves at turn-on compared with avalanche diodes (above 5 V or so) that are commonly called Zeners.  Do you ever look at data sheets before trying things?

[Benta]

My suggestion would be a BAS70-04 (or similar) dual Schottky diode connected to a +/- 2 V bias circuit. Very fast, very low capacitance.

[LoveLaika]

Look at the data sheet I-V curves for any diode:  there is no "turn-on" voltage threshold, as found in Zener diodes, for a signal diode.
A good graph shows the voltage on a linear scale vs. the current on a logarithmic scale.  As I stated above, it is better to treat the current as the independent variable and the voltage as the dependent variable when computing quantitative values.  See  https://datasheet.octopart.com/1N4148-ON-Semiconductor-datasheet-42765246.pdf   for the Fairchild 1N4148 (which, at 4 ns trr, is not fast enough for your 200 MHz case).
Figures 3 to 6 show this for different conditions (current ranges and temperature).  In Fig 4, at 25 C, 200 uA corresponds to about 535 mV.  If you want less voltage at that current, you need a Schottky diode (e.g., 1N5711), which is also faster.  By the way, in your Spice calculation, is 5 V truly the peak-to-peak in your specification of input voltage?  In regular Spice, the voltage parameter in an independent voltage source "V" is the amplitude, which is the peak voltage.  This is important in a .TRAN calculation for non-linear devices, as opposed to a linearized .AC calculation for linear passive devices and small-signal behavior of active devices.

Sorry, turns out I misspoke. My input signal is a simple sine wave with an amplitude of 5 volts (so peak-to-peak voltage is 10 volts); also, frequency has been reduced down to 20 MHz which allows a bit more flexibility. Also, just running a simple DC voltage of 2.5 volts through my circuit, test reults show it draws around 1.45 mA of current at the source; so, for this test circuit's sake, that 9k resistor should be 1.37k-ohms

Going by what you said, most of the diode datasheets I've seen seem to limit it down to 100 mA or so on the diode's IV curve. All the graphs show that as the current gets lower, the voltage across the diode gets smaller; since the diodes are there to clip the voltage, the diode voltage has to be at a certain value at that specific current in my application. Putting aside other factors for now, unless I'm totally misunderstanding how this works, that would require the diode to have its max forward voltage at a low max current value. Assuming a linear graph (I know IV curves are logarithmic, but assuming linear relationship for rough simplicity), that would get me the voltage that I need at that specific current.

[TimFox]

The linear graph is very inaccurate at low current.  If you only see graphs down to 100 mA, you must be looking at rectifier diodes (e.g. 1N4007).  They are also too slow, even at 20 MHz, due to their reverse recovery time, a non-linear and time-dependent effect.  When you forward-bias a diode, some charge is stored in the junction;  thereafter, when you reverse-bias the diode, instead of shutting off instantaneously, current flows in the “reverse” direction for a finite time until that charge has been “swept out” of the junction.  For the 1N4148, that is 4 ns (very large compared with your original 200 MHz, not trivial compared with 20 MHz).
  Try signal diodes (e.g. 1N4148 at my link) or, better for speed Schottky diodes (e.g. 1N5712).  The Schottky construction has much less stored charge than the PN diode construction.

[LoveLaika]

Thank you for your reply. I appreciate your help and patience in all of this. I must say though, by the time I read your message, I found one diode that fits my needs: the MMBD330T1G by ON SEMI. Running tests, it successfully clips the voltage (when the diodes have been biased), the capacitance is low enough so that it doesn't affect the rest of my circuit, and it reacts fast enough at a 20 MHz input signal.
I found a similar product in one section of Digikey's site, but the quantity was low. Then, I saw that there was a subsection of diodes meant for RF applications. I found that mildly annoying.

[Siwastaja]

Up to what frequency do you actually need? This pretty much defines the whole design regarding what diode types you can use, but also other components, their parasitics, and even PCB layout.

I.e., if the final signal ends up being some audio at 20kHz max, then simulating and designing for 20MHz, let alone 200MHz is waste of effort.

[TimFox]

If you can use bias, that changes the whole question.  The best clipping designs use symmetrical bias to keep the diodes reverse-biased when not clipping, thus reducing their capacitance.  Capacitance of a diode at low voltages is a strong function of the voltage, increasing dramatically at zero.
The ONSEMI part you found is a good signal-level Schottky diode.  My suggestion for a biased clipper would be to use several 1N4148 forward-biased diodes in series to ground, fed approximately 1 mA DC by a resistor, to develop each of two balanced DC voltages, and connect each Schottky to the relevant resistor-diode node.  If a large current is applied to the Schottky, it will pass through to the series string of 1N4148s and raise the voltage slightly.

[LoveLaika]

Thanks for your suggestion. At this point, I found that adding a bias would be the simplest, straightforward solution to my issues.

My initial thought of adding the bias was to use a 10k-pot to apply the bias, but running sims show that doing so, in this case, would not achieve the results I want. With regards to your idea about using 1N4148 diodes, that seems like a much better idea than mine. The way you described it, is it similar to the attached picture? Like this, I can add almost any voltage bias (within reason) to the resistor end and bias the diodes in a much more steady way than simply using a pot whereas the bias voltage seen by the Schottky diode is determined by the number of diodes in series (at least that's what it seems like to me, assuming a constant fwd-voltage drop).

[TimFox]

Yes, that drawing represents my suggestion.
However, to forward bias two 1N4148s through a resistor, you probably want more than 1.5 V;  I would suggest at least +/-3V, depending on what's available.  You don't need much bias current, since a large current through the Schottkys will just add (never subtract) from the bias current, and the 1N4148s never shut off (so Trr is not so important).