jon_sree@world.std.com (Jon Sreekanth) (01/22/91)
Hi, A rather curious query : what is the audio frequency impedance of a reverse biased diode ? Off-hand, one would say, it's high, but just how high ? Specifically, for a small signal diode like 1N4148, between 0 - 70 C, with about 2.5 V or more of reverse bias, is the small signal audio impedance at least > 1 Meg ? 10 Meg ? I don't need an exact number, just a minimum, to see if it interferes with other circuitry. Where does one look for such info ? Diodes and transistors are just there, one dosen't usually look at data books for them, so I realize I have many many chip data books, but none for discretes. Thanks, / Jon Sreekanth Assabet Valley Microsystems Fax and PC products 346 Lincoln St #722, Marlboro, MA 01752 508-562-0722 jon_sree@world.std.com
mark@mips.COM (Mark G. Johnson) (01/22/91)
In article <JON_SREE.91Jan21215810@world.std.com> jon_sree@world.std.com (Jon Sreekanth) writes: > >A rather curious query : what is the audio frequency impedance of >a reverse biased diode ? Off-hand, one would say, it's high, but >just how high ? > >Specifically, for a small signal diode like 1N4148, between 0 - 70 C, >with about 2.5 V or more of reverse bias, is the small signal audio >impedance at least > 1 Meg ? 10 Meg ? TI's datasheet (1966, revised 1973) for the 1N4148 sez that the max capacitance is 4 pF and the max leakage current at 100C is 3 microamperes (with 20V of reverse bias). Leakage current at room temp is 0.025 microamps. So a crude model would be a 6.7Megohm resistor in parallel with a 4pF capacitor. At 20kHz, 4pF has a capacitive reactance of 2 Megohms. -- -- Mark Johnson MIPS Computer Systems, 930 E. Arques M/S 2-02, Sunnyvale, CA 94086 (408) 524-8308 mark@mips.com {or ...!decwrl!mips!mark}
tomb@hplsla.HP.COM (Tom Bruhns) (01/23/91)
jon_sree@world.std.com (Jon Sreekanth) writes: >A rather curious query : what is the audio frequency impedance of >a reverse biased diode ? >Off-hand, one would say, it's high, but just how high ? >Specifically, for a small signal diode like 1N4148, between 0 - 70 C, >with about 2.5 V or more of reverse bias, is the small signal audio >impedance at least > 1 Meg ? 10 Meg ? I don't need an exact number, >just a minimum, to see if it interferes with other circuitry. You can calculate it from the diode equation by measuring the saturation current for the diode in question--that will get you the resistive component. I think you will find that the most significant component at audio frequencies, however, is the capacitance, which will be in the low-megohms region above a kilohertz. Diode equation: I=Isat * (e^(qV/kT)-1) Differentiate with respect to V to find dI/dV; invert for small signal resistive component. Isat you can measure as essentially the current at about half the reverse breakdown voltage, but as you can see from the equation, the exact voltage you measure it at isn't critical. (V is negative at that point, and will be many times larger than kT/q) This should be fine for most signal and power diodes, but watch your step for things like tunnel or PIN diodes... In practice, resistive leakage paths will far outweigh the contribution of the diode equation, also, at the voltages you mention. V of -2.6 volts is about 100 times larger than kT/q at room temp, so you get an e^-100 which is pretty small... It may well come down to how clean your construction is, as far as resistive components go.
verive@tellabs.com (Jeff Verive) (01/23/91)
In article <JON_SREE.91Jan21215810@world.std.com> jon_sree@world.std.com (Jon Sreekanth) writes: > >A rather curious query : what is the audio frequency impedance of > >Specifically, for a small signal diode like 1N4148, between 0 - 70 C, >with about 2.5 V or more of reverse bias, is the small signal audio >impedance at least > 1 Meg ? 10 Meg ? I don't need an exact number, >just a minimum, to see if it interferes with other circuitry. I checked my National Semiconductor data book, so here goes. The reverse capacitance for the 1N4148 is about 1.5 pF at 2.5V (don't know the measurement frequency - probably 1 MHz.) Assuming an upper frequency limit of 20kHz for audio, the impedance will be approximately equal to the capacitive reactance in parallel with the leakage impedance. From Xc = 1/(2*pi*f*C), Xc is about 5.3 Meg. Leakage current is harder to determine (graph only goes down to 5 Volts, so I'll use that value.) At 25 degrees C, leakage current is about 8 nA (the graph actually shows 8 mA, but the 'm' in 'mA' is a typo!). The value decreases at lower temperatures and increases at higher temperatures. If we allow the current to double every 10 degrees, then from 25 to 75 degrees, the current will be increased by a factor of 32 (this is a common approx. to the temperature coefficient of *reverse* biased silicon semi's), to a value of approx 250 nA (32*8=256). The impedance due to reverse leakage is then, from Ohm's Law, E/I = 2.5V/250 nA = 10 Meg. Applying basic complex math, the dynamic impedance is about 3.7 Meg. Based on the assumptions above, this should be somewhat of a lower limit, so this can be treated as a minimum. It is interesting to note that the diode's capacitance is so low that the circuitry around it (wiring, etc.) is far more likely to affect the reactance of the circuit's nodes. -- **************************************************************************** ** Jeff Verive | If they ever stop making those little candy flowers ** ** 259371048378 | for birthday cakes, I shall lose my will to live. ** ****************************************************************************
grege@gold.gvg.tek.com (Greg Ebert) (01/23/91)
A reverse-biased diode will have an appreciable impedance at audio frequencies, because the main mechanism is junction capacitance, which decreases as more reverse-bias is applied. You can measure it by connecting a DC source, audio generator, diode, and a variable resistor, all in series. Set the audio generator to, say, 0.1v p-p, and adjust the resistor until the voltage across the resistor is 0.05v p-p. Measure the resistance; it equals the diode's impedance *at the measured frequency*. Try varying the frequency. I've never measured the impedance of diodes before, but I would expect it to be near a megohm at 1Khz. Now forward-bias the diode and repeat the experiment. Neat eh ? Diodes make nifty (but sometimes noisy) analog switches. ----> NOTE ! Use a scope with a 10X probe to reduce loading effects. You should parallel the variable resistor with the scope (1M or 10M) when you measure it. Otherwise, you'll have to do some calculations (not hard).
jon_sree@world.std.com (Jon Sreekanth) (01/24/91)
In article <5170101@hplsla.HP.COM> tomb@hplsla.HP.COM (Tom Bruhns) writes:
Thanks to Tom and to others who replied in this newsgroup. I also
received several replies by email. The summary is
that, from data books, and from theoretical calculation, the
impedance is better than a few megohms.
The leakage current is, according to one email response I got,
25nA max at 20 volts, 5000 at 75 volts (from a National Semi databook)
Assuming a linear slope, the dynamic impedance is 55V/5uA = 11 Mohm.
The actual dynamic impedance will be better because the exponential
curve will lie below the straight line approximation.
The capacitance of the reverse biased junction, according to one posted
response, is 1.5 pF at 2.5V. In that case, Xc is 5.3Mohm, for 20KHz.
Actually I only care about telephone line bandwidth (4KHz, max),
so the impedance is greater.
In my application there are 2 of these reverse biased diodes, in
parallel, sharing voltage with a 220K resistor. So it seems that the
parallel combination of two 1N4148's is negligible compared to 220K.
Thanks again,
/ Jon Sreekanth
Assabet Valley Microsystems Fax and PC products
346 Lincoln St #722, Marlboro, MA 01752 508-562-0722
jon_sree@world.std.com