#### Chuck D. Bones

##### Circuit Wizard

Like Germanium transistors, Germanium diodes are leaky. They behave like they have a resistor in parallel and in many cases, that resistor is less than 50K. The higher the leakage, the lower the resistance. Two questions that keep popping up are:

How much leakage is too much?

How do we measure leakage?

The answer to the first question is pretty simple: it depends on the circuit. If that leakage resistance is larger than all of the resistors that connect to the diode in the circuit, then those diodes are probably ok. If the Ge diodes are being used in a low impedance circuit, such as a hard clipper, then we can tolerate a lot of leakage because the circuit resistance is smaller. If the Ge diodes are used as soft clippers in a BMP circuit, such as the Cornish N-2, then they need to be very low leakage because they are in parallel with a 470K resistor.

The answer to the second question is pretty simple too because we can measure either the leakage current or leakage resistance with a DMM.

Let's look at a diode V-I curve. The scales on this plot are distorted to highlight what's going on in both the forward biased and reverse biased regions. In reality, there is no kink in the curve as it passes thru zero. For this discussion, we'll focus on the blue region.

At the low voltages and currents we see in pedal circuits, the relationship between diode voltage and current can be described by

Where Id is the diode current, Is is a scale factor called saturation current, Vd is the diode voltage, n is a "shape constant" that is between 1 and 2, usually closer to 2, Vt is the thermal voltage which is 26mV at 27°C. For this discussion, we'll let n = 2.

When Vd is a negative value much larger than nVt, then Id = Is. In other words, the leakage current is approximately equal to Is.

The slope of the curve at any given point is the diode conductance. Conductance is the reciprocal of resistance. With a little calculus, we can get the diode resistance at any point on the curve

We're interested in the diode resistance near zero volts. When Vd = 0,

and since Is is pretty close to Ilkg, the diode resistance is:

As we increase the reverse voltage on the diode, Rd gets larger, but if the leakage is too large then we can never develop enough reverse voltage in the first place. Let's look at a practical example, the Cornish N-2. The distortion stages are basically a Big Muff with Ge clipping diodes. Here's an LTSpice model of the N-2 without the buffers. The 1N34A diodes in the LTSpice library have Is = 1.2uA, which gives us an equivalent resistance of 43K for one diode. The red trace is the current in D6, one of the clippers in the 3rd stage. The green trace is the current in R34, the feedback resistor in the 3rd stage. Notice that the diode current goes almost 2uA negative. If this was a silicon diode, the reverse current would be a few nA. As it is, D6's reverse current is huge compared to the signal current in feedback resistor R34. D5 and D6 have effectively paralleled R34 with two 43K resistors, killing the gain of that stage. The same thing happens with the 2nd stage. Using typical 1N34A diodes, we're lucky if we get much more than unit gain from the distortion stages.

Now suppose we replace those generic 1N34A diodes with Ge diodes that have been selected for low leakage. In this next sim, I've replace all four Ge diodes with low leakage diodes that have a leakage resistance of 1Meg. Note: the 1N34B diode is not in the LTSpice library, I made it by copying a 1N34A and then setting the Is parameter to 52nA. Notice two things: The minimum D6 current is now very close to 0uA. The maximum D6 current is now 28uA compared to 6uA on the previous chart. The max current went up because D5 is not leaking current on the other 1/2 cycle. The gain is much higher, but is still not as high as it would be if the diodes were 1N4148s because their leakage current is a few nA.

So now let's look at two methods for measuring the diode leakage current & resistance.

DMMs measures resistance by applying a precise constant current and measuring the voltage drop. Modern meters keep the voltage drop low, well under 1V to protect semiconductors if we try to measure their resistance. My Fluke 115 applies 100nA when it's in the 6Meg range. A 1Meg resistor has 100mV across it during the measurement. If I use the Fluke to measure the input resistance of my Extech 411, I get a reading of 9.98Meg on the Fluke and 496mV on the Extech. Therefore, the Fluke is applying 50nA when it's on the 60Meg range. This means that when we're trying to measure reverse leakage of a diode, the DMM applies some current, measures voltage and converts that measurement to resistance. If the reading is too high or too low, the autoranging feature kicks in and switches the DMM to another range. This works fine for resistors but usually goes haywire when measuring non-linear devices like diodes. We will have you use manual ranging.

When I measure the leakage resistance of a GD012 Ge diode, the reverse voltage is 8.3mV (measured by the Extech) and the resistance reading is 83K. This tells us that the Fluke is applying 100nA because 8.3mV / 83K = 100nA. The only two ranges on the Fluke that give a reading are the 6Meg and 60Meg ranges and they more-or-less agree.

In this method, we apply 9V thru a large resistor, say 100K, and measure the voltage drop in that resistor. The current in that resistor is the voltage drop divided by 100K. Using the same GD012 diode, the voltage drop in the 100K resistor is 83mV. 83mV / 100K = 830nA. This is measured with 9V across the diode, At that voltage Ilkg is so close to Is that we can say they are the same. Knowing Is, we can calculate Rd at any Vd we like. First, let's calculate Rd at Vd = 0V.

Rd = nVt / Is = 52mV / 830nA = 62K.

When we measured Rd using the Resistance Method, Vd was -8.3mV. If we calculate Rd for Vd = -8.3mV using Equation 2, then we get 73.5K, which is pretty close to the 83K we measured on the Fluke using the Resistance Method.

In the N-2, the peak reverse diode voltage can get up to around -200mV, depending on the diodes. With the GD012, Rd = 2.9Meg when Vd = -200mV. This assumes we have enough drive to overcome the leakage and get to -200mV.

We can measure diode leakage and resistance using either of the two methods described above and get roughly the same result.

Leaky diodes will kill the gain in a BMP circuit; we have to select Ge diodes for maximum off resistance. With soft clippers, the diode resistance must be much larger than the feedback resistor (typically 470K in the BMP).

How much leakage is too much?

How do we measure leakage?

The answer to the first question is pretty simple: it depends on the circuit. If that leakage resistance is larger than all of the resistors that connect to the diode in the circuit, then those diodes are probably ok. If the Ge diodes are being used in a low impedance circuit, such as a hard clipper, then we can tolerate a lot of leakage because the circuit resistance is smaller. If the Ge diodes are used as soft clippers in a BMP circuit, such as the Cornish N-2, then they need to be very low leakage because they are in parallel with a 470K resistor.

The answer to the second question is pretty simple too because we can measure either the leakage current or leakage resistance with a DMM.

Let's look at a diode V-I curve. The scales on this plot are distorted to highlight what's going on in both the forward biased and reverse biased regions. In reality, there is no kink in the curve as it passes thru zero. For this discussion, we'll focus on the blue region.

At the low voltages and currents we see in pedal circuits, the relationship between diode voltage and current can be described by

*Equation 1*:Where Id is the diode current, Is is a scale factor called saturation current, Vd is the diode voltage, n is a "shape constant" that is between 1 and 2, usually closer to 2, Vt is the thermal voltage which is 26mV at 27°C. For this discussion, we'll let n = 2.

When Vd is a negative value much larger than nVt, then Id = Is. In other words, the leakage current is approximately equal to Is.

The slope of the curve at any given point is the diode conductance. Conductance is the reciprocal of resistance. With a little calculus, we can get the diode resistance at any point on the curve

*(Equation 2)*:We're interested in the diode resistance near zero volts. When Vd = 0,

and since Is is pretty close to Ilkg, the diode resistance is:

As we increase the reverse voltage on the diode, Rd gets larger, but if the leakage is too large then we can never develop enough reverse voltage in the first place. Let's look at a practical example, the Cornish N-2. The distortion stages are basically a Big Muff with Ge clipping diodes. Here's an LTSpice model of the N-2 without the buffers. The 1N34A diodes in the LTSpice library have Is = 1.2uA, which gives us an equivalent resistance of 43K for one diode. The red trace is the current in D6, one of the clippers in the 3rd stage. The green trace is the current in R34, the feedback resistor in the 3rd stage. Notice that the diode current goes almost 2uA negative. If this was a silicon diode, the reverse current would be a few nA. As it is, D6's reverse current is huge compared to the signal current in feedback resistor R34. D5 and D6 have effectively paralleled R34 with two 43K resistors, killing the gain of that stage. The same thing happens with the 2nd stage. Using typical 1N34A diodes, we're lucky if we get much more than unit gain from the distortion stages.

Now suppose we replace those generic 1N34A diodes with Ge diodes that have been selected for low leakage. In this next sim, I've replace all four Ge diodes with low leakage diodes that have a leakage resistance of 1Meg. Note: the 1N34B diode is not in the LTSpice library, I made it by copying a 1N34A and then setting the Is parameter to 52nA. Notice two things: The minimum D6 current is now very close to 0uA. The maximum D6 current is now 28uA compared to 6uA on the previous chart. The max current went up because D5 is not leaking current on the other 1/2 cycle. The gain is much higher, but is still not as high as it would be if the diodes were 1N4148s because their leakage current is a few nA.

So now let's look at two methods for measuring the diode leakage current & resistance.

**Resistance Method**DMMs measures resistance by applying a precise constant current and measuring the voltage drop. Modern meters keep the voltage drop low, well under 1V to protect semiconductors if we try to measure their resistance. My Fluke 115 applies 100nA when it's in the 6Meg range. A 1Meg resistor has 100mV across it during the measurement. If I use the Fluke to measure the input resistance of my Extech 411, I get a reading of 9.98Meg on the Fluke and 496mV on the Extech. Therefore, the Fluke is applying 50nA when it's on the 60Meg range. This means that when we're trying to measure reverse leakage of a diode, the DMM applies some current, measures voltage and converts that measurement to resistance. If the reading is too high or too low, the autoranging feature kicks in and switches the DMM to another range. This works fine for resistors but usually goes haywire when measuring non-linear devices like diodes. We will have you use manual ranging.

When I measure the leakage resistance of a GD012 Ge diode, the reverse voltage is 8.3mV (measured by the Extech) and the resistance reading is 83K. This tells us that the Fluke is applying 100nA because 8.3mV / 83K = 100nA. The only two ranges on the Fluke that give a reading are the 6Meg and 60Meg ranges and they more-or-less agree.

**Leakage Current Method**In this method, we apply 9V thru a large resistor, say 100K, and measure the voltage drop in that resistor. The current in that resistor is the voltage drop divided by 100K. Using the same GD012 diode, the voltage drop in the 100K resistor is 83mV. 83mV / 100K = 830nA. This is measured with 9V across the diode, At that voltage Ilkg is so close to Is that we can say they are the same. Knowing Is, we can calculate Rd at any Vd we like. First, let's calculate Rd at Vd = 0V.

Rd = nVt / Is = 52mV / 830nA = 62K.

When we measured Rd using the Resistance Method, Vd was -8.3mV. If we calculate Rd for Vd = -8.3mV using Equation 2, then we get 73.5K, which is pretty close to the 83K we measured on the Fluke using the Resistance Method.

In the N-2, the peak reverse diode voltage can get up to around -200mV, depending on the diodes. With the GD012, Rd = 2.9Meg when Vd = -200mV. This assumes we have enough drive to overcome the leakage and get to -200mV.

**Conclusions**We can measure diode leakage and resistance using either of the two methods described above and get roughly the same result.

Leaky diodes will kill the gain in a BMP circuit; we have to select Ge diodes for maximum off resistance. With soft clippers, the diode resistance must be much larger than the feedback resistor (typically 470K in the BMP).

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