# Can ADL5304 be used without internal bias on pin 4, still function as a Log-Amp?

Hello,

I'm trying to use  ADL5304 as a log-amp without having bias on pin 4, I need to measure 10pA-3mA!

Thank you.
Shawn

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• Hi Shawn,

I'm not exactly sure what you mean here. If you want to use the device at zero Volts input, the answer is YES, see datasheet section on Using a Negative Supply. With negative supply, now VSMx pins can be tied to ground. An example is shown in Figure 50. Output voltage will be proportional to log(InputCurrent). With a series resistor added, output voltage will be proportional to log(InputVoltage).

Note that log converters such as ADL5304 ALWAYS need some positive input current. Remember, the log of zero or a negative number is undefined. If that presents a problem, your application needs to externally inject a small amount of leakage current into the input pin, to keep input always positive current. That small leakage current error can be subtracted in floating point math externally, after making the measurement with the ADC.

Hope that helps answer your question.  -Bruce H.

• Hi Bruce,

That is exactly what I was looking for but in a single supply, is it possible to measure 1pA with dual supply?

Thank you for your help.

Best regards,
Shawn

• Hi Shawn,

No, it is not normal to need feedback capacitor as you've described to stop any oscillation.

The problem might be too much capacitance from input pin INUM to ground. Is there long wire(s) connected? Could there be a lot of stray capacitance to ground? If so, try a series resistor nearby to the INUM pin 4. Use just the minimum amount of resistance, so that the voltage drop up at the highest input current doesn't cause problems.

Remember internal to the IC, the output of the JFET-input op-amp feeds back to the input through the logging transistor, so loop gain can be very high. Input capacitance adds extra phase shift to that path, which means a delayed feedback, which could be the root cause of the oscillation you describe. External resistance (or a large inductance) in series with INUM pin reduces the feedback phase shift.

-Bruce H

• Hi Bruce,

Thank you again for the tips and suggestions, I have a question regarding dual supply operation with figure 44 configuration, is it okay to use ground as shielding throughout the circuit including around pin 4 input?

BTW, I tried using 100 ohms through 2.5k resistor on the input, no improvement, what would be your suggestion on the value of an inductor to start? we do have hi capacitive load with long lines on the input!

Best regards,
Shawn

• Hi Shawn,

1. With VSMx at ground, in dual-supply configuration, the voltage at INUM will normally servo to ground potential also, so a zero volt guard is fine.

2. Unless you observed 2500 Ohms showing some improvement, it's doubtful that an inductor would help.

To help isolate the problem, you might try completely disconnecting the long input lines right near the ADL5304. Instead, supply a positive input current derived simply from a resistor between INUM and either 2VLT or VPOS. Keep the resistor lead connecting to INUM short or shielded (or both), to minimize noise pickup. With external influence from long wires removed, verify that VLOG is relatively free of noise and oscillation and the correct DC value. Then try a few other resistor values that span a large range of input current, again confirming that without external influence of long wires, the circuit is stable as-built on the PCB. If instability persists, this way we know the problem is on the PC board somewhere. If problem goes away, we know to look closer at the long interconnecting wires arrangement.

• Hi Bruce,

I hope that all is well with you, back on the LogAmp final design, thank you for all of  your help, it seems to be working fine except calibration and accuracy, I would like your help with performing logarithmic aX+b mathimatical calibration on the 8 different ranges? appreciate all the help you can provide. thank you. Shawn

• Hi Shawn,

Good to hear your circuit is basically working.

Calibration is fairly simple. ADL5304 output response is basically governed by two coefficients: log slope and log intercept. Both are specified parameters. Most customers choose to calibrate the device at two or more points on the response curve, because 2 points determine a straight line, like shown in datasheet Figure 3. If more than 2 points, then a least-squares or best-fit algorithm is used. But if left uncalibrated, accuracy may be good enough for your application. If you choose to calibrate, the result should be a slope and intercept that are relatively close to the datasheet values of same. Once you know slope and intercept, datasheet eq. (2) provides Inum in terms of Vlog.

Lets take a simple case of calibration for just 2 points for example:

The log slope is simply the output voltage change per decade of input current. So lets say Vlog=2.5V at 10mA, and also Vlog= 0.7V at 10pA. The log slope = (2.5 - 0.7) / log(10mA/10pA) = 1.8V / 9 = 0.2 volts per decade.

The log intercept is calculated by solving eq. (2) for intercept, and since slope is now a known:

log intercept = I / (10^ (Vlog/slope)  where Vlog and I are either one of the 2 data points.

Hope this helps!

-Bruce H.

Reply
• Hi Shawn,

Good to hear your circuit is basically working.

Calibration is fairly simple. ADL5304 output response is basically governed by two coefficients: log slope and log intercept. Both are specified parameters. Most customers choose to calibrate the device at two or more points on the response curve, because 2 points determine a straight line, like shown in datasheet Figure 3. If more than 2 points, then a least-squares or best-fit algorithm is used. But if left uncalibrated, accuracy may be good enough for your application. If you choose to calibrate, the result should be a slope and intercept that are relatively close to the datasheet values of same. Once you know slope and intercept, datasheet eq. (2) provides Inum in terms of Vlog.

Lets take a simple case of calibration for just 2 points for example:

The log slope is simply the output voltage change per decade of input current. So lets say Vlog=2.5V at 10mA, and also Vlog= 0.7V at 10pA. The log slope = (2.5 - 0.7) / log(10mA/10pA) = 1.8V / 9 = 0.2 volts per decade.

The log intercept is calculated by solving eq. (2) for intercept, and since slope is now a known:

log intercept = I / (10^ (Vlog/slope)  where Vlog and I are either one of the 2 data points.

Hope this helps!

-Bruce H.

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