I really appreciate the convenience offered by microcontrollers that
have a fairly-high-accuracy internal oscillator. But these oscillators
are never as precise as a crystal, so there is always a lingering
doubt—is the internal oscillator accurate enough for all the timing
functionality that I might need?

One of the most common timing-sensitive tasks is UART communication. The interface lacks an external clock, and the Tx and Rx devices can reliably share data only when their internal baud rates are equal.

But of course “equal” is not really an engineering word—we always have to account for noise and error and variation, and in the case of a microcontroller’s internal oscillator, we’re not talking about a few parts per million. In my experience, microcontrollers rarely offer an internal oscillator that has accuracy better than ±1.5%. Two microcontrollers with ±1.5% clock sources will have baud rates that are different by as much as 3%. Is that good enough?

The first thing to keep in mind is that the only relevant error is the

The following diagram shows the basic timing procedure involved in UART reception. If the receiver baud rate perfectly matches the transmitter baud rate and the first bit is sampled at the exact middle of the bit period, the last data bit will also be sampled at the exact middle of the bit period.

If the baud rates (and thus the bit periods) are different, each sampling point will move gradually closer to the bit transition. In other words, the last data bit is the one that is most severely affected by baud rate discrepancy.

The bit period for the Tx device is defined as follows:

If we use E to refer to the absolute value of the percentage
difference between the baud rate of the receiver and the baud rate of
the transmitter, we have the following for the additional bit-period
time on the Rx side:

For this first analysis, we’ll simplify by assuming that the timing
procedure begins in the exact middle of the start bit. So for an
eight-data-bit interface, the sampling time for the last data bit will
be off by 8Î”T

because we said that the acceptable sampling window for the last data
bit is from 50% (the ideal value) to 80% (as close to the transition as
we are willing to go), and 80% – 50% = 30% = 0.3.

We can solve for maximum allowable E as follows:

So, based on this simplified analysis, eight-data-bit UART
communication should be reliable as long as the difference between
transmitter baud rate and receiver baud rate is less than 3.75%.

To do this, we will incorporate the following:

Now our final-bit-sampling margin is a variable; we’ll use M
(previously we used 20%). As you can see from the following equation, M
must be entered as a decimal, not a percentage.

We still have the 8 out in front of Î”T

The last thing we need is the variable that accounts for the actual
position at which the start bit is sampled. The equation in its current
form assumes 50% (i.e., the exact middle of the bit period). This is the
ideal position. Any deviation from the ideal can cause the sampling
position of later bits to be closer to the transition (and thus closer
to an error). We will incorporate this into the equation by adding the
following term to the left-hand side:

This term refers to the deviation from the ideal start-bit sampling
position, expressed as a decimal multiplied by the receiver’s bit
period. So, for example, if the start bit is sampled 60% of the way
through the bit period, this term would be |(50% – 60%)|T

Now our equation is

If we recall that the Rx bit period is equal to the Tx bit period
plus the additional time corresponding to the error percentage, our
final equation is as follows:

I hope this article helps you with any future baud-rate-accuracy concerns, and remember that a careful baud-rate analysis may indicate that you can confidently omit an external crystal, even if your internal oscillator is not as precise as you might like.

One of the most common timing-sensitive tasks is UART communication. The interface lacks an external clock, and the Tx and Rx devices can reliably share data only when their internal baud rates are equal.

But of course “equal” is not really an engineering word—we always have to account for noise and error and variation, and in the case of a microcontroller’s internal oscillator, we’re not talking about a few parts per million. In my experience, microcontrollers rarely offer an internal oscillator that has accuracy better than ±1.5%. Two microcontrollers with ±1.5% clock sources will have baud rates that are different by as much as 3%. Is that good enough?

### Baud Rate, Bit Period, and Sampling

(**Note:**For the remaining analysis we’ll use the Tx device as the reference; in other words, we’ll assume that the Tx device is always at the nominal baud rate and the Rx device is at the inaccurate baud rate.)The first thing to keep in mind is that the only relevant error is the

*difference*between the transmitter’s baud rate and the receiver’s baud rate (as opposed to the difference between an expected baud rate and an actual baud rate). For example, you will have no difficulty transferring data if your two communicating devices are both operating at 9800 baud, even if the expected rate (based on the nominal clock frequency) is 9600 baud.The following diagram shows the basic timing procedure involved in UART reception. If the receiver baud rate perfectly matches the transmitter baud rate and the first bit is sampled at the exact middle of the bit period, the last data bit will also be sampled at the exact middle of the bit period.

If the baud rates (and thus the bit periods) are different, each sampling point will move gradually closer to the bit transition. In other words, the last data bit is the one that is most severely affected by baud rate discrepancy.

**Note:**For convenience, we’ll always assume that the Rx bit period is longer than the Tx bit period; the result would be the same if we assumed a shorter Rx bit period.### Solving for Maximum Allowable Error: The Simpler Version

Based on this observation, we can first propose the following: the baud rates are sufficiently accurate if the last data bit is sampled before the transition from last data bit to stop bit. (We’re assuming here that the system can tolerate an improperly sampled stop bit.) But we don’t want to sample one nanosecond before the expected transition; we need some margin. I think a reasonable margin is 20%—i.e., the last bit must be sampled at least 20% of the bit period*before*the transition from last data bit to stop bit.The bit period for the Tx device is defined as follows:

_{RX}. Thus, our is-it-accurate-enough condition is the following:We can solve for maximum allowable E as follows:

### Maximum Allowable Error: The Complete Analysis

The previous section gives you a rule of thumb for any eight-data-bit UART system that can tolerate frame errors (i.e., errors indicating an improperly sampled stop bit). In this section, we’ll develop a comprehensive equation that can provide a more precise and customized estimate.To do this, we will incorporate the following:

- variable location of start-bit sampling
- variable number of data bits
- variable margin for sampling the final bit
- presence or absence of a parity bit
- frame-error tolerance

_{RX.}We need to change that because now the number of data bits (denoted by N) is a variable. We also will include a variable for the parity bit (P) and the stop bit (S). If you have a parity bit, use 1 for P; if not, use 0. If you want to ensure that the stop bit is sampled correctly, use 1 for S; if not, use 0._{RX}= 0.1T_{RX}. You can’t readily determine an exact value for this term, but if you know the details of your UART’s low-level functionality, you could come up with a reasonable worst-case estimate.Now our equation is

### Example and Conclusion

Let’s consider the following system:- margin is 30%
- worst-case start-bit sampling deviation is 5%
- eight data bits
- no frame errors allowed
- parity bit is used

I hope this article helps you with any future baud-rate-accuracy concerns, and remember that a careful baud-rate analysis may indicate that you can confidently omit an external crystal, even if your internal oscillator is not as precise as you might like.